Exercise 16.3
1.
a.
Soln:
For x ≥ 0, f(x) = x + 2
Right hand limit at x = 0 is
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\ \to\end{array}$ 0 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\ \to\end{array}$ 0 + (x + 2) = 0 + 2 = 0,
For x < 0, f(x) = 4x + 2
Left hand limit at x = 0 is
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 – (4x + 2) = 0 + 2 = 2.
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 – f(x)
So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 f(x) = 2.
b.
Soln:
For x ≥ 1, f(x) = 3x + 2
Right hand limit at x = 1 is
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 + (3x + 2) = 3*1 + 2 = 5,
For x < 1, f(x) = 2x.
Left hand limit at x = 1 is
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 – (2x) = 2 * 1 = 2.
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 + f(x) ≠ x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 – f(x)
So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 f(x) does not exist.
c.
Soln:
For x > 2, f(x) = 5x + 2
Right hand limit at x = 2 is
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + (5x + 2) = 5*2 + 2 = 12,
For x < 2, f(x) = 7x – 2
Left hand limit at x = 2 is
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – (7x – 2) = 7*2 – 2 = 12.
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – f(x)
So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 f(x) = 12.
d.
Soln:
For x ≥ 2, f(x) = 3x + 1
Right hand limit at x = 2 is
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + (3x + 1) = 3*2 + 1 = 7,
For x < 2, f(x) = 2x2 – 1
Left hand limit at x = 2 is
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – (2x2 – 1) = 2 * 4 – 1 = 7.
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – f(x)
So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 f(x) = 7.
e.
Soln:
For x ≥ 1, f(x) = 2x + 1
Right hand limit at x = 1 is
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 + (2x + 1) = 2*1 + 1 = 3.
For x < 1, f(x) = 4x2 – 1.
Left hand limit at x = 1 is
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 – (4x2 – 1) = 4*1 – 1 = 3.
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 – f(x)
So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 f(x) = 3.
f.
Soln:
For x ≥ 1, f(x) = 3x – 2
Right hand limit at x = 2 is
RHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + (3x – 2) = 3*2 – 2 = 4,
For x < 2, f(x) = 2x2 + 1
Left hand limit at x = 2 is
LHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – (2x2 + 1) = 2*4 + 1 = 9.
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + f(x) ≠x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – f(x)
So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 f(x) does not exist.
2.
a.
Soln:
LHL = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 f(2 – h)
= h$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 |2 – h – 2| = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 |h| = 0
RHL = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 f(2 + h)
= h$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 |2 + h – 2| = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 |h| = 0.
LHL = RHL
So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 |x – 2| = 0.
b.
Soln:
RHL = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 f(0 + h)
= h$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\left| {0 + {\rm{h}}} \right|}}{{0 + {\rm{h}}}}$ = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\rm{h}}}{{\rm{h}}}$ = 1.
LHL = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\left| {0 - {\rm{h}}} \right|}}{{0 - {\rm{h}}}}$ = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\rm{h}}}{{ - {\rm{h}}}}$ = –1.
LHL ≠ RHL
So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\left| {\rm{x}} \right|}}{{\rm{x}}}$ does not exist.
Exercise 16.4
1.
(i)
Soln:
f(x) = x2
Left hand limit at x = 4 is
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 – x2 = (4)2 = 16.
Right hand limit at x = 4 is
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 + x2 = (4)2 = 16.
f(4) = (4)2 = 16.
So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 + f(x) = f(4)
So, f(x) is continuous at x = 4.
(ii)
Soln:
f(x) = 2 – 3x2
Left hand limit at x = 0 is
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\ \to\end{array}$ 0 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 – (2 – 3x2) = 2 – 0 = 2.
Right hand limit at x = 0 is
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 + (2 – 3x2) = 2 – 0 = 2.
f(0) = 2 – 3 * 0 = 2.
So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 + f(x) = f(0)
So, f(x) is continuous at x = 0.
(iii)
Soln:
f(x) = 3x2 – 2x + 4.
Left hand limit at x = 1 is
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 – (3x2 – 2x + 4) = 3 * 1 – 2 * 1 + 4 = 5.
Right hand limit at x = 1 is
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 + (3x2 – 2x + 4) = 3 * 1 – 2 * 1 + 4 = 5.
f(1) = 3 * (1)2 – 2 * 1 + 4 = 5.
So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 + f(x) = f(1)
So, f(x) is continuous at x = 1.
(iv)
Soln:
f(x) = $\frac{1}{{2{\rm{x}}}}$.
f(0) = $\frac{1}{{2.0}}$ = $\frac{1}{0}$ = does not exist.
Hence, f(x) is discontinuous at x = 0.
(v)
Soln:
f(0) = $\frac{1}{{{\rm{x}} - 2}}$
Let x = a ≠ 2.
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a –${\rm{\: }}\frac{1}{{{\rm{x}} - 2}}$ = $\frac{1}{{{\rm{a}} - 2}}$
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a + $\frac{1}{{{\rm{x}} - 2}}{\rm{\: }}$= $\frac{1}{{{\rm{a}} - 2}}$.
f(a) = $\frac{1}{{{\rm{a}} - 2}}$.
So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a + f(x) = f(a)
So, f(x) is continuous at x = a ≠ 2.
Note: since a ≠ 2, so a – 2 ≠ 0 and hence $\frac{1}{{{\rm{a}} - 2}}$ is a finite number.
(vi)
f(x) = $\frac{1}{{3{\rm{x}}}}$
Let x = a ≠ 0.
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a${\rm{\: }}\frac{1}{{3{\rm{x}}}}$ = $\frac{1}{{3{\rm{a}}}}$
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a + $\frac{1}{{3{\rm{x}}}}{\rm{\: }}$= $\frac{1}{{3{\rm{a}}}}$.
f(a) = $\frac{1}{{3{\rm{a}}}}$.
So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a + f(x) = f(a)
So,f(x) is continuous at x = a ≠ 0.
