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CBSE Class 11 Mathematics Chapter 12 Limits and Derivatives – NCERT Solutions 2025-26

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Download Free PDF of Limits and Derivatives for Class 11 Maths Chapter 12

Stepping into Class 11 Mathematics, you encounter Chapter 12, where the concept of limits forms the heart of calculus. The primary keyword, NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.1, connects your classroom learning directly with practical, exam-focused answers, crafted to match the latest board requirements. Here, even if terms like "indeterminate form" seem tricky at first, a clear, stepwise approach will guide you past any confusion.

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Many students search for ways to master "class 11 maths limits and derivatives ex 12.1"—because understanding each definition and calculation builds the foundation for all higher Maths. This content prioritizes definitions, common mistake alerts, and easy-to-follow steps so you can identify error-prone calculations and confidently tackle board-style questions within the syllabus.


Since this chapter often carries about 8 marks in the CBSE exam, a focused revision of limit formulas and evaluation steps is essential. All solutions here are designed by trusted Vedantu academic teams, ensuring you build not just answers, but genuine clarity for both final exams and your mathematical journey ahead.

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Access NCERT Solutions for Maths Class 11 Chapter 12 - Limits and Derivatives

Exercise 12.1

1. Evaluate the Given limit.$\underset{x\to 3}{\mathop{\lim }}\,x+3$ 

Ans. Given,

 $\underset{x\to 3}{\mathop{\lim }}\,x+3$

=$3+3$

=$6$


2. Evaluate the Given limit.$\underset{x\to \pi }{\mathop{\lim }}\,\left( x-\frac{22}{7} \right)$ 

Ans. Given,

$\underset{x\to \pi }{\mathop{\lim }}\,\left( x-\frac{22}{7} \right)$

=\[\left( \pi -\frac{22}{7} \right)\]


3. Evaluate the Given limit. $\underset{r\to 1}{\mathop{\lim }}\,\pi {{r}^{2}}$

Ans. Given,

$\underset{r\to 1}{\mathop{\lim }}\,\pi {{r}^{2}}$

=\[\pi ({{1}^{2}})\]

=\[\pi \]


4. Evaluate the Given limit.\[\underset{x\to 1}{\mathop{\lim }}\,\frac{4x+3}{x-2}\]

Ans. Given,

\[\underset{x\to 1}{\mathop{\lim }}\,\frac{4x+3}{x-2}\]

=$\frac{4(4)+3}{4-2}$ 

=$\frac{16+3}{2}$ 

=$\frac{19}{2}$ 


5. Evaluate the Given limit. \[\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{10}}+{{x}^{5}}+1}{x-1}\]

Ans. Given,

\[\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{10}}+{{x}^{5}}+1}{x-1}\]

=$\frac{{{(-1)}^{10}}+{{(-1)}^{5}}+1}{-1-1}$ 

=$\frac{1-1+1}{-2}$ 

=$-\frac{1}{2}$ 


6. Evaluate the Given limit. \[\underset{x\to 0}{\mathop{\lim }}\,\frac{(x+1)_{{}}^{5}-1}{x}\]

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 \[\left[ \underset{x\to a}{\mathop{\lim }}\,\frac{x_{{}}^{n}-{{a}^{n}}}{x-a}=n{{a}^{n-1}} \right]\]

=${{5.1}^{5-1}}$

=$5$


7. Evaluate the Given limit. \[\underset{x\to 2}{\mathop{\lim }}\,\frac{3{{x}^{2}}-x-10}{{{x}^{2}}-4}\]

Ans. At $x=2$

The value of the given rational function takes the form \[\frac{0}{0}\] 

=\[\underset{x\to 2}{\mathop{\lim }}\,\frac{3{{x}^{2}}-x-10}{{{x}^{2}}-4}\]

=\[\underset{x\to 2}{\mathop{\lim }}\,\frac{(x-2)(3x+5)}{(x-2)(x+2)}\]

=\[\underset{x\to 2}{\mathop{\lim }}\,\frac{3x+5}{x+2}\]

=\[\frac{3(2)+5}{2+2}\]

=$\frac{11}{4}$


8. Evaluate the Given limit. \[\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{4}}-81}{2{{x}^{2}}-5x-3}\]

Ans. At $x=2$

The value of the given rational function takes the form \[\frac{0}{0}\] 

=\[\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{4}}-81}{2{{x}^{2}}-5x-3}\]

=\[\underset{x\to 3}{\mathop{\lim }}\,\frac{(x-3)(x+3)({{x}^{2}}+9)}{(x-3)(2x+1)}\]

=\[\underset{x\to 3}{\mathop{\lim }}\,\frac{(x+3)({{x}^{2}}+9)}{(2x+1)}\]

