NCERT Solutions For Class 11 Maths Chapter 13 Exercise 13.1

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Chapter 13 Ex.13.1 Question 1

Evaluate the given limit: \(\mathop {\lim }\limits_{x \to 3} x + 3\)

Solution

\[\begin{align}\mathop {\lim }\limits_{x \to 3} x + 3 &= 3 + 3\\ &= 6\end{align}\]

Chapter 13 Ex.13.1 Question 2

Evaluate the given limit: \(\mathop {\lim }\limits_{x \to \pi } \left( {x - \frac{{22}}{7}} \right)\)

Solution

\[\mathop {\lim }\limits_{x \to \pi } \left( {x - \frac{22}{7}} \right) = \left( {\pi - \frac{22}{7}} \right)\]

Chapter 13 Ex.13.1 Question 3

Evaluate the given limit: \(\mathop {\lim }\limits_{r \to 1} \pi {r^2}\)

Solution

\[\begin{align}\mathop {\lim }\limits_{r \to 1} \pi {r^2} &= \pi {\left( 1 \right)^2}\\&= \pi\end{align}\]

Chapter 13 Ex.13.1 Question 4

Evaluate the given limit: \(\mathop {\lim }\limits_{x \to 4} \frac{{4x + 3}}{{x - 2}}\)

Solution

\[\begin{align}\mathop {\lim }\limits_{x \to 4} \frac{4x + 3}{x - 2} &= \frac{4\left( 4 \right) + 3}{4 - 2}\\&= \frac{16 + 3}{2}\\&= \frac{19}{2}\end{align}\]

Chapter 13 Ex.13.1 Question 5

Evaluate the given limit: \(\mathop {\lim }\limits_{x \to - 1} \frac{{{x^{10}} + {x^5} + 1}}{{x - 1}}\)

Solution

\[\begin{align}\mathop {\lim }\limits_{x \to - 1} \frac{{x^{10} + {x^5} + 1}}{x - 1} &= \frac{{{{\left( - 1 \right)}^{10}} + {{\left(- 1\right)}^5} + 1}}{ - 1 - 1}\\&= \frac{1 - 1 + 1}{ - 2}\\&= - \frac{1}{2}a\end{align}\]

Chapter 13 Ex.13.1 Question 6

Evaluate the given limit: \(\mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {x + 1} \right)}^5} - 1}}{x}\)

Solution

It is given that, \(\mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {x + 1} \right)}^5} - 1}}{x}\)

Put \(x + 1 = y\), so that \(x = y - {\rm{1}}\)

Accordingly,

\[\begin{align} \underset{x\to 0}{\mathop{\lim }}\,\frac{{{\left( x+1 \right)}^{5}}-1}{x}&=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\left( y \right)}^{5}}-1}{y-1} \\& =5\times {{1}^{5-1}} \qquad \left[ \because \underset{x\to 0}{\mathop{\lim }}\,\frac{{x}^{n}}-{{a}^{n}}{x-a}=n{{a}^{n-1}} \right] \\& =5\end{align}\]

Hence, \(\mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {x + 1} \right)}^5} - 1}}{x} = 5\)

Chapter 13 Ex.13.1 Question 7

Evaluate the given limit: \(\mathop {\lim }\limits_{x \to 2} \frac{{3{x^2} - x - 10}}{{{x^2} - 4}}\)

Solution

At \(x = 2\), the value of the given rational function takes the form \(\frac{0}{0}.\)

\[\begin{align}\mathop {\lim }\limits_{x \to 2} \frac{3{x^2} - x - 10}{{x^2} - 4} &= \mathop {\lim }\limits_{x \to 2} \frac{\left( {x - 2} \right)\left( {3x + 5} \right)}{\left( {x - 2} \right)\left( {x + 2} \right)}\\&= \mathop {\lim }\limits_{x \to 2} \frac{{\left(3x + 5\right)}}{\left( {x + 2} \right)}\\&= \frac{3\left( 2 \right) + 5}{2 + 2}\\&= \frac{11}{4}\end{align}\]

Chapter 13 Ex.13.1 Question 8

Evaluate the given limit: \(\mathop {\lim }\limits_{x \to 3} \frac{{{x^4} - 81}}{{2{x^2} - 5x - 3}}\)

Solution

At \(x = 3\), the value of the given rational function takes the form \(\frac{0}{0}.\)

\[\begin{align}\mathop {\lim }\limits_{x \to 3} \frac{{x^4 - 81}}{2{x^2} - 5x - 3}& = \mathop {\lim }\limits_{x \to 3} \frac{{\left(x - 3\right)\left(x + 3\right)\left( {x^2 + 9} \right)}}{{\left(x - 3\right)\left(2x + 1\right)}}\\&= \mathop {\lim }\limits_{x \to 3} \frac{{\left(x + 3\right)\left( {x^2 + 9} \right)}}{{\left(2x + 1\right)}}\\&= \frac{{\left( {3 + 3} \right)\left({3^2} + 9 \right)}}{2\left( 3 \right) + 1}\\&= \frac{6 \times 18}{6 + 1}\\&= \frac{108}{7}\end{align}\]