(vii)
Soln:
f(x) = $\frac{1}{{1 - {\rm{x}}}}$
f(1) = $\frac{1}{{1 - 1}}$ = $\frac{1}{0}$ which does not exist.
So, f(x) is discontinuous at x = 3.
(ix)
Soln:
f(x) = $\frac{{{{\rm{x}}^2} - 9}}{{{\rm{x}} - 3}}$
RHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 $\frac{{{{\rm{x}}^2} - 9}}{{{\rm{x}} - 3}}$
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 $\frac{{\left( {{\rm{x}} + 3} \right)\left( {{\rm{x}} - 3} \right)}}{{{\rm{x}} - 3}}$ = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 (x + 3) = 3 + 3 = 6.
LHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 $\frac{{{{\rm{x}}^2} - 9}}{{{\rm{x}} - 3}}$.
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 – $\frac{{\left( {{\rm{x}} + 3} \right)\left( {{\rm{x}} - 3} \right)}}{{{\rm{x}} - 3}}$ = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 (x + 3) = 3 + 3 = 6.
f(3) = $\frac{{9 - 9}}{{3 - 3}}$ = $\frac{0}{0}$.
So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 – f(x) ≠ f(3)
So, f(x) is continuous at x = 3.
Note: since a ≠ 2, so a – 2 ≠ 0 and hence $\frac{1}{{{\rm{a}} - 2}}$ is a finite number.
(x)
Soln:
f(x) = $\frac{{{{\rm{x}}^2} - 16}}{{{\rm{x}} - 4}}$
LHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 $\frac{{{{\rm{x}}^2} - 16}}{{{\rm{x}} - 4}}$
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 – $\frac{{\left( {{\rm{x}} + 4} \right)\left( {{\rm{x}} - 4} \right)}}{{{\rm{x}} - 4}}$ = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 – (x + 4) = 4 + 4 = 8.
LHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 + $\frac{{\left( {{\rm{x}} + 4} \right)\left( {{\rm{x}} - 4} \right)}}{{{\rm{x}} - 4}}$ = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 (x + 4) + 4 = 8.
f(x) = $\frac{{{{\left( 4 \right)}^2} - 16}}{{4 - 4}}$ = $\frac{0}{0}$.
So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 + f(x) ≠ f(0)
So, f(x) is discontinuous at x = 4.
Note: since a ≠ 2, so a – 2 ≠ 0 and hence $\frac{1}{{{\rm{a}} - 2}}$ is a finite number.
2.
(i)
Soln:
For x ≤ 2, f(x) = 2 – x2.
LHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – (2 – x2) = 2 – 4 = –2.
For x > 2, f(x) = x – 4
RHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + (x – 4) = 2 – 4 = –2.
For, x = 2, f(x) = 2 – x2
f(2) = 2 – (2)2 = –2.
x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + f(x) = f(2)
So, f(x) is continuous at x = 2.
(ii)
Soln:
For x < 2, f(x) = 2x2 + 1.
LHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – (2x2 + 1) = 2 *4 + 1 = 9.
For x > 2, f(x) = 4x + 1.
RHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + (4x + 1) = 4 * 1 + 1 = 9.
For, x = 2, f(x) = 2x2 + 1.
f(2) = 2 * (2)2 + 1 = 9.
x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + f(x) = f(2)
So, f(x) is continuous at x = 2.
(iii)
Soln:
For x < 3, f(x) = 2x.
LHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 – 2x = 2 *3 = 6.
For x > 3, f(x) = 3x – 3.
RHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 + (3x – 3) = 3 * 3 – 3 = 6.
For, x = 3, f(x) = 2x
f(3) = 2 * 3 = 6.
x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 + f(x) = f(3)
So, f(x) is continuous at x = 3.
(iv)
Soln:
For x < 1, f(x) = 2x + 1.
LHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 – (2x + 1) = 2 *1 + 1 = 3.
For x > 1, f(x) = 3x.
RHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 + 3x = 3 * 1 = 3.
For, x = 1, f(1) = 2.
x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 + f(x) ≠ f(1)
So, f(x) is discontinuous at x = 1.
(v)
Soln:
For x < 5, f(x) = x2 = 2.
LHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 5 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 5 – (x2 + 2) = 25 + 2 = 27.
For x > 5, f(x) = 3x + 12.
RHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 5 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 5 + (3x + 12) = 3 * 5 + 12 = 27.
For, x = 5, f(x) = x2 + 2
f(5) = (5)2 + 2 = 27.
x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 5 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 5 + f(x) = f(5)
So, f(x) is continuous at x = 5.
3.
Soln:
For x < 2, f(x) = 2x – 3.
LHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\ \to\end{array}$ 2 – (2x – 3) = 2 *2 – 3 = 1.
For x > 2, f(x) = 3x – 5.
RHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + (3x – 5) = 3 * 2 – 5 = 1.
f (2) = 2.
x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + f(x) ≠ f(2)
So, f(x) is discontinuous at x = 2.
So, the given function will be continuous if f(x) is redefined as follows:
f(x) = $\{ \begin{array}{*{20}{c}}{2{\rm{x}} - 3}&{{\rm{for\: x}} < 2}\\1&{{\rm{for\: x}} = 2}\\{3{\rm{x}} - 5}&{{\rm{for\: x}} > 2}\end{array}{\rm{\: \: }}$
3
(i)
Soln:
LHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 5– f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 5– (x2 + 2) = 25 + 2 = 27.
RHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 5+ f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 5+(3x + 12) = 3.5 + 12 = 27.
Functional value = f(5) = 20.
Here, LHL = RHL ≠ f(5).
Therefore, given function is removable discontinuous at x = 5.
But, the given function can be made continuous by redefining as:
f(x) = $\{ \begin{array}{*{20}{c}}{{{\rm{x}}^2} + 2}&{{\rm{for\: x}} < 5}\\{27}&{{\rm{for\: x}} = 5}\\{3{\rm{x}} + 12}&{{\rm{for\: x}} > 5}\end{array}$
4.