=\[\frac{(3+3)({{3}^{2}}+9)}{(2(3)+1)}\]

=\[\frac{6\times 18}{7}\]

=\[\frac{108}{7}\]


9. Evaluate the Given limit.\[\underset{x\to 0}{\mathop{\lim }}\,\frac{ax+b}{cx+1}\]

Ans. Given,

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{ax+b}{cx+1}\]

=\[\frac{a(0)+b}{c(0)+1}\]

=\[b\]


10. Evaluate the Given limit.\[\underset{z\to 1}{\mathop{\lim }}\,\frac{{{z}^{\frac{1}{3}}}-1}{{{z}^{\frac{1}{6}}}-1}\]

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 \[\left[ \underset{x\to a}{\mathop{\lim }}\,\frac{x_{{}}^{n}-{{a}^{n}}}{x-a}=n{{a}^{n-1}} \right]\]

=${{2.1}^{2-1}}$

=$2$


 11. Evaluate the Given limit. \[\underset{x\to 1}{\mathop{\lim }}\,\frac{a{{x}^{2}}+bx+c}{c{{x}^{2}}+bx+a}\],\[a+b+c\ne 0\]

Ans. Given,

\[\underset{x\to 1}{\mathop{\lim }}\,\frac{a{{x}^{2}}+bx+c}{c{{x}^{2}}+bx+a}\]

=\[\frac{a{{(1)}^{2}}+b(1)+c}{c{{(1)}^{2}}+b(1)+a}\]

=\[\frac{a+b+c}{c+b+a}\]

=$1$                    \[(a+b+c\ne 0)\]


12. Evaluate the Given limit. \[\underset{x\to -2}{\mathop{\lim }}\,\frac{\frac{1}{x}+\frac{1}{2}}{x+2}\]

Ans. Given,

\[\underset{x\to -2}{\mathop{\lim }}\,\frac{\frac{1}{x}+\frac{1}{2}}{x+2}\]

At $x=-2$

The value of the given rational function takes the form \[\frac{0}{0}\] 

\[\underset{x\to -2}{\mathop{\lim }}\,\frac{\frac{1}{x}+\frac{1}{2}}{x+2}\]=\[\underset{x\to -2}{\mathop{\lim }}\,\frac{\left( \frac{2+x}{2x} \right)}{x+2}\]

=\[\underset{x\to -2}{\mathop{\lim }}\,\frac{1}{2x}\]

=\[\frac{1}{2(-2)}\]

=\[-\frac{1}{4}\]


13. Evaluate the Given limit. \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ax}{bx}\]

Ans. Given,

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ax}{bx}\]

At $x=0$

The value of the given rational function takes the form \[\frac{0}{0}\] 

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ax}{bx}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ax}{bx}\times \frac{ax}{ax}\]

=\[\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{\sin ax}{bx} \right)\times \frac{a}{b}\]

=\[\frac{a}{b}\underset{ax\to 0}{\mathop{\lim }}\,\left( \frac{\sin ax}{bx} \right)\]

$x\to 0\Rightarrow ax\to 0$ 

=\[\frac{a}{b}\times 1\]\[\]         $\left[ \underset{x\to 0}{\mathop{\lim }}\,\left( \frac{\sin y}{y} \right) \right]$ 

=\[\frac{a}{b}\]


14. Evaluate the Given limit. \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ax}{\sin bx},a,b\ne 0\]

Ans.Given,

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ax}{\sin bx},a,b\ne 0\]

At $x=0$

The value of the given rational function takes the form \[\frac{0}{0}\] 

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ax}{\sin bx}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( \frac{\sin ax}{ax} \right)\times ax}{\left( \frac{\sin ax}{ax} \right)\times bx}\]

=\[\frac{a}{b}\times \frac{\underset{ax\to 0}{\mathop{\lim }}\,\left( \frac{\sin ax}{ax} \right)}{\underset{bx\to 0}{\mathop{\lim }}\,\left( \frac{\sin ax}{ax} \right)}\]         $\left[ \begin{align} & x\to 0\Rightarrow ax\to 0 \\ & x\to 0\Rightarrow bx\to 0 \\ \end{align} \right]$

=\[\frac{a}{b}\times \frac{1}{1}\]                   $\left[ \underset{x\to 0}{\mathop{\lim }}\,\left( \frac{\sin y}{y} \right)=1 \right]$

=\[\frac{a}{b}\]


15. Evaluate the Given limit. \[\underset{x\to \pi }{\mathop{\lim }}\,\frac{\sin (\pi -x)}{\pi (\pi -x)}\]

Ans. Given,

\[\underset{x\to \pi }{\mathop{\lim }}\,\frac{\sin (\pi -x)}{\pi (\pi -x)}\]