Chapter 13 Ex.13.1 Question 9

Evaluate the given limit: \(\mathop {\lim }\limits_{x \to 0} \frac{{ax + b}}{{cx + 1}}\)

Solution

\[\begin{align}\mathop {\lim }\limits_{x \to 0} \frac{ax + b}{cx + 1} &= \frac{{a\left( 0 \right) + b}}{{c\left( 0 \right) + 1}}\\&= b\end{align}\]

Chapter 13 Ex.13.1 Question 10

Evaluate the Given limit:\(\mathop {\lim }\limits_{z \to 1} \frac{{{z^{\frac{1}{3}}} - 1}}{{{z^{\frac{1}{6}}} - 1}}\)

Solution

At \(z = 1,\) the value of the given function takes the form \(\frac{0}{0}.\)

Put \({z^{\frac{1}{6}}} = x\) so that \(z \to 1\) as \(x \to {\rm{1}}\)

Accordingly,

\[\begin{align}& \underset{z\to 1}{\mathop{\lim }}\,\frac{{{z}^{\frac{1}{3}}}-1}{{{z}^{\frac{1}{6}}}-1}=\underset{x\to 1}{\mathop{\lim }}\,\frac{{x}^{2}-1}{x-1} \\& =\underset{x\to 1}{\mathop{\lim }}\,\frac{{x}^{2}-1}{x-1} \\& =2\times {1}^{2-1} \qquad \left[ \because \underset{x\to a}{\mathop{\lim }}\,\frac{{x}^{n}-{a}^{n}}{x-a}=n{a}^{n-1} \right] \\& =2\end{align}\]

Chapter 13 Ex.13.1 Question 11

Evaluate the given limit: \(\mathop {\lim }\limits_{x \to 1} \frac{{a{x^2} + bx + c}}{{c{x^2} + bx + a}},\;a + b + c \ne 0\)

Solution

\[\begin{align}\mathop {\lim }\limits_{x \to 1} \frac{a{x^2} + bx + c}{c{x^2} + bx + a}& = \frac{{a{{\left( 1 \right)}^2} + b\left( 1 \right) + c}}{{c{{\left( 1 \right)}^2} + b\left( 1 \right) + a}}\\&= \frac{a + b + c}{a + b + c}\\&= 1 \qquad \qquad \left [ {a + b + c \ne 0} \right]\end{align}\]

Chapter 13 Ex.13.1 Question 12

Evaluate the Given limit: \(\mathop {\lim }\limits_{x \to - 2} \frac{{\frac{1}{x} + \frac{1}{2}}}{{x + 2}}\)

Solution

At \(x = - 2\), the value of the given function takes the form \(\frac{0}{0}.\)

Now,

\[\begin{align}\mathop {\lim }\limits_{x \to - 2} \frac{{\frac{1}{x} + \frac{1}{2}}}{x + 2} &= \mathop {\lim }\limits_{x \to - 2} \frac{{\left( {\frac{{2 + x}}{{2x}}} \right)}}{x + 2}\\&= \mathop {\lim }\limits_{x \to - 2} \frac{1}{{2x}}\\&= \frac{1}{{2\left(- 2 \right)}}\\&= - \frac{1}{4}\end{align}\]

Chapter 13 Ex.13.1 Question 13

Evaluate the Given limit: \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{bx}}\)

Solution

\(\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{bx}}\)

At \(x = 0\), the value of the given function takes the form \(\frac{0}{0}.\)

Now,

\[\begin{align}\mathop {\lim }\limits_{x \to 0} \frac{\sin ax}{bx}& = \mathop {\lim }\limits_{x \to 0} \frac{\sin ax}{ax} \times \frac{ax}{bx}\\&= \mathop {\lim }\limits_{x \to 0} \left( {\frac{\sin ax}{ax}} \right) \times \frac{a}{b}\\&= \frac{a}{b}\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin ax}}{{ax}}} \right) \qquad \left[ {x \to 0 \Rightarrow ax \to 0} \right]\\&= \frac{a}{b} \times 1 \qquad \left[\because {\mathop {\lim }\limits_{y \to 0} \left( {\frac{{\sin y}}{y}} \right) = 1} \right]\\&= \frac{a}{b}\end{align}\]

Chapter 13 Ex.13.1 Question 14

Evaluate the Given limit:\(\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{\sin bx}},\;a\;,b \ne 0\)

Solution

At \(x = 0\), the value of the given function takes the form \(\frac{0}{0}.\)