(i)
Soln:
For x < 2, f(x) = kx + 3.
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 (3x – 1) = 3.2 – 1 = 5.
Being continuous,
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – f(x)
2k + 3 = 5.
So, k = 1.
(ii)
Soln:
For x > 3 and x < 3, f(x) = $\frac{{2{{\rm{x}}^2} - 19}}{{{\rm{x}} - 13}}$.
= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 $\frac{{2{{\rm{x}}^2} - 18}}{{{\rm{x}} - 3}}$ = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 $\frac{{2\left( {{\rm{x}} - 3} \right)\left( {{\rm{x}} + 3} \right)}}{{{\rm{x}} - 3}}$.
= 2(3 + 3) = 12.
For, x = 3, f(x) = k, i.e. f(3) = k.
Being continuous,
12 = k
So, k = 12.
Additional Questions (Limit)
1.
Soln:
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\ \to\end{array}$ -2/3 $\frac{2}{{2 + 3{\rm{x}}}}$ = $\frac{2}{{2 + 3\left( { - \frac{2}{3}} \right)}}$ = $\frac{2}{{2 - 2}}$ = $\frac{2}{0}$ which does not exist.
2.
(i)
Soln:
Or, f(x) = $\frac{{{\rm{x}} - 1}}{{{\rm{x}} + 1}}$.
Or, f(1) = $\frac{{1 - 1}}{{1 + 2}}$ = $\frac{0}{3}$ = 0 which exists.
So, f(x) is defined for x = 1.
(ii)
Soln:
Or, f(x) = $\frac{{{{\rm{x}}^3} + 1}}{{{\rm{x}} - 1}}$.
Or, f(1) = $\frac{{{{\left( 1 \right)}^3} + 1}}{{1 - 1}}$ = $\frac{2}{0}$ which does not exist.
So, f(x) is not defined for x = 1.
3.
Soln:
Let f(x) be a function and x = 1 be a point. The value of f(x) which it approaches when x approaches a from the left side is known as the left hand limit of f(x) at x = 1 and is denoted by x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a – 0 f(x) or x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a – f(x) or, h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 f(a – h).
Also the value of f(x) which is approaches when x approaches a from right side is known as the right hand limit of f(x) at x = a and is denoted by x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a + 0 f(x) or x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a + f(x).
The following are the conditions for the existence of the limit at x = a for the function f(x) or h$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 f(a + h).
x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a – f(x) and x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a + f(x) are finite.x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a + f(x)
For the limit: x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 |x| = 0LHL = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 |0 – h| = 0RHL = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 |0 + h| = 0
LHL = RHL
So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 |x| = 0
For the limit: x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\left| {\rm{x}} \right|}}{{\rm{x}}}$.
LHL = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\left| {0 - {\rm{h}}} \right|}}{{0 - {\rm{h}}}}$ = $\frac{{\rm{h}}}{{ - {\rm{h}}}}$ = -1
RHL = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\left| {0 + {\rm{h}}} \right|}}{{0 + {\rm{h}}}}$ = $\frac{{\rm{h}}}{{\rm{h}}}$ = 1.
LHL ≠ RHL
So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\left| {\rm{x}} \right|}}{{\rm{x}}}$ does not exist.
4.
Soln:
The value L which the function f(x) approaches when x = a is known as the limit of f(x). It is denoted by x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a f(x) = L. The value of f(x) when x = a is substituted in it is known as the value of the function. It is denoted by f(a).
Or, f(x) = $\frac{{{\rm{ax}} + {\rm{b}}}}{{{\rm{x}} + 1}}$ …..(i)
Or, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{{\rm{ax}} + {\rm{b}}}}{{{\rm{x}} + 1}}$.
Or, 2 = $\frac{{0 + {\rm{b}}}}{{0 + 1}}$.
Again, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ ∞ f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ ∞ $\frac{{{\rm{ax}} + {\rm{b}}}}{{{\rm{x}} + 1}}$.
Or, 1 = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ ∞ $\frac{{{\rm{ax}} + {\rm{b}}}}{{{\rm{x}} + 1}}$
Or, 1 = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ ∞ $\frac{{\frac{{{\rm{ax}} + {\rm{b}}}}{{\rm{x}}}}}{{\frac{{{\rm{x}} + 1}}{{\rm{x}}}}}$.
Or, 1 = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ ∞ $\frac{{{\rm{a}} + \frac{{\rm{b}}}{{\rm{x}}}}}{{1 + \frac{1}{{\rm{x}}}}}$
Or, 1 = $\frac{{{\rm{a}} + 0}}{{1 + 0}}$
So, a = 1.
Substituting the values of a and b in (i), we get,
Or, f(x) = $\frac{{1.{\rm{x}} + 2}}{{{\rm{x}} + 1}}$ = $\frac{{{\rm{x}} + 2}}{{{\rm{x}} + 1}}$.
Or, f(-2) = $\frac{{\left( { - 2 + 2} \right)}}{{2 + 1}}$ = 0.
5.
Soln:
For the first part see Q.no.4
Or, f(x) = $\frac{{{\rm{x}} + 6}}{{{\rm{cx}} - {\rm{d}}}}$ …(i)
Or, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{{\rm{x}} + 6}}{{{\rm{cx}} - {\rm{d}}}}$.
Or, -6 = $\frac{6}{{ - {\rm{d}}}}$.
Again, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ ∞ f(x) $\frac{{{\rm{x}} + 6}}{{{\rm{cx}} - {\rm{d}}}}$.
Or, $\frac{1}{3}$ = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ ∞ $\frac{{\frac{{{\rm{x}} + 6}}{{\rm{x}}}}}{{\frac{{{\rm{cx}} - {\rm{d}}}}{{\rm{x}}}}}$
Or, $\frac{1}{3}$ = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ ∞ $\frac{{1 + \frac{6}{{\rm{x}}}}}{{{\rm{c}} - \frac{{\rm{d}}}{{\rm{x}}}}}$.
