Q1Written answer
Evaluate the Given limit.$\underset{x\to 3}{\mathop{\lim }}\,x+3$
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Worked solution
Ans. Given,
$\underset{x\to 3}{\mathop{\lim }}\,x+3$
=$3+3$
=$6$
NCERT · Class 11 · Maths
Chapter 13: Limits and Derivatives Exercise 13.1
Practice each question, then open the answer or worked solution.
Exercise 13.1
32 questionsQ2Written answer
Evaluate the Given limit.$\underset{x\to \pi }{\mathop{\lim }}\,\left( x-\frac{22}{7} \right)$
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Worked solution
Ans. Given,
$\underset{x\to \pi }{\mathop{\lim }}\,\left( x-\frac{22}{7} \right)$
=\[\left( \pi -\frac{22}{7} \right)\]
Q3Written answer
Evaluate the Given limit. $\underset{r\to 1}{\mathop{\lim }}\,\pi {{r}^{2}}$
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Worked solution
Ans. Given,
$\underset{r\to 1}{\mathop{\lim }}\,\pi {{r}^{2}}$
=\[\pi ({{1}^{2}})\]
=\[\pi \]
Q4Written answer
Evaluate the Given limit.\[\underset{x\to 1}{\mathop{\lim }}\,\frac{4x+3}{x-2}\]
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Worked solution
Ans. Given,
\[\underset{x\to 1}{\mathop{\lim }}\,\frac{4x+3}{x-2}\]
=$\frac{4(4)+3}{4-2}$
=$\frac{16+3}{2}$
=$\frac{19}{2}$
Q5Written answer
Evaluate the Given limit. \[\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{10}}+{{x}^{5}}+1}{x-1}\]
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Worked solution
Ans. Given,
\[\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{10}}+{{x}^{5}}+1}{x-1}\]
=$\frac{{{(-1)}^{10}}+{{(-1)}^{5}}+1}{-1-1}$
=$\frac{1-1+1}{-2}$
=$-\frac{1}{2}$
Q6Written answer
Evaluate the Given limit. \[\underset{x\to 0}{\mathop{\lim }}\,\frac{(x+1)_{{}}^{5}-1}{x}\]
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Worked solution
Ans. Given,
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{(x+1)_{{}}^{5}-1}{x}\]
Put $x+1=y$ So,
$y\to 1$ as $x\to 0$
Than, \[\underset{x\to 0}{\mathop{\lim }}\,\frac{(x+1)_{{}}^{5}-1}{x}\]
=\[\underset{x\to 1}{\mathop{\lim }}\,\frac{(y)_{{}}^{5}-1}{y-1}\]
Using \[\left[ \underset{x\to a}{\mathop{\lim }}\,\frac{x_{{}}^{n}-{{a}^{n}}}{x-a}=n{{a}^{n-1}} \right]\]
=${{5.1}^{5-1}}$
=$5$
Q7Written answer
Evaluate the Given limit. \[\underset{x\to 2}{\mathop{\lim }}\,\frac{3{{x}^{2}}-x-10}{{{x}^{2}}-4}\]
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Worked solution
Ans. At $x=2$
The value of the given rational function takes the form \[\frac{0}{0}\]
=\[\underset{x\to 2}{\mathop{\lim }}\,\frac{3{{x}^{2}}-x-10}{{{x}^{2}}-4}\]
=\[\underset{x\to 2}{\mathop{\lim }}\,\frac{(x-2)(3x+5)}{(x-2)(x+2)}\]
=\[\underset{x\to 2}{\mathop{\lim }}\,\frac{3x+5}{x+2}\]
=\[\frac{3(2)+5}{2+2}\]
=$\frac{11}{4}$
Q8Written answer
Evaluate the Given limit. \[\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{4}}-81}{2{{x}^{2}}-5x-3}\]
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Worked solution
Ans. At $x=2$
The value of the given rational function takes the form \[\frac{0}{0}\]
=\[\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{4}}-81}{2{{x}^{2}}-5x-3}\]
=\[\underset{x\to 3}{\mathop{\lim }}\,\frac{(x-3)(x+3)({{x}^{2}}+9)}{(x-3)(2x+1)}\]
=\[\underset{x\to 3}{\mathop{\lim }}\,\frac{(x+3)({{x}^{2}}+9)}{(2x+1)}\]
=\[\frac{(3+3)({{3}^{2}}+9)}{(2(3)+1)}\]
=\[\frac{6\times 18}{7}\]
=\[\frac{108}{7}\]
Q9Written answer
Evaluate the Given limit.\[\underset{x\to 0}{\mathop{\lim }}\,\frac{ax+b}{cx+1}\]
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Worked solution
Ans. Given,
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{ax+b}{cx+1}\]
=\[\frac{a(0)+b}{c(0)+1}\]
=\[b\]
Q10Written answer
Evaluate the Given limit.\[\underset{z\to 1}{\mathop{\lim }}\,\frac{{{z}^{\frac{1}{3}}}-1}{{{z}^{\frac{1}{6}}}-1}\]
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Worked solution
Ans. At $z=1$
The value of the given rational function takes the form \[\frac{0}{0}\]
Put ${{z}^{\frac{1}{6}}}=x$ So,
$z\to 1$ as $x\to 1$
Than,\[\underset{z\to 1}{\mathop{\lim }}\,\frac{{{z}^{\frac{1}{3}}}-1}{{{z}^{\frac{1}{6}}}-1}\]
=\[\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{2}}-1}{{{x}^{{}}}-1}\]
Using \[\left[ \underset{x\to a}{\mathop{\lim }}\,\frac{x_{{}}^{n}-{{a}^{n}}}{x-a}=n{{a}^{n-1}} \right]\]
=${{2.1}^{2-1}}$
=$2$