Now,

\[\begin{align}\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{\sin bx}} &= \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\frac{{\sin ax}}{{ax}}} \right) \times ax}}{{\left( {\frac{{\sin bx}}{{bx}}} \right) \times bx}}\\ &= \frac{a}{b} \times \frac{{\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin ax}}{{ax}}} \right)}}{{\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin bx}}{{bx}}} \right)}} \qquad \left[ \begin{array}{l} x \to 0 \Rightarrow ax \to 0\\and\;x \to 0 \Rightarrow bx \to 0\end{array} \right]\\& = \frac{a}{b} \times \frac{1}{1} \qquad \left[ \because{\mathop {\lim }\limits_{y \to 0} \left( {\frac{{\sin y}}{y}} \right) = 1} \right]\\& = \frac{a}{b}\end{align}\]

Chapter 13 Ex.13.1 Question 15

Evaluate the Given limit: \(\mathop {\lim }\limits_{x \to \pi } \frac{{\sin \left( {\pi - x} \right)}}{{\pi \left( {\pi - x} \right)}}\)

Solution

It can be seen that \({\rm{ }}x \to \pi \Rightarrow \left( {\pi - x} \right) \to 0\)

Therefore,

\[\begin{align} \underset{x\to \pi }{\mathop{\lim }}\,\frac{\sin \left( \pi -x \right)}{\pi \left( \pi -x \right)}&=\frac{1}{\pi }\underset{(\pi -x)\to 0}{\mathop{\lim }}\,\frac{\sin \left( \pi -x \right)}{\pi \left( \pi -x \right)} \\& \text{=}\frac{1}{\pi }\times 1 \qquad \left[ \because \underset{y\to 0}{\mathop{\lim }}\,\frac{\sin y}{y}=1 \right] \\& \text{=}\frac{1}{\pi }\end{align}\]

Chapter 13 Ex.13.1 Question 16

Evaluate the Given limit: \(\mathop {\lim }\limits_{x \to 0} \frac{{\cos x}}{{\pi - x}}\)

Solution

\[\begin{align}\mathop {\lim }\limits_{x \to 0} \frac{\cos x}{\pi - x} &= \frac{\cos 0}{\pi - 0}\\ &= \frac{1}{\pi }\end{align}\]

Chapter 13 Ex.13.1 Question 17

Evaluate the Given limit: \(\mathop {\lim }\limits_{x \to 0} \frac{\cos 2x - 1}{\cos x - 1}\)

Solution

\[\mathop {\lim }\limits_{x \to 0} \frac{{\cos 2x - 1}}{{\cos x - 1}}\]

At \(x = 0\), the value of the given function takes the form \(\frac{0}{0}.\)

Now,

\[\begin{align} \underset{x\to 0}{\mathop{\lim }}\,\frac{\cos 2x-1}{\cos x-1}&=\underset{x\to 0}{\mathop{\lim }}\,\frac{1-2{{\sin }^{2}}x-1}{1-2{{\sin }^{2}}\frac{x}{2}-1} \qquad \left[ \because \cos x=1-2{{\sin }^{2}}\frac{x}{2} \right] \\& \text{ }=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{2}}x}{{{\sin }^{2}}\frac{x}{2}} \\& =\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( \frac{{{\sin ^2}}x}{{x^2}} \right)\times {x^2}}{\left( \frac{{\sin }^{2}\frac{x}{2}}{\left( \frac{x}{2} \right)} \right)\times \frac{x^2}{4}}\\\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos 2x-1}{\cos x-1}&=4\frac{\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{{\sin^2}x}{{x^2}} \right)}{\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{{\sin ^2}\frac{x}{2}}{{{\left( \frac{x}{2} \right)}^{2}}} \right)} \\& =4\frac{{{\left( \underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x} \right)}^{2}}}{{{\left( \underset{x\to 0}{\mathop{\lim }}\,\frac{\sin \frac{x}{2}}{{{\left( \frac{x}{2} \right)}^{2}}} \right)^2}}}\text{ }\left[ x\to 0\Rightarrow \frac{x}{2}\to 0 \right] \\& =4\times \frac{{1^2}}{{1^2}}\qquad \left[ \because \underset{y\to 0}{\mathop{\lim }}\,\frac{\sin y}{y}=1 \right] \\& =4\end{align}\]

Chapter 13 Ex.13.1 Question 18

Evaluate the given limit: \(\mathop {\lim }\limits_{x \to 0} \frac{{ax + x\cos x}}{{b\sin x}}\)

Solution

At \(x = 0\), the value of the given function takes the form \(\frac{0}{0}.\)

