Examples (Revised) - Chapter 12 Limits & Derivatives class 11 ncert solutions Maths - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Chapter 12: Limits & Derivatives - NCERT Solutions for Class 11 Maths

Example 1

Find the limits:
(i) $\lim _{x \rightarrow 1}\left[x^3-x^2+1\right]$
(ii) $\lim _{x \rightarrow 3}[x(x+1)]$
(iii) $\lim _{x \rightarrow-1}\left[1+x+x^2+\ldots+x^{10}\right]$.

Solution

The required limits are all limits of some polynomial functions. Hence the limits are the values of the function at the prescribed points. We have
(i) $\lim _{x \rightarrow 1}\left[x^3-x^2+1\right]=1^3-1^2+1=1$
(ii) $\lim _{x \rightarrow 3}[x(x+1)]=3(3+1)=3(4)=12$
(iii) $\lim _{x \rightarrow-1}\left[1+x+x^2+\ldots+x^{10}\right]=1+(-1)+(-1)^2+\ldots+(-1)^{10}$ $=1-1+1 \ldots+1=1$.

Example 2

Find the limits:
(i) $\lim _{x \rightarrow 1}\left[\frac{x^2+1}{x+100}\right]$
(ii) $\lim _{x \rightarrow 2}\left[\frac{x^3-4 x^2+4 x}{x^2-4}\right]$
(iii) $\lim _{x \rightarrow 2}\left[\frac{x^2-4}{x^3-4 x^2+4 x}\right]$
(iv) $\lim _{x \rightarrow 2}\left[\frac{x^3-2 x^2}{x^2-5 x+6}\right]$
(v) $\lim _{x \rightarrow 1}\left[\frac{x-2}{x^2-x}-\frac{1}{x^3-3 x^2+2 x}\right]$.

Solution

All the functions under consideration are rational functions. Hence, we first evaluate these functions at the prescribed points. If this is of the form $\frac{0}{0}$, we try to rewrite the function cancelling the factors which are causing the limit to be of $\text { the form } \frac{0}{0} \text {. }$

(i) We have $\lim _{x \rightarrow 1} \frac{x^2+1}{x+100}=\frac{1^2+1}{1+100}=\frac{2}{101}$
(ii) Evaluating the function at 2 , it is of the form $\frac{0}{0}$.

Hence
$
\begin{aligned}
\lim _{x \rightarrow 2} \frac{x^3-4 x^2+4 x}{x^2-4} & =\lim _{x \rightarrow 2} \frac{x(x-2)^2}{(x+2)(x-2)} \\
& =\lim _{x \rightarrow 2} \frac{x(x-2)}{(x+2)} \quad \text { as } x \neq 2 \\
& =\frac{2(2-2)}{2+2}=\frac{0}{4}=0 .
\end{aligned}
$
(iii) Evaluating the function at 2, we get it of the form $\frac{0}{0}$.

Hence
$
\begin{aligned}
\lim _{x \rightarrow 2} \frac{x^2-4}{x^3-4 x^2+4 x} & =\lim _{x \rightarrow 2} \frac{(x+2)(x-2)}{x(x-2)^2} \\
& =\lim _{x \rightarrow 2} \frac{(x+2)}{x(x-2)}=\frac{2+2}{2(2-2)}=\frac{4}{0}
\end{aligned}
$
which is not defined.
(iv) Evaluating the function at 2, we get it of the form $\frac{0}{0}$.

$\text { Hence } \quad \begin{aligned}
\lim _{x \rightarrow 2} \frac{x^3-2 x^2}{x^2-5 x+6} & =\lim _{x \rightarrow 2} \frac{x^2(x-2)}{(x-2)(x-3)} \\
& =\lim _{x \rightarrow 2} \frac{x^2}{(x-3)}=\frac{(2)^2}{2-3}=\frac{4}{-1}=-4 .
\end{aligned}$

(v) First, we rewrite the function as a rational function.
$
\begin{aligned}
{\left[\frac{x-2}{x^2-x}-\frac{1}{x^3-3 x^2+2 x}\right] } & =\left[\frac{x-2}{x(x-1)}-\frac{1}{x\left(x^2-3 x+2\right)}\right] \\
& =\left[\frac{x-2}{x(x-1)}-\frac{1}{x(x-1)(x-2)}\right] \\
& =\left[\frac{x^2-4 x+4-1}{x(x-1)(x-2)}\right] \\
& =\frac{x^2-4 x+3}{x(x-1)(x-2)}
\end{aligned}
$

Evaluating the function at 1 , we get it of the form $\frac{0}{0}$.
Hence
$
\begin{aligned}
\lim _{x \rightarrow 1}\left[\frac{x^2-2}{x^2-x}-\frac{1}{x^3-3 x^2+2 x}\right] & =\lim _{x \rightarrow 1} \frac{x^2-4 x+3}{x(x-1)(x-2)} \\
& =\lim _{x \rightarrow 1} \frac{(x-3)(x-1)}{x(x-1)(x-2)} \\
& =\lim _{x \rightarrow 1} \frac{x-3}{x(x-2)}=\frac{1-3}{1(1-2)}=2 .
\end{aligned}
$

We remark that we could cancel the term $(x-1)$ in the above evaluation because $x \neq 1$.