Or, $\frac{1}{3}$ = $\frac{{1 + 0}}{{{\rm{c}} - 0}}$
So, c = 3.
From (i), f(x) = $\frac{{{\rm{x}} + 6}}{{3{\rm{x}} - 1}}$.
f(13) = $\frac{{13 + 6}}{{3{\rm{*}}13 - 1}}$ = $\frac{{19}}{{38}}$ = $\frac{1}{2}$.
6.
Soln:
If the value of the function f(x) at x = a in the form $\frac{0}{0}$ which is meaningless but indicates only that the value in the numerator and denominator are each zero is known as an indeterminate form. The other indeterminate form are $\frac{{\rm{\infty }}}{{\rm{\infty }}}$, ∞ - ∞, etc.
=x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ ∞ $\sqrt {\rm{x}} \left( {\sqrt {\rm{x}} - \sqrt {{\rm{x}} - {\rm{a}}} } \right)$ (∞ - ∞ form)
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ ∞ $\sqrt {\rm{x}} \left( {\sqrt {\rm{x}} - \sqrt {{\rm{x}} - {\rm{a}}} } \right)$*$\frac{{\sqrt {\rm{x}} + \sqrt {{\rm{x}} - {\rm{a}}} }}{{\sqrt {\rm{x}} + \sqrt {{\rm{x}} - {\rm{a}}} }}$.
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ ∞$\frac{{\sqrt {\rm{x}} \left( {{\rm{x}} - {\rm{x}} + {\rm{a}}} \right)}}{{\sqrt {\rm{x}} + \sqrt {{\rm{x}} - {\rm{a}}} }}$ = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ ∞ $\frac{{{\rm{a}}\sqrt {\rm{x}} }}{{\sqrt {\rm{x}} + \sqrt {{\rm{x}} - {\rm{a}}} }}$.
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ ∞ $\left( {\frac{{\frac{{{\rm{a}}\sqrt {\rm{x}} }}{{\sqrt {\rm{x}} }}}}{{\frac{{\sqrt {\rm{x}} + \sqrt {{\rm{x}} - {\rm{a}}} }}{{\sqrt {\rm{x}} }}}}} \right)$.
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ ∞ $\frac{{\rm{a}}}{{1 + \sqrt {1 - \frac{{\rm{a}}}{{\rm{x}}}} {\rm{\: }}}}{\rm{\: \: }}$= $\frac{{\rm{a}}}{{1 + \sqrt {1 + 0} }}$ = $\frac{{\rm{a}}}{2}$.
7.
Soln:
For x is not an integer, f(x) = 0.
x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 + f(x) = 0
So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 f(x) = 0When x is an integer, then f(x) = x
So, f(1) = 1
=, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 f(x) ≠ f(1).
8.
(i)
Soln:
LHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 $\left( {\frac{1}{{{\rm{x}} - 3}} - \frac{9}{{{{\rm{x}}^3} - 3{{\rm{x}}^2}}}} \right)$(∞ - ∞ form)
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 $\left( {\frac{1}{{{\rm{x}} - 3}} - \frac{9}{{{{\rm{x}}^2}\left( {{\rm{x}} - 3} \right)}}} \right)$ = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 $\frac{{{{\rm{x}}^2} - 9}}{{{{\rm{x}}^2}\left( {{\rm{x}} - 3} \right)}}$.
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 $\frac{{\left( {{\rm{x}} + 3} \right)\left( {{\rm{x}} - 3} \right)}}{{{{\rm{x}}^2}\left( {{\rm{x}} - 3} \right)}}$ = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 $\frac{{{\rm{x}} + 3}}{{{{\rm{x}}^2}}}$ = $\frac{{3 + 3}}{9}$ = $\frac{2}{3}$.
(ii)
Soln:
LHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 $\left( {\frac{{{{\rm{x}}^2} + 9}}{{{{\rm{x}}^2} - 9}} - \frac{3}{{{\rm{x}} - 3}}} \right)$(∞ - ∞ form)
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3${\rm{\: }}\frac{{{\rm{\: }}{{\rm{x}}^2} + 9 - 3\left( {{\rm{x}} + 3} \right)}}{{\left( {{\rm{x}} + 3} \right)\left( {{\rm{x}} - 3} \right)}}$ = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 $\frac{{{\rm{\: }}{{\rm{x}}^2} + 9 - 3{\rm{x}} - 9}}{{\left( {{\rm{x}} + 3} \right)\left( {{\rm{x}} - 3} \right)}}$.
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 $\frac{{\rm{x}}}{{{\rm{x}} + 3}}$ = $\frac{3}{{3 + 3}}$ = $\frac{1}{2}$.
9.
(i)
Soln:
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 $\frac{{{{\rm{x}}^{ - 3}} - {2^{ - 3}}}}{{{\rm{x}} - 2}}$$\left[ {\frac{0}{0}{\rm{form}}} \right]$.
Since, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a $\frac{{{{\rm{x}}^{\rm{n}}} - {{\rm{a}}^{\rm{n}}}}}{{{\rm{x}} - {\rm{a}}}}$ = nan-1, so
=x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 $\frac{{{{\rm{x}}^{ - 3}} - {2^{ - 3}}}}{{{\rm{x}} - 2}}$ = -3.(2)-3-1 = $ - \frac{3}{{16}}$.
(ii)
Soln:
=x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ ∞ $\frac{{{{\left( {2{\rm{x}} - 1} \right)}^6}{{\left( {3{\rm{x}} - 1} \right)}^4}}}{{{{\left( {2{\rm{x}} + 1} \right)}^{10}}}}$$\left( {\frac{{\rm{\infty }}}{{\rm{\infty }}}{\rm{form}}} \right)$.