$\left[ x\to \pi \Rightarrow (\pi -x)\to 0 \right]$

\[\underset{x\to \pi }{\mathop{\lim }}\,\frac{\sin (\pi -x)}{\pi (\pi -x)}=\frac{1}{\pi }\underset{(\pi -x)\to 0}{\mathop{\lim }}\,\frac{\sin (\pi -x)}{(\pi -x)}\]

$=\frac{1}{\pi }\times 1$                      $\left[ \underset{x\to 0}{\mathop{\lim }}\,\left( \frac{\sin y}{y} \right)=1 \right]$

$=\frac{1}{\pi }$


16. Evaluate the Given limit. \[\underset{x\to 0}{\mathop{\lim }}\,\frac{cosx}{\pi -x}\]

Ans. Given,

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\operatorname{cosx}}{\pi -x}\]

=\[\frac{\cos 0}{\pi -0}\]

=\[\frac{1}{\pi }\]


17. Evaluate the Given limit. \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos 2x-1}{\cos x-1}\]

Ans. Given,

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

At $x=0$

The value of the given rational function takes the form \[\frac{0}{0}\] 

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos 2x-1}{\cos x-1}=\underset{x\to 0}{\mathop{\lim }}\,\frac{1-2{{\sin }^{2}}x-1}{1-2{{\sin }^{2}}\frac{x}{2}-1}\]

=\[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{2}}x}{{{\sin }^{2}}\frac{x}{2}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( \frac{{{\sin }^{2}}x}{{{x}^{2}}} \right)\times {{x}^{2}}}{\left( \frac{{{\sin }^{2}}\frac{x}{2}}{{{\left( \frac{x}{2} \right)}^{2}}} \right)\times \frac{{{x}^{2}}}{4}}\]

=\[4\frac{\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{{{\sin }^{2}}x}{{{x}^{2}}} \right)}{\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{{{\sin }^{2}}\frac{x}{2}}{{{\left( \frac{x}{2} \right)}^{2}}} \right)}\]

=\[4\frac{{{\left( \underset{x\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{2}}x}{{{x}^{2}}} \right)}^{2}}^{{}}}{{{\left( \underset{x\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{2}}\frac{x}{2}}{{{\left( \frac{x}{2} \right)}^{2}}} \right)}^{2}}}\]

$\left[ x\to 0\Rightarrow \frac{x}{2}\to 0 \right]$

=\[4\frac{{{1}^{2}}^{{}}}{{{1}^{2}}}\]         $\left[ \underset{x\to 0}{\mathop{\lim }}\,\left( \frac{\sin y}{y} \right)=1 \right]$

=\[4\]   


18. Evaluate the Given limit. \[\underset{x\to 0}{\mathop{\lim }}\,\frac{ax+x\cos x}{b\sin x}\]

Ans.Given,

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{ax+x\cos x}{b\sin x}\]

At $x=0$

The value of the given rational function takes the form \[\frac{0}{0}\] 

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{ax+x\cos x}{b\sin x}=\frac{1}{b}\underset{x\to 0}{\mathop{\lim }}\,\frac{x(a+\cos x)}{\sin x}\]

\[\frac{1}{b}\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{x}{\sin x} \right)\times \underset{x\to 0}{\mathop{\lim }}\,\left( a+\cos x \right)\]

=\[\frac{1}{b}\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1}{\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{\sin x}{x} \right)} \right)\times \underset{x\to 0}{\mathop{\lim }}\,\left( a+\cos x \right)\]

=\[\frac{1}{b}\times \left( a+\cos 0 \right)\]               $\left[ \underset{y\to 0}{\mathop{\lim }}\,\left( \frac{\operatorname{sinx}}{x} \right)=1 \right]$

  =\[\frac{a+1}{b}\]         


19. Evaluate the Given limit. \[\underset{x\to 0}{\mathop{\lim }}\,x\sec x\]

Ans. Given,

\[\underset{x\to 0}{\mathop{\lim }}\,x\sec x\]

  =\[\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{\cos x}\]        

  =\[\frac{0}{\cos 0}\]

=\[\frac{0}{1}\]

=\[0\]


20. Evaluate the Given limit. \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ax+bx}{ax+\sin bx}\]\[a,b,a+b\ne 0\]

Ans. At $x=0$

The value of the given rational function takes the form \[\frac{0}{0}\] 

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ax+bx}{ax+\sin bx}\]

  =\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( \frac{\sin ax}{ax} \right)ax+bx}{ax+bx\left( \frac{\sin bx}{bx} \right)}\]        