Evaluation of an important limit which will be used in the sequel is given as a theorem below.

Theorem 2 For any positive integer $n$,
$
\lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}=n a^{n-1} .
$

Example 3

Evaluate:
(i) $\lim _{x \rightarrow 1} \frac{x^{15}-1}{x^{10}-1}$
(ii) $\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{x}$

Solution

(i) We have
$
\begin{aligned}
\lim _{x \rightarrow 1} \frac{x^{15}-1}{x^{10}-1} & =\lim _{x \rightarrow 1}\left[\frac{x^{15}-1}{x-1} \div \frac{x^{10}-1}{x-1}\right] \\
& =\lim _{x \rightarrow 1}\left[\frac{x^{15}-1}{x-1}\right] \div \lim _{x \rightarrow 1}\left[\frac{x^{10}-1}{x-1}\right] \\
& =15(1)^{14} \div 10(1)^9 \text { (by the theorem above) } \\
& =15 \div 10=\frac{3}{2}
\end{aligned}
$
(ii) Put $y=1+x$, so that $y \rightarrow 1$ as $x \rightarrow 0$.

Then
$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{x} & =\lim _{y \rightarrow 1} \frac{\sqrt{y}-1}{y-1} \\
& =\lim _{y \rightarrow 1} \frac{y^{\frac{1}{2}}-1^{\frac{1}{2}}}{y-1} \\
& =\frac{1}{2}(1)^{\frac{1}{2}-1} \text { (by the remark above) }=\frac{1}{2}
\end{aligned}
$

Example 4

Evaluate:
(i) $\lim _{x \rightarrow 0} \frac{\sin 4 x}{\sin 2 x}$
(ii) $\lim _{x \rightarrow 0} \frac{\tan x}{x}$

Solution

(i)
$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\sin 4 x}{\sin 2 x} & =\lim _{x \rightarrow 0}\left[\frac{\sin 4 x}{4 x} \cdot \frac{2 x}{\sin 2 x} \cdot 2\right] \\
& =2 \cdot \lim _{x \rightarrow 0}\left[\frac{\sin 4 x}{4 x}\right] \div\left[\frac{\sin 2 x}{2 x}\right] \\
& =2 . \lim _{4 x \rightarrow 0}\left[\frac{\sin 4 x}{4 x}\right] \div \lim _{2 x \rightarrow 0}\left[\frac{\sin 2 x}{2 x}\right] \\
& =2.1 .1=2(\text { as } x \rightarrow 0,4 x \rightarrow 0 \text { and } 2 x \rightarrow 0)
\end{aligned}
$

(ii) We have $\lim _{x \rightarrow 0} \frac{\tan x}{x}=\lim _{x \rightarrow 0} \frac{\sin x}{x \cos x}=\lim _{x \rightarrow 0} \frac{\sin x}{x} \cdot \lim _{x \rightarrow 0} \frac{1}{\cos x}=1.1=1$

A general rule that needs to be kept in mind while evaluating limits is the following.
Say, given that the limit $\lim _{x \rightarrow a} \frac{f(x)}{g(x)}$ exists and we want to evaluate this. First we check the value of $f(a)$ and $g(a)$. If both are 0 , then we see if we can get the factor which is causing the terms to vanish, i.e., see if we can write $f(x)=f_1(x) f_2(x)$ so that $f_1(a)=0$ and $f_2(a) \neq 0$. Similarly, we write $g(x)=g_1(x) g_2(x)$, where $g_1(a)=0$ and $g_2(a) \neq 0$. Cancel out the common factors from $f(x)$ and $g(x)$ (if possible) and write
$
\frac{f(x)}{g(x)}=\frac{p(x)}{q(x)}, \text { where } q(x) \neq 0 .
$

Then
$
\lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{p(a)}{q(a)}
$

Example 5

Find the derivative at $x=2$ of the function $f(x)=3 x$.
Solution

We have
$
\begin{aligned}
f^{\prime}(2) & =\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}=\lim _{h \rightarrow 0} \frac{3(2+h)-3(2)}{h} \\
& =\lim _{h \rightarrow 0} \frac{6+3 h-6}{h}=\lim _{h \rightarrow 0} \frac{3 h}{h}=\lim _{h \rightarrow 0} 3=3 .
\end{aligned}
$