Now,

\[\begin{align} \underset{x\to 0}{\mathop{\lim }}\,\frac{ax+x\cos x}{b\sin x}&=\frac{1}{b}\underset{x\to 0}{\mathop{\lim }}\,\frac{x\left( a+\cos x \right)}{\sin x} \\& =\frac{1}{b}\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{x}{\sin x} \right)\times \underset{x\to 0}{\mathop{\lim }}\,\left( a+\cos x \right) \\& =\frac{1}{b}\times \frac{1}{\left( \underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x} \right)}\times \underset{x\to 0}{\mathop{\lim }}\,\left( a+\cos x \right) \\& =\frac{1}{b}\times \left( a+\cos 0 \right) \qquad \left[ \because \underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}\text{=1} \right] \\& =\frac{a+1}{b}\end{align}\]

Chapter 13 Ex.13.1 Question 19

Evaluate the Given limit: \(\mathop {\lim }\limits_{x \to 0} x\sec x\)

Solution

\[\begin{align}\mathop {\lim }\limits_{x \to 0} x\sec x &= \mathop {\lim }\limits_x \to 0 \frac{x}{\cos x}\\&= \frac{0}{\cos 0}\\&= \frac{0}{1}\\&= 0\end{align}\]

Chapter 13 Ex.13.1 Question 20

Evaluate the Given limit: \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax + bx}}{{ax + \sin bx}}a,b,a + b \ne 0\)

Solution

At \(x = 0\), the value of the given function takes the form \(\frac{0}{0}.\)

Now,

\[\begin{align}\mathop {\lim }\limits_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx} &= \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\frac{\sin ax}{ax}} \right)ax + bx}}{{ax + bx\left( {\frac{\sin bx}{bx}} \right)}}\\&= \frac{{\left( {\mathop {\lim }\limits_{x \to 0} \frac{\sin ax}{ax}} \right) \times \mathop {\lim }\limits_{x \to 0} \left( {ax} \right) + \mathop {\lim \left( {bx} \right)}\limits_{x \to 0} }}{{\mathop {\lim }\limits_{x \to 0} ax + \mathop {\lim }\limits_{x \to 0} bx\left( {\mathop {\lim }\limits_{x \to 0} \frac{\sin bx}{bx}} \right)}} \\& \qquad \qquad \left[ {{\rm{As\; }}x \to 0 \Rightarrow ax \to 0{\text{ and b}}x \to 0} \right]\\&= \frac{{\mathop {\lim }\limits_{x \to 0} \left( {ax} \right) + \mathop {\lim }\limits_{x \to 0} bx}}{{\mathop {\lim }\limits_{x \to 0} ax + \mathop {\lim }\limits_{x \to 0} bx}} \qquad \left[ {\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} = 1} \right]\\&= \frac{{\mathop {\lim }\limits_{x \to 0} \left( {ax + bx} \right)}}{{\mathop {\lim }\limits_{x \to 0} \left( ax + bx \right)}}\\&= \mathop {\lim }\limits_{x \to 0} \left( 1 \right)\\&= 1\end{align}\]

Chapter 13 Ex.13.1 Question 21

Evaluate the Given limit: \(\mathop {\lim }\limits_{x \to 0} \left( {{\rm{cosec}}\;x - \cot x} \right)\)

Solution

At \(x = 0\), the value of the given function takes the form \(\infty - \infty \)

Now,

\[\begin{align}\mathop {\lim }\limits_{x \to 0} \left( {{\rm{cosec}}\;x - \cot x} \right) &= \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{\sin x}} - \frac{{\cos x}}{{\sin x}}} \right)\\ &= \mathop {\lim }\limits_{x \to 0} \left( {\frac{{1 - \cos x}}{{\sin x}}} \right)\\ &= \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\frac{{1 - \cos x}}{x}} \right)}}{{\left( {\frac{{\sin x}}{x}} \right)}}\\ &= \frac{{\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{x}}}{{\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x}}}{\rm{       }}\\ &= \frac{0}{1} \qquad \quad  \left[\because {\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{x} = 0{\rm{ and }}\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1} \right]\\ &= 0\end{align}\]

Chapter 13 Ex.13.1 Question 22

Evaluate the Given limit: \(\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\tan 2x}}{{x - \frac{\pi }{2}}}\)

Solution

At \(x = \frac{\pi }{2}\), the value of the given function takes the form \(\frac{0}{0}.\)

Now, put \(x - \frac{\pi }{2} = y\) so that \(x \to \frac{\pi }{2},y \to 0\)