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ ∞ $\frac{{{{\rm{x}}^6}{{\left( {2 - \frac{1}{{\rm{x}}}} \right)}^6}.{{\rm{x}}^4}{{\left( {3 - \frac{1}{{\rm{x}}}} \right)}^4}}}{{{{\rm{x}}^{10}}{{\left( {2 + \frac{1}{{\rm{x}}}} \right)}^{10}}}}$.
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ ∞ $\frac{{{{\left( {2 - \frac{1}{{\rm{x}}}} \right)}^6}{{\left( {3 - \frac{1}{{\rm{x}}}} \right)}^4}}}{{{{\left( {2 + \frac{1}{{\rm{x}}}} \right)}^{10}}}}$ = $\frac{{{2^6}{{.3}^4}}}{{{2^{10}}}}$ = $\frac{{{3^4}}}{{{2^4}}}$ = $\frac{{81}}{{16}}$.
(iii)
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\left( {{{\left( {1 + {\rm{x}}} \right)}^6} - 1} \right)}}{{{{\left( {1 + {\rm{x}}} \right)}^2} - 1}}$$\left[ {\frac{0}{0}{\rm{form}}} \right]$
Put 1 + x = y when x→0, y →1.
Or, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0$\frac{{\left( {{{\left( {1 + {\rm{x}}} \right)}^6} - 1} \right)}}{{{{\left( {1 + {\rm{x}}} \right)}^2} - 1}}$ = y $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 $\frac{{{{\rm{y}}^6} - 1}}{{{{\rm{y}}^2} - 1}}$.
= y $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 $\frac{{{{\left( {{{\rm{y}}^2}} \right)}^3} - {{\left( 1 \right)}^3}}}{{{{\rm{y}}^2} - 1}}$ = y $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 $\frac{{\left( {{{\rm{y}}^2} - 1} \right)\left( {{{\rm{y}}^4} + {{\rm{y}}^2} + 1} \right)}}{{{{\rm{y}}^2} - 1}}$.
= y $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 (y4 + y2 + 1) = 1 + 1 + 1 = 3.
(iv)
Soln:
= y $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a $\frac{{{{\left( {{\rm{x}} + 2} \right)}^{\frac{5}{2}}} - {{\left( {{\rm{a}} + 2} \right)}^{\frac{5}{2}}}}}{{{\rm{x}} - {\rm{a}}}}$$\left[ {\frac{0}{0}{\rm{form}}} \right]$.
= (x + 2) $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0${\rm{\: \: }}$(a + 2) $\frac{{{{\left( {{\rm{x}} + 2} \right)}^{\frac{5}{2}}} - {{\left( {{\rm{a}} + 2} \right)}^{\frac{5}{2}}}}}{{\left( {{\rm{x}} + 2} \right) - \left( {{\rm{a}} + 2} \right)}}$
= $\frac{5}{2}{\left( {{\rm{a}} + 2} \right)^{\frac{5}{2} - 1}}$$\left[ {{\rm{x\: }}\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}{\rm{\: a}}\frac{{{{\rm{x}}^{\rm{n}}} - {{\rm{a}}^{\rm{n}}}}}{{{\rm{x}} - {\rm{a}}}} = {\rm{n}}{{\rm{a}}^{{\rm{n}} - 1}}} \right]$.
= $\frac{5}{2}$(a + 2)3/2.
10.
Soln:
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a $\frac{{{{\rm{x}}^3} - {{\rm{a}}^3}}}{{{\rm{x}} - {\rm{a}}}}$ = 27.
Or, 3.a3-1 = 27
Or, a2 = 9
So, a = ± 3.
11.
(i)
Soln:
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{{\rm{sin}}{{\rm{x}}^0}}}{{\rm{x}}}$ = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\sin \frac{{{\rm{\pi x}}}}{{180}}}}{{\rm{x}}}$.
= x $\begin{array}{*{20}{c}}{{\rm{lim}}180}\\\to\end{array}$ 0 = $\frac{{\frac{{\sin \frac{{{\rm{\pi x}}}}{{180}}}}{{\rm{x}}}}}{{\frac{{{\rm{\pi x}}}}{{180}}}}$.$\frac{{\rm{\pi }}}{{180}}$ = 1.$\frac{{\rm{\pi }}}{{180}}$ = $\frac{{\rm{\pi }}}{{180}}$.
(ii)
Soln:
= θ $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{1 - {\rm{cos}}4\theta }}{{1 - {\rm{cos}}6\theta }}$$\left( {\frac{0}{0}{\rm{form}}} \right)$
= θ $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{2{{\sin }^2}2\theta }}{{2{{\sin }^2}3\theta }}$ = θ $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{{{\sin }^2}2\theta }}{{{{\sin }^2}3\theta }}$ = θ $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 ${\left( {\frac{{{\rm{sin}}2\theta }}{{{\rm{sin}}3\theta }}} \right)^2}$.
= θ $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\left( {\frac{{\frac{{{\rm{sin}}2\theta }}{{2\theta }}.2\theta }}{{\frac{{{\rm{sin}}3\theta }}{{3\theta }}.3\theta }}} \right){\rm{\: }}$= θ $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\left( {\frac{{2{\rm{sin}}2\theta }}{{2\theta }}.\frac{2}{{\frac{{{\rm{sin}}3\theta }}{{3\theta }}.3}}} \right)$
(iii)
Soln:
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ π/2 $\frac{{{\rm{cosx}}}}{{\frac{{\rm{\pi }}}{2} - {\rm{x}}}}$$\left( {\frac{0}{0}{\rm{form}}} \right)$
Put x = $\frac{{\rm{\pi }}}{2}$ – h. When x →$\frac{{\rm{\pi }}}{2}$, h →0.
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ π/2 $\frac{{{\rm{cosx}}}}{{\frac{{\rm{\pi }}}{2} - {\rm{x}}}}{\rm{\: }}$= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\cos \left( {\frac{{\rm{\pi }}}{2} - {\rm{h}}} \right)}}{{\frac{{\rm{\pi }}}{2} - \left( {\frac{{\rm{\pi }}}{2} - {\rm{h}}} \right)}}$.