The derivative of the function $3 x$ at $x=2$ is 3 .
Example 6

Find the derivative of the function $f(x)=2 x^2+3 x-5$ at $x=-1$. Also prove that $f^{\prime}(0)+3 f^{\prime}(-1)=0$.
Solution

We first find the derivatives of $f(x)$ at $x=-1$ and at $x=0$. We have
$
\begin{aligned}
f^{\prime}(-1) & =\lim _{h \rightarrow 0} \frac{f(-1+h)-f(-1)}{h} \\
& =\lim _{h \rightarrow 0} \frac{\left[2(-1+h)^2+3(-1+h)-5\right]-\left[2(-1)^2+3(-1)-5\right]}{h} \\
& =\lim _{h \rightarrow 0} \frac{2 h^2-h}{h}=\lim _{h \rightarrow 0}(2 h-1)=2(0)-1=-1
\end{aligned}
$
and
$
f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}
$

$=\lim _{h \rightarrow 0} \frac{\left[2(0+h)^2+3(0+h)-5\right]-\left[2(0)^2+3(0)-5\right]}{h}$

$
=\lim _{h \rightarrow 0} \frac{2 h^2+3 h}{h}=\lim _{h \rightarrow 0}(2 h+3)=2(0)+3=3
$

Clearly $\quad f^{\prime}(0)+3 f^{\prime}(-1)=0$

Example 7

Find the derivative of $\sin x$ at $x=0$.
Solution

Let $f(x)=\sin x$. Then
$
\begin{aligned}
f^{\prime}(0) & =\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h} \\
& =\lim _{h \rightarrow 0} \frac{\sin (0+h)-\sin (0)}{h}=\lim _{h \rightarrow 0} \frac{\sin h}{h}=1
\end{aligned}
$

Example 8

Find the derivative of $f(x)=3$ at $x=0$ and at $x=3$.
Solution

Since the derivative measures the change in function, intuitively it is clear that the derivative of the constant function must be zero at every point. This is indeed, supported by the following computation.
$
f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{3-3}{h}=\lim _{h \rightarrow 0} \frac{0}{h}=0 .
$

Similarly $\quad f^{\prime}(3)=\lim _{h \rightarrow 0} \frac{f(3+h)-f(3)}{h}=\lim _{h \rightarrow 0} \frac{3-3}{h}=0$.

We now present a geometric interpretation of derivative of a function at a point. Let $y=f(x)$ be a function and $\operatorname{let} \mathrm{P}=(a, f(a))$ and $\mathrm{Q}=(a+h, f(a+h)$ be two points close to each other on the graph of this function. The Fig 12.11 is now self explanatory.

We know that $f^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$
From the triangle $\mathrm{PQR}$, it is clear that the ratio whose limit we are taking is precisely equal to $\tan (\mathrm{QPR})$ which is the slope of the chord $\mathrm{PQ}$. In the limiting process, as $h$ tends to 0 , the point $\mathrm{Q}$ tends to $\mathrm{P}$ and we have
$
\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}=\lim _{\mathrm{Q} \rightarrow \mathrm{P}} \frac{\mathrm{QR}}{\mathrm{PR}}
$

This is equivalent to the fact that the chord $\mathrm{PQ}$ tends to the tangent at $\mathrm{P}$ of the curve $y=f(x)$. Thus the limit turns out to be equal to the slope of the tangent. Hence
$
f^{\prime}(a)=\tan \psi \text {. }
$

For a given function $f$ we can find the derivative at every point. If the derivative exists at every point, it defines a new function called the derivative of $f$. Formally, we define derivative of a function as follows.

Definition 2 Suppose $f$ is a real valued function, the function defined by
$
\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}
$
wherever the limit exists is defined to be the derivative of $f$ at $x$ and is denoted by $f^{\prime}(x)$. This definition of derivative is also called the first principle of derivative.

Thus
$
f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}
$

Clearly the domain of definition of $f^{\prime}(x)$ is wherever the above limit exists. There are different notations for derivative of a function. Sometimes $f^{\prime}(x)$ is denoted by $\frac{d}{d x}(f(x))$ or if $y=f(x)$, it is denoted by $\frac{d y}{d x}$. This is referred to as derivative of $f(x)$ or $y$ with respect to $x$. It is also denoted by $\mathrm{D}(f(x))$. Further, derivative of $f$ at $x=a$ is also denoted by $\left.\frac{d}{d x} f(x)\right|_a$ or $\left.\frac{d f}{d x}\right|_a$ or even $\left(\frac{d f}{d x}\right)_{x=a}$.

Example 9

Find the derivative of $f(x)=10 x$.
Solution

Since $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$
\begin{aligned}
& =\lim _{h \rightarrow 0} \frac{10(x+h)-10(x)}{h} \\
& =\lim _{h \rightarrow 0} \frac{10 h}{h}=\lim _{h \rightarrow 0}(10)=10 .
\end{aligned}
$

Example 10

Find the derivative of $f(x)=x^2$.