Therefore,

\[\begin{align} \underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,\frac{\tan 2x}{x-\frac{\pi }{2}}&=\underset{y\to 0}{\mathop{\lim }}\,\frac{\tan 2\left( y+\frac{\pi }{2} \right)}{y} \\& =\underset{y\to 0}{\mathop{\lim }}\,\frac{\tan \left( \pi +2y \right)}{y} \\& =\underset{y\to 0}{\mathop{\lim }}\,\frac{\tan 2y}{y} \qquad \quad \left[ \because \tan \left( \pi +2y \right)=\tan 2y \right] \\& =\underset{y\to 0}{\mathop{\lim }}\,\frac{\sin 2y}{y\cos 2y} \\& =\underset{y\to 0}{\mathop{\lim }}\,\left( \frac{\sin 2y}{2y}\times \frac{2}{\cos 2y} \right) \\& =\left( \underset{y\to 0}{\mathop{\lim }}\,\frac{\sin 2y}{2y} \right)\times \underset{y\to 0}{\mathop{\lim }}\,\left( \frac{2}{\cos 2y} \right)\text{ }\left[ y\to 0\Rightarrow 2y\to 0 \right]\end{align}\]

\[\begin{align} \underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,\frac{\tan 2x}{x-\frac{\pi }{2}}&=1\times \frac{2}{\cos 0} \qquad \left[ \because \underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}=1 \right] \\& =1\times \frac{2}{1} \\& =2\end{align}\]

Chapter 13 Ex.13.1 Question 23

Find \(\mathop {\lim }\limits_{x \to 0} f\left( x \right)\) and \(\mathop {\lim }\limits_{x \to 1} f\left( x \right),\) where \({\rm{ }}f\left( x \right) = \left\{ \begin{array}{l}2x + 3,\;\;\;\;\;\;\;\;\;x \le 0\\3\left( {x + 1} \right),\;\;\;\;\;\;x > 0\end{array} \right.\)

Solution

The given function is \({\rm{ }}f\left( x \right) = \left\{ \begin{array}{l}2x + 3,\;\;\;\;\;\;\;\;\;x \le 0\\3\left( {x + 1} \right),\;\;\;\;\;\;x > 0\end{array} \right.\)

Now,

\[\begin{align}\mathop {\lim }\limits_{x \to 0^ - } f\left( x \right) &= \mathop {\lim }\limits_{x \to 0} \left[ {2x + 3} \right]\\ &= 2\left( 0 \right) + 3\\ &= 3\\\mathop {\lim }\limits_{x \to 0^ + } f\left( x \right)& = \mathop {\lim }\limits_x \to 0 {\rm{3}}\left( x{\rm{ + 1}} \right)\\ &= 3\left( {0 + 1} \right)\\& = 3\end{align}\]

Hence,

\[\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to 0^ + } f\left( x \right) = \mathop {\lim }\limits_x \to 0 f\left( x \right) = 3\]

Now,

\[\begin{align}\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) &= \mathop {\lim }\limits_{x \to 1} \left( {x + 1} \right)\\ &= 3\left( 1 + 1 \right)\\& = 6\end{align}\]

Hence,

\[\mathop {\lim }\limits_{x \to 1^ + } f\left( x \right) = \mathop {\lim }\limits_{x \to 1^ - } f\left( x \right) = \mathop {\lim }\limits_{x \to 1} f\left( x \right) = 6\]

Chapter 13 Ex.13.1 Question 24

Find \(\mathop {\lim }\limits_{x \to 1} f\left( x \right),\)where \(f\left( x \right) = \left\{ \begin{array}{l}{x^2} - 1, \quad x \le 1\\- x - {\rm{1,}}\quad x{\rm{ > 1}}\end{array} \right.\)

Solution

The given function is \(f\left( x \right) = \left\{ \begin{array}{l}{x^2} - 1,\;\;\;\;\;\;\;\;x \le 1\\- x - {\rm{1,}}\;\;\;\;\;\;\;x{\rm{ > 1}}\end{array} \right.\)

Now,

\[\begin{align}\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right)& = \mathop {\lim }\limits_{x \to 1} \left[ {{x^2} - 1} \right]\\&= {1^2} - 1\\&= 1 - 1\\&= 0\end{align}\]

It is observed that \(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to 1} f\left( x \right)\)

Hence, \(\mathop {\lim }\limits_{x \to 1} f\left( x \right)\) does not exist.

Chapter 13 Ex.13.1 Question 25

Evaluate \(\mathop {\lim }\limits_{x \to 0} f\left( x \right),\) where \(f\left( x \right) = \left\{ \begin{array}{l}\frac{{\left| x \right|}}{x},\;\;\;\;\;\;x \ne 0{\rm{ }}\\{\rm{ }}0,\;\;\;\;\;\;\;x = 0\end{array} \right.\)

Solution

The given function is \(f\left( x \right) = \left\{ \begin{align}\frac{{\left| x \right|}}{x}, \quad x \ne 0{\rm{ }}\\{\rm{ }}0,\quad x = 0\end{align} \right.\)