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{{\rm{sinh}}}}{{\rm{h}}}$ = 1$.$
(iv)
Soln:
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{1{\rm{tan}}2{\rm{x}} - {\rm{x}}}}{{3{\rm{x}} - {\rm{sinx}}}}$$\left( {\frac{0}{0}{\rm{form}}} \right)$
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\frac{{{\rm{sin}}2{\rm{x}}}}{{{\rm{cos}}2{\rm{x}}}} - {\rm{x}}}}{{3{\rm{x}} - {\rm{sinx}}}}$ = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{{\rm{sin}}2{\rm{x}} - {\rm{xcos}}2{\rm{x}}}}{{{\rm{cos}}2{\rm{x}}\left( {3{\rm{x}} - {\rm{sinx}}} \right)}}$ = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\frac{{{\rm{sin}}2{\rm{x}}}}{{2{\rm{x}}}}.2 - {\rm{cos}}2{\rm{x}}}}{{{\rm{cos}}2{\rm{x}}\left( {3 - \frac{{{\rm{sinx}}}}{{\rm{x}}}} \right)}} = \frac{{1.2 - 1}}{{1.\left( {3 - 1} \right)}}$ = $\frac{1}{2}$.
(v)
Soln:
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ π $\frac{{1 - {\rm{sinx}}/2}}{{{{\left( {{\rm{\pi }} - {\rm{x}}} \right)}^2}}}$$\left( {\frac{0}{0}{\rm{form}}} \right)$
Put x =${\rm{\: }}$π – h. When x →π, h →0.
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ π $\frac{{1 - {\rm{sinx}}/2}}{{{{\left( {{\rm{\pi }} - {\rm{x}}} \right)}^2}}}{\rm{\: }}$= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{1 - \sin \left( {\frac{{{\rm{\pi }} - {\rm{h}}}}{2}} \right)}}{{\frac{{\rm{\pi }}}{2} - \left( {\frac{{\rm{\pi }}}{2} - {\rm{h}}} \right)}}$.
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{1 - \sin \left( {\frac{{\rm{\pi }}}{2} - \frac{{\rm{h}}}{2}} \right)}}{{\frac{{\rm{\pi }}}{2} - \left( {\frac{{\rm{\pi }}}{2} - {\rm{h}}} \right)}}$ = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{1 - \cos \frac{{\rm{h}}}{2}}}{{{{\rm{h}}^2}}}$
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{2{{\sin }^2}\frac{{\rm{h}}}{4}}}{{{{\rm{h}}^2}}}$ = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{2{{\left( {\sin \frac{{\rm{h}}}{4}} \right)}^2}}}{{{{\rm{h}}^2}}}{\rm{\: }}$
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $2{\left( {\frac{{\sin \frac{{\rm{h}}}{4}}}{{\frac{{\rm{h}}}{4}.4}}} \right)^2}$
= 2.1.$\frac{1}{{16}}$ = $\frac{1}{8}$.
(vi)
Soln:
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ π/2 $\frac{{1 + {\rm{cos}}2{\rm{x}}}}{{{{\left( {{\rm{\pi }} - 2{\rm{x}}} \right)}^2}}}$$\left( {\frac{0}{0}{\rm{form}}} \right)$
Put x =${\rm{\: }}\frac{{\rm{\pi }}}{2} - {\rm{h}}$. When x →$\frac{{\rm{\pi }}}{2}$, h →0.
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{1 - {\rm{cos}}2{\rm{h}}}}{{{{\left( {{\rm{\pi }} - 2{\rm{x}}} \right)}^2}}}{\rm{\: }}$= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{1 - \cos 2\left( {\frac{{\rm{\pi }}}{2} - {\rm{h}}} \right)}}{{{{\left( {{\rm{\pi }} - {\rm{\pi }} + 2{\rm{h}}} \right)}^2}}}$.
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{1 - {\rm{cos}}2{\rm{h}}}}{{4{{\rm{h}}^2}}}$ = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{2{{\sin }^2}{\rm{h}}}}{{4{{\rm{h}}^2}}}$
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{1}{2}.{\left( {\frac{{{\rm{sinh}}}}{{\rm{h}}}} \right)^2}$ = $\frac{1}{2}.{\left( 1 \right)^2}$ = $\frac{1}{2}$.
(vii)
Soln:
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\sin \frac{1}{{\rm{x}}}$
If x →0, $\frac{1}{{\rm{x}}}$→-∞ and if x →0 + $\frac{1}{{\rm{x}}}$→∞. In both cases, sin $\frac{1}{{\rm{x}}}$ can have any value between -1 and +1, so x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 sin $\frac{1}{{\rm{x}}}$ does not exist.
(viii)
Soln:
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 xsin $\frac{1}{{\rm{x}}}$.
For x →0+, sin $\frac{1}{{\rm{x}}}$ can have a finite number between -1 and +1.
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0+ xsin $\frac{1}{{\rm{x}}}$ = 0 * finite number = 0
Also, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 – xsin $\frac{1}{{\rm{x}}}$ = 0 * finite number = 0
So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 xsin $\frac{1}{{\rm{x}}}$ = 0.
(ix)
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a $\frac{{{\rm{sinx}} - {\rm{sina}}}}{{\sqrt {\rm{x}} - \sqrt {\rm{a}} }}$$\left( {\frac{0}{0}{\rm{form}}} \right)$.
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a $\frac{{{\rm{sinx}} - {\rm{sina}}}}{{\sqrt {\rm{x}} - \sqrt {\rm{a}} }}{\rm{*}}\frac{{\sqrt {\rm{x}} + \sqrt {\rm{a}} }}{{\sqrt {\rm{x}} + \sqrt {\rm{a}} }}$.