Solution

We have,
$
\begin{aligned}
& f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
& =\lim _{h \rightarrow 0} \frac{(x+h)^2-(x)^2}{h}=\lim _{h \rightarrow 0}(h+2 x)=2 x
\end{aligned}
$

Example 11

Find the derivative of the constant function $f(x)=a$ for a fixed real number $a$.

Solution

We have
$
\begin{aligned}
& f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
& =\lim _{h \rightarrow 0} \frac{a-a}{h}=\lim _{h \rightarrow 0} \frac{0}{h}=0 \text { as } h \neq 0
\end{aligned}
$

Example 12

Find the derivative of $f(x)=\frac{1}{x}$
Solution

We have $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$\begin{aligned}
& =\lim _{h \rightarrow 0} \frac{\frac{1}{(x+h)}-\frac{1}{x}}{h} \\
& =\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{x-(x+h)}{x(x+h)}\right] \quad=\lim _{h \rightarrow 0} \frac{-1}{x(x+h)}=-\frac{1}{x^2}
\end{aligned}$

Example 13

Compute the derivative of $6 x^{100}-x^{55}+x$.
Solution

A direct application of the above theorem tells that the derivative of the above function is $600 x^{99}-55 x^{54}+1$.

Example 14

Find the derivative of $f(x)=1+x+x^2+x^3+\ldots+x^{50}$ at $x=1$.
Solution

A direct application of the above Theorem 6 tells that the derivative of the above function is $1+2 x+3 x^2+\ldots+50 x^{49}$. At $x=1$ the value of this function equals
$
1+2(1)+3(1)^2+\ldots+50(1)^{49}=1+2+3+\ldots+50=\frac{(50)(51)}{2}=1275 \text {. }
$

Example 15

Find the derivative of $f(x)=\frac{x+1}{x}$
Solution

Clearly this function is defined everywhere except at $x=0$. We use the quotient rule with $u=x+1$ and $v=x$. Hence $u^{\prime}=1$ and $v^{\prime}=1$. Therefore
$
\frac{d f(x)}{d x}=\frac{d}{d x}\left(\frac{x+1}{x}\right)=\frac{d}{d x}\left(\frac{u}{v}\right)=\frac{u^{\prime} v-u v^{\prime}}{v^2}=\frac{1(x)-(x+1) 1}{x^2}=-\frac{1}{x^2}
$

Example 16

Compute the derivative of $\sin x$.
Solution

Let $f(x)=\sin x$. Then
$
\begin{aligned}
\frac{d f(x)}{d x} & =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim _{h \rightarrow 0} \frac{\sin (x+h)-\sin (x)}{h} \\
& =\lim _{h \rightarrow 0} \frac{2 \cos \left(\frac{2 x+h}{2}\right) \sin \left(\frac{h}{2}\right)}{h} \text { (using formula for } \sin \mathrm{A}-\sin \mathrm{B} \text { ) } \\
& =\lim _{h \rightarrow 0} \cos \left(x+\frac{h}{2}\right) \cdot \lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}}=\cos x \cdot 1=\cos x
\end{aligned}
$

Example 17

Compute the derivative of $\tan x$.
Solution

Let $f(x)=\tan x$. Then
$
\begin{aligned}
\frac{d f(x)}{d x} & =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim _{h \rightarrow 0} \frac{\tan (x+h)-\tan (x)}{h} \\
& =\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{\sin (x+h)}{\cos (x+h)}-\frac{\sin x}{\cos x}\right]
\end{aligned}
$

$
\begin{aligned}
& =\lim _{h \rightarrow 0}\left[\frac{\sin (x+h) \cos x-\cos (x+h) \sin x}{h \cos (x+h) \cos x}\right] \\
& =\lim _{h \rightarrow 0} \frac{\sin (x+h-x)}{h \cos (x+h) \cos x} \text { (using formula for sin (A+B)) } \\
& =\lim _{h \rightarrow 0} \frac{\sin h}{h} \cdot \lim _{h \rightarrow 0} \frac{1}{\cos (x+h) \cos x} \\
& =1 \cdot \frac{1}{\cos ^2 x}=\sec ^2 x .
\end{aligned}
$

Example 18

Compute the derivative of $f(x)=\sin ^2 x$.
Solution

We use the Leibnitz product rule to evaluate this.
$
\begin{aligned}
\frac{d f(x)}{d x} & =\frac{d}{d x}(\sin x \sin x) \\
& =(\sin x)^{\prime} \sin x+\sin x(\sin x)^{\prime} \\
& =(\cos x) \sin x+\sin x(\cos x) \\
& =2 \sin x \cos x=\sin 2 x
\end{aligned}
$