  =\[\frac{\left( \underset{x\to \infty }{\mathop{\lim }}\,\frac{\sin ax}{ax} \right)\times \underset{x\to 0}{\mathop{\lim }}\,(ax)+\underset{x\to 0}{\mathop{\lim }}\,(bx)}{\underset{x\to 0}{\mathop{\lim }}\,ax+\underset{x\to 0}{\mathop{\lim }}\,bx\left( \underset{x\to 0}{\mathop{\lim }}\,\frac{\sin bx}{bx} \right)}\]

\[\left[ x\to \pi \Rightarrow ax\to 0 \right]\] and \[\left[ bx\to 0 \right]\]

  =\[\frac{\underset{x\to 0}{\mathop{\lim }}\,(ax)+\underset{x\to 0}{\mathop{\lim }}\,(bx)}{\underset{x\to 0}{\mathop{\lim }}\,ax+\underset{x\to 0}{\mathop{\lim }}\,bx}\]         $\left[ \underset{y\to 0}{\mathop{\lim }}\,\left( \frac{\operatorname{sinx}}{x} \right)=1 \right]$

  =\[\frac{\underset{x\to 0}{\mathop{\lim }}\,(ax+bx)}{\underset{x\to 0}{\mathop{\lim }}\,(ax+bx)}\]   

  =\[\underset{x\to 0}{\mathop{\lim }}\,(1)\]         

  =\[1\]         


21. Evaluate the Given limit. \[\underset{x\to 0}{\mathop{\lim }}\,(\operatorname{cosecx}-cotx)\]

Ans. At $x=0$

The value of the given rational function takes the form \[\infty \to \infty \] 

\[\underset{x\to 0}{\mathop{\lim }}\,(\operatorname{cosecx}-cotx)\]

=\[\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1}{\sin x}-\frac{\cos x}{\sin x} \right)\]         

  =\[\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1-\cos x}{\sin x} \right)\]       

  =\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( \frac{1-\cos x}{x} \right)}{\left( \frac{\sin x}{x} \right)}\]         

  =\[\frac{\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1-\cos x}{x} \right)}{\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{\sin x}{x} \right)}\]   

$\left[ \underset{y\to 0}{\mathop{\lim }}\,\frac{1-\cos x}{x}=0 \right]$ and $\left[ \underset{y\to 0}{\mathop{\lim }}\,\left( \frac{\operatorname{sinx}}{x} \right)=1 \right]$

  =\[\frac{0}{1}\]        

  =\[0\]               


22. Evaluate the Given limit. \[\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,\frac{\tan 2x}{x-\frac{\pi }{2}}\]

Ans. Given, 

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan 2x}{x-\frac{\pi }{2}}\]

At $x=\frac{\pi }{2}$

The value of the given rational function takes the form \[\frac{0}{0}\] 

Put $x-\frac{\pi }{2}=y$ 

So,\[\left[ x\to \frac{\pi }{2},y\to 0 \right]\]

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan 2x}{x-\frac{\pi }{2}}=\underset{y\to 0}{\mathop{\lim }}\,\frac{\tan 2\left( y+\frac{\pi }{2} \right)}{y}\]

  =\[\underset{y\to 0}{\mathop{\lim }}\,\frac{\tan (\pi +2y)}{y}\]       

  =\[\underset{y\to 0}{\mathop{\lim }}\,\frac{\tan 2y}{y}\]    \[\left[ \tan (\pi +2y)=\tan 2y \right]\]

  =\[\underset{y\to 0}{\mathop{\lim }}\,\frac{\sin 2y}{y\cos 2y}\]

  =\[\underset{y\to 0}{\mathop{\lim }}\,\left( \frac{\sin 2y}{2y}\times \frac{2}{\cos 2y} \right)\]

  =\[\underset{y\to 0}{\mathop{\lim }}\,\left( \frac{\sin 2y}{2y} \right)\times \underset{y\to 0}{\mathop{\lim }}\,\left( \frac{2}{\cos 2y} \right)\]  \[\left[ y\to 0\Rightarrow 2y\to 0 \right]\]

  =\[1\times \frac{2}{\cos 0}\]    \[\left[ \underset{x\to 0}{\mathop{\lim }}\,\left( \frac{\operatorname{sinx}}{x} \right)=1 \right]\]

  =\[1\times \frac{2}{1}\]

  =\[2\]


23. Find $\underset{x\to 0}{\mathop{\lim }}\,f(x)$and $\underset{x\to 1}{\mathop{\lim }}\,f(x)$, where \[f(x)=\left\{ \begin{align} & 2x+3 \\ & 3(x+1) \\ \end{align} \right.\]  $\begin{align} & x\le 0 \\ & x>0 \\ \end{align}$