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a $\left( {\frac{{2\cos \frac{{{\rm{x}} + {\rm{a}}}}{2}.\sin \frac{{{\rm{x}} - {\rm{a}}}}{2}}}{{{\rm{x}} - {\rm{a}}}}} \right).\left( {\sqrt {\rm{x}} + \sqrt {\rm{a}} } \right)$.
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a 2cos$\frac{{{\rm{x}} + {\rm{a}}}}{2}\left( {\sqrt {\rm{x}} + \sqrt {\rm{a}} } \right)$ = x – a $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\left( {\frac{{\sin \frac{{{\rm{x}} - {\rm{a}}}}{2}}}{{\frac{{{\rm{x}} - {\rm{a}}}}{2}}}} \right)$
= 2cos $\frac{{2{\rm{a}}}}{2}\left( {\sqrt {\rm{a}} + \sqrt {\rm{a}} } \right).\frac{1}{2}$ = $2\sqrt {\rm{a}} $cosa.
(x)
Soln:
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a (a – x)tan $\frac{{{\rm{\pi x}}}}{{2{\rm{a}}}}$$\left( {\frac{0}{0}{\rm{form}}} \right)$
Put x =${\rm{\: }}$a – h. When x →a, h →0.
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a (a – x) tan $\frac{{{\rm{\pi x}}}}{{2{\rm{a}}}}{\rm{\: }}$= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 h.tan$\frac{{\rm{\pi }}}{{2{\rm{a}}}}$(a – h).
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 h.tan$\left( {\frac{{\rm{\pi }}}{2} - \frac{{{\rm{\pi h}}}}{{2{\rm{a}}}}} \right)$ = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 h.cot $\frac{{{\rm{\pi h}}}}{{2{\rm{a}}}}$.
= h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 h.$\frac{{\cos \frac{{{\rm{\pi h}}}}{{2{\rm{a}}}}}}{{\sin \frac{{{\rm{\pi h}}}}{{\rm{a}}}}}$ = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\cos \frac{{{\rm{\pi h}}}}{{2{\rm{a}}}}}}{{\frac{{\left( {\sin \frac{{{\rm{\pi h}}}}{{2{\rm{a}}}}} \right)}}{{\frac{{{\rm{\pi h}}}}{{2{\rm{a}}}}}}.\frac{{\rm{\pi }}}{{2{\rm{a}}}}}}$ = $\frac{1}{{1.\frac{{\rm{\pi }}}{{2{\rm{a}}}}}}$ = $\frac{{2{\rm{a}}}}{{\rm{\pi }}}$.${\rm{\: \: }}$
(xi)
Soln:
= y $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\left( {{\rm{x}} + {\rm{y}}} \right)\sec \left( {{\rm{x}} + {\rm{y}}} \right) - {\rm{xsecx}}}}{{\rm{y}}}$$\left( {\frac{0}{0}{\rm{form}}} \right)$.
= y $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{{\rm{ysec}}\left( {{\rm{x}} + {\rm{y}}} \right) + {\rm{x}}\left[ {\sec \left( {{\rm{x}} + {\rm{y}}} \right) - {\rm{secx}}} \right]}}{{\rm{y}}}$
= y $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\left[ {\sec \left( {{\rm{x}} + {\rm{y}}} \right) + \frac{{\rm{x}}}{{\rm{y}}}\left\{ {\frac{1}{{\cos \left( {{\rm{x}} + {\rm{y}}} \right)}} - \frac{1}{{{\rm{cosx}}}}} \right\}} \right]$
= y $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\left[ {\sec \left( {{\rm{x}} + {\rm{y}}} \right) + \frac{{\rm{x}}}{{\rm{y}}}.\frac{{{\rm{cosx}} - {\rm{cosx}}\left( {{\rm{x}} + {\rm{y}}} \right)}}{{\cos \left( {{\rm{x}} + {\rm{y}}} \right).{\rm{cosx}}}}} \right]$
= y $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\left[ {\sec \left( {{\rm{x}} + {\rm{y}}} \right) + \frac{{\frac{{\rm{x}}}{{\rm{y}}}\left( {2\sin \frac{{{\rm{x}} + {\rm{x}} + {\rm{y}}}}{2}.\sin \frac{{{\rm{x}} + {\rm{y}} - {\rm{x}}}}{2}} \right)}}{{\cos \left( {{\rm{x}} + {\rm{y}}} \right).{\rm{cosx}}}}} \right]$
= y $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\left[ {\sec \left( {{\rm{x}} + {\rm{y}}} \right) + \frac{{\rm{x}}}{{\rm{y}}}.\frac{{2\sin \left( {{\rm{x}} + \frac{{\rm{y}}}{2}} \right){\rm{\: }}\sin \frac{{\rm{y}}}{2}}}{{\cos \left( {{\rm{x}} + {\rm{y}}} \right).{\rm{cosx}}}}} \right]$
= y $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $[\sec \left( {{\rm{x}} + {\rm{y}}} \right) + \frac{{{\rm{x}}.2\sin \left( {{\rm{x}} + \frac{{\rm{y}}}{2}} \right).\sin \frac{{\rm{y}}}{2}}}{{\cos \left( {{\rm{x}} + {\rm{y}}} \right).{\rm{cosx}}.\frac{{\rm{y}}}{2}.2}}$
= secx + $\frac{{{\rm{x}}.2{\rm{sinx}}}}{{{\rm{cosx}}.{\rm{cosx}}}}.\frac{1}{2}$ = secx + xtanx.secx.
12.
(i)
Soln:
For x < 1, f(x) = 3x2 + 2
LHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 – (3x2 + 2) = 3*1 + 2 = 5.
For, x > 1, f(x) = 2x + 3.
RHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 + (2x + 3) = 2*1 + 3 = 5.
LHL = RHL
So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 f(x) = 5.
(ii)
Soln:
For x < 1, f(x) = 3x2 + 2
LHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 – (3x2 + 2) = 3*1 + 2 = 5.
For, x > 1, f(x) = 2x + 3.
RHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 + (2x + 3) = 2*1 + 3 = 5.
LHL = RHL
So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 f(x) = 5.