Now,

\[\begin{align}\mathop {\lim }\limits_{x \to 0^ - } f\left( x \right) &= \mathop {\lim }\limits_{x \to 0^ - } \left[ {\frac{{\left| x \right|}}{x}} \right]= \mathop {\lim }\limits_{x \to 0} \left( {\frac{ - x}{x}} \right) \qquad \left[ {{\rm{When \;}}x{\text{ is negative, }}\left| x \right| = - x} \right]\\&=\mathop {\lim }\limits_{x \to 0} \left( { - 1} \right)\\& = - 1\\\\\mathop {\lim }\limits_{x \to 0^ + } f\left( x \right)& = \mathop {\lim }\limits_{x \to {0^ + }} \left[ {\frac{{\left| x \right|}}{x}} \right]\\&=\mathop {\lim }\limits_{x \to 0} \left( {\frac{x}{x}} \right) \qquad \left[ {{\text{When }}x{\text{ is positive, }}\left| x \right| = x} \right]\\ &=\mathop {\lim }\limits_{x \to 0} \left( 1 \right)\\& = 1\end{align}\]

It is observed that \(\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)\)

Hence, \(\mathop {\lim }\limits_{x \to 0} f\left( x \right)\) does not exist.

Chapter 13 Ex.13.1 Question 26

Find \(\mathop {\lim }\limits_{x \to 0} f\left( x \right),\) where \({\rm{ }}f\left( x \right) = \left\{ \begin{array}{l}\frac{x}{{\left| x \right|}},\;\;\;\;\;x \ne 0{\rm{ }}\\{\rm{ }}0,\;\;\;\;\;\;x = 0\end{array} \right.\)

Solution

The given function is \({\rm{ }}f\left( x \right) = \left\{ \begin{array}{l}\frac{x}{{\left| x \right|}},\;\;\;\;\;x \ne 0{\rm{ }}\\{\rm{ }}0,\;\;\;\;\;\;x = 0\end{array} \right.\)

\[\begin{align}\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) &=\mathop {\lim }\limits_{x \to {0^ - }} \left[ {\frac{x}{{\left| x \right|}}} \right]\\&= \mathop {\lim }\limits_{x \to 0} \left( {\frac{x}{{ - x}}} \right) \qquad \left[ {{\rm{when \,}}x < 0,\;\left| x \right| = - x} \right]\\&=\mathop {\lim }\limits_{x \to 0} \left( { - 1} \right)\\& = - 1\\\\\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) &=\mathop {\lim }\limits_{x \to {0^ + }} \left[ {\frac{x}{{\left| x \right|}}} \right]\&=\mathop {\lim }\limits_{x \to 0} \left( {\frac{x}{x}} \right) \qquad \left[ {{\rm{when\; }}x > 0,\;\left| x \right| = x} \right]\\&=\mathop {\lim }\limits_{x \to 0} \left( 1 \right)\\& = 1\end{align}\]

It is observed that \(\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right).\)

Hence, \(\mathop {\lim }\limits_{x \to 0} f\left( x \right)\) does not exist.

Chapter 13 Ex.13.1 Question 27

Find \(\mathop {\lim }\limits_{x \to 5} f\left( x \right)\), where\({\rm{ }}f\left( x \right) = \left| x \right| - 5\)

Solution

The given function is \(f\left( x \right) = \left| x \right| - 5\)

\[\begin{align}\mathop {\lim }\limits_{x \to {5^ - }} f\left( x \right)&=\mathop {\lim }\limits_{x \to {5^ - }} \left( {\left| x \right| - 5} \right)\\&=\mathop {\lim }\limits_{x \to 5} \left( {x - 5} \right) \qquad \left[ {{\rm{when\; }}x > 0,\;\left| x \right| = x} \right]\\& = 5 - 5\\& = 0\\\\ \mathop {\lim }\limits_{x \to {5^ + }} f\left( x \right) &=\mathop {\lim }\limits_{x \to {5^ + }} \left( {\left| x \right| - 5} \right)\\&=\mathop {\lim }\limits_{x \to 5} \left( {x - 5} \right)\qquad \left[ {{\rm{when \;}}x > 0,\;\left| x \right| = x} \right]\\& = 5 - 5\\ &= 0\end{align}\]

Therefore,

\[\mathop {\lim }\limits_{x \to {5^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {5^ + }} f\left( x \right) = 0\]

Hence, \(\mathop {\lim }\limits_{x \to 5} f\left( x \right) = 0\).

Chapter 13 Ex.13.1 Question 28

Suppose \(f\left( x \right) = \left\{ \begin{array}{l}a - bx,\;\;\;\;x < 0\\4,\;\;\;\;\;\;\;\;\;\;\;x = 1{\rm{ }}\\b - ax,\;\;\;\;x > 1\end{array} \right.\) and if \(\mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( 1 \right)\) what are possible values of \(a\) and \(b\)?