Ans. Given.

\[f(x)=\left\{ \begin{align} & 2x+3 \\ & 3(x+1) \\ \end{align} \right.\]    $\begin{align} & x\le 0 \\ & x>0 \\ \end{align}$

\[\underset{x\to 0}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,\left[ 2x+3 \right]\] 

  =\[2(0)+3=3\]

\[\underset{x\to 0}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,\left[ 3x+1 \right]\] 

  =\[3(0+1)=3\]

Therefore,

$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,f(x)=3$

$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{1}^{{}}}}{\mathop{\lim }}\,(x+1)=3(1+1)=6$

Therefore,

$\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to 1}{\mathop{\lim }}\,f(x)=6$


24. Find $\underset{x\to 1}{\mathop{\lim }}\,f(x)$, where \[f(x)=\left\{ \begin{align} & {{x}^{2}}-1 \\ & -x-1 \\ \end{align} \right.\]   $\begin{align} & x\le 1 \\ & x>1 \\ \end{align}$

Ans. Given, 

\[f(x)=\left\{ \begin{align} & {{x}^{2}}-1 \\ & -x-1 \\ \end{align} \right.\]     $$\begin{align} & x\le 1 \\ & x>1 \\ \end{align}$

$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{1}^{{}}}}{\mathop{\lim }}\,({{x}^{2}}-1)={{1}^{2}}-1=1-1=0$

So, it is observed that 

$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)\ne \underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)$ 

Hence, $\underset{x\to {{1}^{{}}}}{\mathop{\lim }}\,f(x)$ does not exist.


25. Find $\underset{x\to 0}{\mathop{\lim }}\,f(x)$, where \[f(x)=\left\{ \begin{align} & \frac{\left| x \right|}{x} \\ & 0 \\ \end{align} \right.\]    $\begin{align} & x\ne 0 \\ & x=0 \\ \end{align}$

Ans. Given

\[f(x)=\left\{ \begin{align} & \frac{\left| x \right|}{x} \\ & 0 \\ \end{align} \right.\] $\begin{align} & x\ne 0 \\ & x=0 \\ \end{align}$

$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left[ \frac{\left| x \right|}{x} \right]$

=\[\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{-x}{x} \right)\]    when x is negative, $\left| x \right|=-x$ 

=\[\underset{x\to 0}{\mathop{\lim }}\,(-1)\]

=\[-1\]

$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left[ \frac{\left| x \right|}{x} \right]$     

=$\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{x}{x} \right)$

 when x is positive, $\left| x \right|=x$ 

=$\underset{x\to 0}{\mathop{\lim }}\,(1)$

=\[1\]

$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)\ne \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)$

Hence, $\underset{x\to {{0}^{{}}}}{\mathop{\lim }}\,f(x)$ does not exist.


26. Find $\underset{x\to 0}{\mathop{\lim }}\,f(x)$, where \[f(x)=\left\{ \begin{align} & \frac{x}{\left| x \right|} \\ & 0 \\ \end{align} \right.\] $\begin{align} & x\ne 0 \\ & x=0 \\ \end{align}$

Ans. Given

\[f(x)=\left\{ \begin{align} & \frac{x}{\left| x \right|} \\ & 0 \\ \end{align} \right.\] $\begin{align} & x\ne 0 \\ & x=0 \\ \end{align}$ 

$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left[ \frac{x}{\left| x \right|} \right]$

=\[\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{-x}{x} \right)\]    when x is negative, $\left| x \right|=-x$ 

=\[\underset{x\to 0}{\mathop{\lim }}\,(-1)\]

=\[-1\]

 $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left[ \frac{x}{\left| x \right|} \right]$   
=$\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{x}{x} \right)$

 when x is positive, $\left| x \right|=x$ 

=\[\underset{x\to 0}{\mathop{\lim }}\,(1)\]

=\[1\]

$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)\ne \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)$

Hence, $\underset{x\to {{0}^{{}}}}{\mathop{\lim }}\,f(x)$ does not exist.