(ii)
Soln:
For x < 0, f(x) = 3 + 2x
LHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 – (3 + 2x) = 3 + 2*0 = 3.
For, x > 0, f(x) = 3 – 2x.
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 (3 – 2x) = 3 – 2*0 = 3.
So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 f(x) = 3.
Again, For x <$\frac{3}{2}$, f(x) = 3 – 2x
LHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$$\frac{3}{2}$ – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$$\frac{3}{2}$ (3 – 2x) = 3 – 2.$\frac{3}{2}$ = 0.
RHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$$\frac{3}{2}$ + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$$\frac{3}{2}$ (-3 – 2x) = -3 – 2.$\frac{3}{2}$ = -6.
So, LHL ≠ RHL
So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$$\frac{3}{2}$+ f(x) does not exist.
13.
a.
Soln:
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{{{\rm{e}}^{{\rm{px}}}} - 1}}{{{{\rm{e}}^{{\rm{qx}}}} - 1}},\frac{0}{0}$ = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\left( {\frac{{\frac{{{{\rm{e}}^{{\rm{px}}}} - 1}}{{{\rm{px}}}}}}{{\frac{{{{\rm{e}}^{{\rm{qx}}}} - 1}}{{{\rm{qx}}}}}}} \right)$.$\frac{{\rm{p}}}{{\rm{q}}}$ = $\frac{1}{1}$.$\frac{{\rm{p}}}{{\rm{q}}}$ = $\frac{{\rm{p}}}{{\rm{q}}}$.
b.
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{{{\rm{e}}^{\rm{x}}} - {{\rm{e}}^{ - {\rm{x}}}} + {\rm{x}}}}{{\rm{x}}}$,$\left( {\frac{0}{0}} \right)$ = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\left( {\frac{{{{\rm{e}}^{\rm{x}}} - 1 - {{\rm{e}}^{ - {\rm{x}}}} + 1}}{{\rm{x}}} + 1} \right)$.
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\left( {\frac{{{{\rm{e}}^{\rm{x}}} - 1}}{{\rm{x}}}} \right)$ - x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\left( {\frac{{{{\rm{e}}^{ - {\rm{x}}}} - 1}}{{\rm{x}}}} \right)$ + 1 = 1 + 1 + 1 = 3.
c.
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{{{\rm{a}}^{\rm{x}}} - 1}}{{{{\rm{b}}^{\rm{x}}} - 1}}$, $\left( {\frac{0}{0}} \right)$ = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\left( {\frac{{\frac{{\frac{{{{\rm{a}}^{\rm{x}}} - 1}}{{\rm{x}}}}}{{{{\rm{b}}^{\rm{x}}} - 1}}}}{{\rm{x}}}} \right)$ = $\frac{{{\rm{loga}}}}{{{\rm{logb}}}}$.
14.
Soln:
a.
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{{2^{\rm{x}}} - 1}}{{{\rm{sinx}}}}$, $\left( {\frac{0}{0}} \right)$ = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\left( {\frac{{\frac{{{2^{\rm{x}}} - 1}}{{\rm{x}}}}}{{\frac{{{\rm{sinx}}}}{{\rm{x}}}}}} \right)$ = $\frac{{{\rm{x\: }}\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}{\rm{\: }}0\frac{{{2^{\rm{x}}} - 1}}{{\rm{x}}}}}{{{\rm{x\: }}\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}{\rm{\: }}0\frac{{{\rm{sinx}}}}{{\rm{x}}}}}$ = log2/
b.
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{{{\rm{e}}^{{\rm{sinx}}}} - {\rm{sinx}} - 1}}{{\rm{x}}}$, $\left( {\frac{0}{0}} \right)$ = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\left( {\frac{{{{\rm{e}}^{{\rm{sinx}}}} - 1}}{{\rm{x}}}} \right)$ – x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\left( {\frac{{{\rm{sinx}}}}{{\rm{x}}}} \right)$ = 1 – 1 = 0.
c.
Soln:
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ π/2 $\frac{{{{\rm{e}}^{{\rm{cosx}}}} - 1}}{{\frac{{\rm{\pi }}}{2} - {\rm{x}}}}$, $\left( {\frac{0}{0}} \right)$.
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ π/2 $\frac{{{{\rm{e}}^{\sin \left( {\frac{{\rm{\pi }}}{2} - {\rm{x}}} \right)}} - 1}}{{\frac{{\rm{\pi }}}{2} - {\rm{x}}}}$ = 1 = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ π/2 $\frac{{{{\rm{e}}^{\sin \left( {\frac{{\rm{\pi }}}{2} - {\rm{x}}} \right)}} - 1}}{{ - \left( {{\rm{x}} - \frac{{\rm{\pi }}}{2}} \right)}}$ = 1.
[OR put x – $\frac{{\rm{\pi }}}{2}$ = y, then evaluate]
d.
= x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ e $\frac{{{\rm{logx}} - 1}}{{{\rm{x}} - {\rm{e}}}}$, $\left( {\frac{0}{0}} \right)$.
Let x →e = y →x = y + e.
So, x →e = y ⇒ x = y + e.
So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ e $\frac{{{\rm{logx}} - 1}}{{{\rm{x}} - {\rm{e}}}}$ = y $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\log \left( {{\rm{y}} + {\rm{e}}} \right) - {\rm{loge}}}}{{\rm{y}}}$ [log e = 1]
= y $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 .$\frac{1}{{\rm{y}}}$ log$\left( {\frac{{{\rm{y}} + {\rm{e}}}}{{\rm{e}}}} \right)$ = y $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\left( {\frac{{\log \left( {1 + \frac{{\rm{y}}}{{\rm{e}}}} \right)}}{{\frac{{\rm{y}}}{{\rm{e}}}}}} \right)$.$\frac{1}{{\rm{e}}}$ = 1.$\frac{1}{{\rm{e}}}$ = $\frac{1}{{\rm{e}}}$.
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