Solution

The given function is \(f\left( x \right) = \left\{ \begin{array}{l}a - bx,\;\;\;\;x < 0\\4,\;\;\;\;\;\;\;\;\;\;\;x = 1{\rm{ }}\\b - ax,\;\;\;\;x > 1\end{array} \right.\)

Therefore,

\[\begin{align}\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) &= \mathop {\lim }\limits_{x \to 1} \left( {a + bx} \right)\\ &= a + b\\\\\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) &= \mathop {\lim }\limits_{x \to 1} \left( {b - ax} \right)\\ &= b - a\\f\left( 1 \right) &= 4\end{align}\]

It is given that \(\mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( 1 \right)\)

Therefore, \(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( 1 \right)\)

Hence, \(a + b = 4\) and \(b - a = 4\)

On solving these two equations, we obtain

\(a = 0\) and \(b = 4\)

Thus, the respective possible values of \(a\) and \(b\) are \(0\) and \(4.\)

Chapter 13 Ex.13.1 Question 29

Let \({a_{_1}},{a_2}, \ldots ,{a_n}\) be fixed real numbers and define a function

\[f\left( x \right) = \left( {x - {a_1}} \right)\left( {x - {a_2}} \right) \ldots \left( {x - {a_n}} \right)\]

What is \(\mathop {\lim }\limits_{x \to {a_1}} f\left( x \right)\)? For some \(a \ne {a_{_1}},{a_2}, \ldots ,{a_n},\) compute \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\)

Solution

The given function is \(f\left( x \right) = \left( {x - {a_1}} \right)\left( {x - {a_2}} \right) \ldots \left( {x - {a_n}} \right)\)

\[\begin{align}\mathop {\lim }\limits_{x \to {a_1}} f\left( x \right) &= \mathop {\lim }\limits_{x \to {a_1}} \left[ {\left( {x - {a_1}} \right)\left( {x - {a_2}} \right) \ldots \left( {x - {a_n}} \right)} \right]\\&= \left( {{a_1} - {a_1}} \right)\left( {{a_1} - {a_2}} \right) \ldots \left( {{a_1} - {a_n}} \right)\\\mathop {\lim }\limits_{x \to {a_1}} f\left( x \right) &= 0\end{align}\]

Now,

\[\begin{align}\mathop {\lim }\limits_{x \to a} f\left( x \right) &= \mathop {\lim }\limits_{x \to a} \left[ {\left( {x - {a_1}} \right)\left( {x - {a_2}} \right) \ldots \left( {x - {a_n}} \right)} \right]\\&= \left( {a - {a_1}} \right)\left( {a - {a_2}} \right) \ldots \left( {a - {a_n}} \right)\end{align}\]

Thus, \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = \left( {a - {a_1}} \right)\left( {a - {a_2}} \right) \ldots \left( {a - {a_n}} \right)\)

Chapter 13 Ex.13.1 Question 30

If \(f\left( x \right) = \left\{ \begin{array}{l}\left| x \right| + 1,\;\;\;\;\;\;x < 0\\0,\;\;\;\;\;\;\;\;\;\;\;\;x = 0\\\left| x \right| - 1,\;\;\;\;\;\;x > 1\end{array} \right.\)

For what value (s) of does \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\) exists?

Solution

The given function is \({\rm{If }}f\left( x \right) = \left\{ \begin{array}{l}\left| x \right| + 1,\;\;\;\;\;\;x < 0\\{\rm{ 0,}}\;\;\;\;\;\;\;\;\;\;\;x = {\rm{0}}\\\left| x \right| - 1,\;\;\;\;\;\;x > 1\end{array} \right.\)

When \(a = 0\)

\[\begin{align}\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) &= \mathop {\lim }\limits_{x \to {0^ - }} f\left( {\left| x \right| + 1} \right)\\&= \mathop {\lim }\limits_{x \to 0} \left( { - x + 1} \right)\qquad\left[ {If{\rm{ }}x < 0,\left| x \right| = - x} \right]\\&= 0 + 1\\&= 1\\\\\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)& = \mathop {\lim }\limits_{x \to {0^ + }} \left( {\left| x \right| - 1} \right)\\&= \mathop {\lim }\limits_{x \to 0} \left( {x - 1} \right) \qquad \left[ {If{\rm{ }}x > 0,\left| x \right| = x} \right]\\&= 0 - 1\\&= - 1\end{align}\]

Here it is observed that \(\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)\)

Therefore, \(\mathop {\lim }\limits_{x \to 0} f\left( x \right)\) does not exist at \(x = a\) where \(a = 0\).