27. Find $\underset{x\to 5}{\mathop{\lim }}\,f(x),$ where $f(x)=\left| x \right|-5$ 

Ans. Given,

 $f(x)=\left| x \right|-5$

$\underset{x\to {{5}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{5}^{-}}}{\mathop{\lim }}\,(\left| x \right|-5)$ 

= $\underset{x\to 5}{\mathop{\lim }}\,(x-5)$    when x is positive, $\left| x \right|=x$ 

= $5-5$ 

= $0$ 

$\underset{x\to {{5}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{5}^{+}}}{\mathop{\lim }}\,(\left| x \right|-5)$ 

= $\underset{x\to 5}{\mathop{\lim }}\,(x-5)$ 

when x is positive, $\left| x \right|=x$ 

= $5-5$ 

= $0$ 

$\underset{x\to {{5}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{5}^{+}}}{\mathop{\lim }}\,f(x)$

Hence, $\underset{x\to 5}{\mathop{\lim }}\,f(x)=0$ 


28. Suppose $f(x)=\left\{ \begin{align} & a+bx \\ & 4 \\ & b-ax \\ \end{align} \right.$   $\begin{align} & x<0 \\ & x=1 \\ & x>1 \\ \end{align}$and if $\underset{x\to 1}{\mathop{\lim }}\,f(x)=f(1)$what are possible values of a and b?

Ans. The given function is

$f(x)=\left\{ \begin{align} & a+bx \\ & 4 \\ & b-ax \\ \end{align} \right.$ $\begin{align} & x<0 \\ & x=1 \\ & x>1 \\ \end{align}$

$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to 1}{\mathop{\lim }}\,f(ax+bx)=a+b$

$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to 1}{\mathop{\lim }}\,f(b-ax)=b-a$

$f(1)=4$

Given

$\underset{x\to 1}{\mathop{\lim }}\,f(x)=f(1)$ 

$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{1}^{{}}}}{\mathop{\lim }}\,f(x)=f(1)$

$a+b=4$and $b-a=4$

On solving,we get

$a=0$and $b=4$


29. Let a1,a2,……an be fixed real number and define a fuction

$f(x)=(x-{{a}_{1}})(x-{{a}_{2}})....(x-{{a}_{n}})$ 

What is $\underset{x\to {{a}_{1}}}{\mathop{\lim }}\,f(x)$ ? For some a$\ne $ a1, a2,…..,an. Compute $\underset{x\to a}{\mathop{\lim }}\,f(x)$.

Ans. Given,

$f(x)=(x-{{a}_{1}})(x-{{a}_{2}})....(x-{{a}_{n}})$

$\underset{x\to {{a}_{1}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}_{1}}}{\mathop{\lim }}\,[(x-{{a}_{1}})(x-{{a}_{2}})....(x-{{a}_{n}})]$

=\[({{a}_{1}}-{{a}_{1}})({{a}_{1}}-{{a}_{2}})......({{a}_{1}}-{{a}_{n}})=0\]

Therefore,

$\underset{x\to {{a}_{1}}}{\mathop{\lim }}\,f(x)=0$

Now, $\underset{x\to a}{\mathop{\lim }}\,f(x)=\underset{x\to a}{\mathop{\lim }}\,[(x-{{a}_{1}})(x-{{a}_{2}})....(x-{{a}_{n}})]$

=\[({{a}_{1}}-{{a}_{1}})({{a}_{1}}-{{a}_{2}})......({{a}_{1}}-{{a}_{n}})\]

Therefore,

$\underset{x\to a}{\mathop{\lim }}\,f(x)=({{a}_{1}}-{{a}_{1}})({{a}_{1}}-{{a}_{2}})......({{a}_{1}}-{{a}_{n}})$


30. If \[f(x)=\left\{ \begin{align} & \left| x \right|+1 \\ & 0 \\ & \left| x \right|-1 \\ \end{align} \right.\] $\begin{align} & x<0 \\ & x=0 \\ & x>1 \\ \end{align}$

For what value (s) of does $\underset{x\to a}{\mathop{\lim }}\,f(x)$  exists? 

Ans. Given,

 \[f(x)=\left\{ \begin{align} & \left| x \right|+1 \\ & 0 \\ & \left| x \right|-1 \\ \end{align} \right.\] $\begin{align} & x<0 \\ & x=0 \\ & x>1 \\ \end{align}$ 

When $a=0$ 

=\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left( \left| x \right|+1 \right)\]

=\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left( -x+1 \right)\]     when x is negative, $\left| x \right|=-x$ 

=$0+1$

=$1$ 

=\[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left( \left| x \right|+1 \right)\]

=\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left( x-1 \right)\]     when x is positive, $\left| x \right|=x$ 

=$0-1$

=$-1$ 

Here, $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)\ne \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)$ 

$\underset{x\to 0}{\mathop{\lim }}\,f(x)$ does not exist.