When \(a < 0\)

\[\begin{align}\mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) &= \mathop {\lim }\limits_{x \to {a^ - }} \left( {\left| x \right| + 1} \right)\\&= \mathop {\lim }\limits_{x \to a} \left( { - x + 1} \right){\rm{ }}\left[ {x < a < 0 \Rightarrow \left| x \right| = - x} \right]\\&= - a + 1\\\\\mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) &= \mathop {\lim }\limits_{x \to {a^ + }} \left( {\left| x \right| + 1} \right)\\&= \mathop {\lim }\limits_{x \to a} \left( { - x + 1} \right){\rm{ }}\left[ {a < x < 0 \Rightarrow \left| x \right| = - x} \right]\\&= - a + 1\end{align}\]

Here, \(\mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = - a + 1\)

Thus, limit of \(f\left( x \right)\) exists at \(x = a\) where \(a < 0\)

When \(a > 0\)

\[\begin{align}\mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) &= \mathop {\lim }\limits_{x \to {a^ - }} \left( {\left| x \right| + 1} \right)\\&= \mathop {\lim }\limits_{x \to a} \left( {x - 1} \right){\rm{ }}\left[ {0 < x < a \Rightarrow \left| x \right| = x} \right]\\&= a - 1\\\\\mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) &= \mathop {\lim }\limits_{x \to {a^ + }} \left( {\left| x \right| - 1} \right)\\&= \mathop {\lim \left( {x - 1} \right)}\limits_{x \to a} {\rm{ }}\left[ {0 < x < a \Rightarrow \left| x \right| = x} \right]\\&= a - 1\end{align}\]

Here, \(\mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = a - 1\)

Thus, limit of \(f\left( x \right)\) exists at \(x = a\) where \(a > 0\)

Hence, \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\) exists for all \(a \ne 0.\)

Chapter 13 Ex.13.1 Question 31

If the function \(f\left( x \right)\) satisfies \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 2}}{{{x^2} - 1}} = \pi ,\) evaluate \(\mathop {\lim }\limits_{x \to 1} f\left( x \right).\)

Solution

It is given that the function \(f\left( x \right)\) satisfies \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 2}}{{{x^2} - 1}} = \pi ,\)

\[\begin{align}\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 2}}{x^2 - 1} &= \frac{{\mathop {\lim }\limits_{x \to 1} \left( {f\left( x \right) - 2} \right)}}{{\mathop {\lim }\limits_{x \to 1} \left( {{x^2} - 1} \right)}} = \pi \\&\Rightarrow \; \mathop {\lim }\limits_{x \to 1} \left( {f\left( x \right) - 2} \right) = \pi \mathop {\lim }\limits_{x \to 1} \left( {{x^2} - 1} \right)\\&\Rightarrow \;\mathop {\lim }\limits_{x \to 1} \left( {f\left( x \right) - 2} \right) = \pi \left( {{1^2} - 1} \right)\\&\Rightarrow \;\mathop {\lim }\limits_{x \to 1} \left( {f\left( x \right) - 2} \right) = 0\\&\Rightarrow \mathop {\lim }\limits_{x \to 1} f\left( x \right) - \mathop {\lim }\limits_{x \to 1} 2 = 0\\&\Rightarrow \;\mathop {\lim }\limits_{x \to 1} f\left( x \right) - 2 = 0\\&\Rightarrow\; \mathop {\lim }\limits_{x \to 1} f\left( x \right) = 2\end{align}\]

Hence, \(\mathop {\lim }\limits_{x \to 1} f\left( x \right) = 2\)

Chapter 13 Ex.13.1 Question 32

If \(f\left( x \right) = \left\{ \begin{array}{l}m{x^2} + n,\;\;\;\;\;x < 0\\nx + m,{\rm{ }}\;\;\;\;\;0 \le x \le 1\\n{x^3} + m,\;\;\;\;\;x > 1\end{array} \right.\). For what integers \(m\) and \(n\) does both \(\mathop {\lim }\limits_{x \to 0} f\left( x \right)\) and \(\mathop {\lim }\limits_{x \to 1} f\left( x \right)\) exists?

Solution

It is given that \(f\left( x \right) = \left\{ \begin{array}{l}m{x^2} + n,\;\;\;\;\;x < 0\\nx + m,{\rm{ }}\;\;\;\;\;0 \le x \le 1\\n{x^3} + m,\;\;\;\;\;x > 1\end{array} \right.\)

Therefore,

\[\begin{align}\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right)& = \mathop {\lim }\limits_{x \to 0} \left( {m{x^2} + n} \right)\\&= m{\left( 0 \right)^2} + n\\&= n\\\\\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) &= \mathop {\lim }\limits_{x \to 0} \left( {nx + m} \right)\\&= n\left( 0 \right) + m\\&= m\end{align}\]

Thus, \(\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)\) exists if \(m = n\)

Now,

\[\begin{align}\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) &= \mathop {\lim }\limits_{x \to 1} \left( {nx + m} \right)\\ &= n\left( 1 \right) + m\\ &= m + n\\\\\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) &= \mathop {\lim }\limits_{x \to 1} \left( {n{x^3} + m} \right)\\ &= n{\left( 1 \right)^3} + m\\ &= m + n\end{align}\]

Therefore, \(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} f\left( x \right)\)

Thus, \(\mathop {\lim }\limits_{x \to 1} f\left( x \right)\) exists for any integral value of \(m\) and \(n\).

  
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