When\[\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,(\left| x \right|+1)\] $a<0$

\[\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,(\left| x \right|+1)\]

=\[\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,(-x+1)\]           $[x<a<0\Rightarrow \left| x \right|=-x]$

=\[-a+1\]

$\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f(x)=-a+1$ 

Thus, limit exists at $x=a$, where \[a<0\]

When \[a>0\]

$\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,(\left| x \right|+1)$

=$\underset{x\to a}{\mathop{\lim }}\,(-x-1)$                         $[0<x<a\Rightarrow \left| x \right|=x]$

=$-a-1$

$\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,(\left| x \right|+1)$

=$\underset{x\to a}{\mathop{\lim }}\,(-x-1)$                         $[0<x<a\Rightarrow \left| x \right|=x]$

=$a-1$

$\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f(x)=a-1$

Thus, limit exists at $x=a$, where \[a>0\]

Thus $\underset{x\to a}{\mathop{\lim }}\,f(x)$ exists for all $a\ne 0$$\underset{x\to 1}{\mathop{\lim }}\,\frac{f(x)-2}{{{x}^{2}}-1}=\pi $.


31. If function f(x) satisfies, \[\underset{x\to 1}{\mathop{\lim }}\,\frac{f(x)-2}{{{x}^{2}}-1}=\pi \], evaluate $\underset{x\to 1}{\mathop{\lim }}\,f(x)$

Ans. Given, 

\[\underset{x\to 1}{\mathop{\lim }}\,\frac{f(x)-2}{{{x}^{2}}-1}=\pi \]

= $\frac{\underset{x\to 1}{\mathop{\lim }}\,(f(x)-2)}{\underset{x\to 1}{\mathop{\lim }}\,({{x}^{2}}-1)}=\pi $

= $\underset{x\to 1}{\mathop{\lim }}\,(f(x)-2)=\pi \underset{x\to 1}{\mathop{\lim }}\,({{x}^{2}}-1)$

=$\underset{x\to 1}{\mathop{\lim }}\,(f(x)-2)=\pi ({{1}^{2}}-1)$

=$\underset{x\to 1}{\mathop{\lim }}\,(f(x)-2)=0$

=$\underset{x\to 1}{\mathop{\lim }}\,f(x)-\underset{x\to 1}{\mathop{\lim }}\,2=0$

=$\underset{x\to 1}{\mathop{\lim }}\,f(x)-2=0$

$\underset{x\to 1}{\mathop{\lim }}\,f(x)=2$


32. If $f(x)=\left\{ \begin{align} & m{{x}^{2}}+n \\ & nx+m \\ & n{{x}^{3}}+m \\ \end{align} \right.$ $\begin{align} & x<0 \\ & 0\le x\le 1 \\ & x>1 \\ \end{align}$

For what integers m and n does $\underset{x\to 0}{\mathop{\lim }}\,f(x)$ and $\underset{x\to 1}{\mathop{\lim }}\,f(x)$ exists?

Ans. Given,  

$f(x)=\left\{ \begin{align} & m{{x}^{2}}+n \\ & nx+m \\ & n{{x}^{3}}+m \\ \end{align} \right.$ $\begin{align} & x<0 \\ & 0\le x\le 1 \\ & x>1 \\ \end{align}$\[\underset{x\to 0}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,(m{{x}^{2}}+n)\]

=$m{{(0)}^{2}}+n$

=$n$

$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,(nx+m)$

=$n{{(0)}^{{}}}+m$

=$m$

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

$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to 1}{\mathop{\lim }}\,(nx+m)$

=$n(1)+m$

=$m+n$

$\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to 1}{\mathop{\lim }}\,(n{{x}^{3}}+m)$

=$n{{(1)}^{3}}+m$

=$m+n$

$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to 1}{\mathop{\lim }}\,f(x)$

Thus, $\underset{x\to 1}{\mathop{\lim }}\,f(x)$exists for any internal value of  m and n


Conclusion 

Calculating limits for polynomial and rational functions involves substitution and simplifying expressions. Understanding Left Hand Limits (LHL) and Right Hand Limits (RHL) is crucial for analysing function behaviour at specific points. Limits for trigonometric functions uses geometric proof of the important inequalities relating trigonometric functions. The Sandwich Theorem helps find limits by bounding a function between two known limits. Focus on mastering these methods, as they are fundamental for calculus. Previous year question papers typically have 3-4 questions on limits, emphasizing their importance in exams. Practice these concepts thoroughly to excel in your studies.


Class 11 Maths Chapter 12: Exercises Breakdown

Exercise

Number of Questions

Exercise 12.2

11 Questions and Solutions

Miscellaneous Exercise

30 Questions and Solutions


CBSE Class 11 Maths Chapter 12 Other Study Materials


Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.



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FAQs on CBSE Class 11 Mathematics Chapter 12 Limits and Derivatives – NCERT Solutions 2025-26

1. What are the main steps for solving limits in Class 11 Maths Exercise 12.1?

The main steps for solving limits in Class 11 Maths Exercise 12.1 are as follows:

  • Understand the problem statement and identify the type of limit (polynomial, rational, trigonometric, etc.).
  • Simplify the function using algebraic techniques such as factoring or rationalising.
  • Check for indeterminate forms (0/0 or ∞/∞) and use suitable methods to resolve them.
  • Apply limit laws and standard results as per the syllabus.
  • Substitute the limiting value and verify the result step-by-step.
Tip: Always write each step clearly to avoid calculation errors and ensure you follow NCERT-recommended solution formats.

2. Why is Exercise 12.1 important for future calculus studies?

Exercise 12.1 is crucial because it builds the foundation for advanced calculus topics.

  • Introduces limits, a fundamental concept required for understanding derivatives and continuity.
  • Helps develop logical and stepwise problem-solving skills necessary for higher-level mathematics.
  • Prepares students for board exams, JEE and NEET by emphasizing exam-relevant concepts and methods.
  • Promotes understanding of indeterminate forms and algebraic manipulation, which recur in later classes.

3. Where can I download NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.1 PDF?

You can download the NCERT Solutions PDF for Class 11 Maths Chapter 12 Exercise 12.1 from trusted educational platforms offering downloadable resources.

  • Look for official syllabus-aligned PDF guides.
  • Ensure the solutions cover step-by-step methods and are updated for the 2025 CBSE pattern.
  • Prefer sources that also offer revision notes, answer keys, and mind maps for Limits and Derivatives.

4. How do I deal with indeterminate forms in Limits and Derivatives?

To handle indeterminate forms (like 0/0 or ∞/∞) in Limits and Derivatives, follow these steps:

  • Factorise or simplify the numerator and denominator if possible.
  • Rationalise using conjugate methods for surds or square roots.
  • Use standard limits or algebraic rules as recommended in the NCERT syllabus.
  • Always show step-by-step calculations for full marks in exams.

5. Are these Class 11 Maths solutions aligned with the latest CBSE syllabus?

Yes, these Class 11 Maths solutions for Limits and Derivatives are fully aligned with the latest 2025 CBSE syllabus.

  • All topics, solution methods, and examples are curated as per official curriculum requirements.
  • Covers limit calculation, derivative basics, indeterminate forms, and standard NCERT pattern exercises.
  • Each answer is peer-reviewed for syllabus accuracy and conceptual clarity.

6. What mistakes do students commonly make in Exercise 12.1?

Common mistakes in Exercise 12.1 include:

  • Confusing different types of indeterminate forms.
  • Skipping steps in algebraic simplification, leading to calculation errors.
  • Applying incorrect limit formulas.
  • Not verifying the final answer by substituting back the limiting value.
  • Copying shortcuts without understanding the underlying logic.
Always write complete steps and double-check your calculation for CBSE exams.

7. What topics are covered in Class 11 Maths Chapter 12 – Limits and Derivatives as per the new syllabus?

Class 11 Maths Chapter 12 covers the following key topics based on the CBSE 2025 syllabus:

  • Intuitive idea of a limit and evaluating limits algebraically
  • Standard limits and indeterminate forms
  • First principle of derivatives
  • Application of limits and derivatives to simple problems
  • Stepwise solution approach for each exercise

8. How can I quickly revise formulas for Limits and Derivatives?

You can quickly revise the main formulas for Limits and Derivatives using a formula sheet or bullet summary:

  • Standard limit 1: lim (x→a) (x^n – a^n)/(x – a) = n·a^(n-1)
  • Standard limit 2: lim (x→0) (sin x)/x = 1
  • First principle: lim (h→0) (f(x+h) – f(x))/h = f'(x)
  • Include key algebraic identities and trigonometric forms as per NCERT.
Keep a revision table handy for last-minute exam prep.

9. How to avoid calculation errors while solving limit questions in Exercise 12.1?

To avoid calculation errors in limit questions:

  • Write every algebraic step separately.
  • Always check the form (indeterminate or direct substitution) before solving.
  • Use standard limits and rules provided in NCERT.
  • After solving, substitute the limit value again to confirm your result.
  • Review common mistakes from the answer key or expert solutions.

10. What is the best approach for practicing Class 11 Maths Exercise 12.1 solutions?

The best approach includes:

  • Attempt each question yourself before looking at the solution.
  • Check each step against the NCERT solutions to identify gaps in logic or algebra.
  • Practice solving without shortcuts to build conceptual clarity for limits and derivatives.
  • Use downloadable revision notes and mind maps for a quick overview.
  • Clarify doubts using the PDF or expert-verified answers.