Chapter 12: Limits and Derivatives

We know that the limit of a polynomial function is the value of the function at the prescribed point. i.e., if $f(x)$ is

a polynomial function, then $\lim _{x \rightarrow a} f(x)=f(a)$, obtained by writing $a$ for $x$ in the function.

$\lim _{x \rightarrow \pi}\left(x-\frac{22}{7}\right)=\left(\pi-\frac{22}{7}\right)$

Remark. $\pi \neq \frac{22}{7}$ since $\pi$ is irrational whereas $\frac{22}{7}$ is rational. However, $\frac{22}{7}$ is an approximate value of $\pi$. $\frac{355}{113}$ is another approximate value of $\pi$.

$\lim _{r \rightarrow 1} \pi r^2=\pi \times 1^2=\pi$.

$\lim _{x \rightarrow 4} \frac{4 x+3}{x-2}=\frac{4(4)+3}{4-2}=\frac{19}{2}$.

$\lim _{x \rightarrow-1} \frac{x^{10}+x^5+1}{x-1}=\frac{(-1)^{10}+(-1)^5+1}{(-1)-1}=\frac{1-1+1}{-2}=-\frac{1}{2}$.

On putting $x=0$, we get $\frac{1^5-1}{0}=\frac{0}{0}$ which is an indeterminate form. Put $x+1=y$, i.e., $\quad x=y-1$ so that $y \rightarrow 1$ as $x \rightarrow 0$.

$\lim _{x \rightarrow 2} \frac{3 x^2-x-10}{x^2-4} \quad\left(\right.$ Putting $x=2$, we get the Form $\left.\frac{0}{0}\right)$

Factorising numerator and denominator:

$\lim _{x \rightarrow 3} \frac{x^4-81}{2 x^2-5 x-3} \quad\left(\right.$ Putting $x=3$, we get the Form $\left.\frac{0}{0}\right)$

Factorising both numerator and denominator:

$\lim _{x \rightarrow 0} \frac{a x+b}{c x+1}=\frac{a(0)+b}{c(0)+1}=b$.

On putting $z=1$, we get the form $\left(\frac{0}{0}\right)$

Put $z^{1 / 6}=y$ so that $y \rightarrow 1$ as $z \rightarrow 1$. Then

**Alternative approach:**

Dividing both numerator and denominator by ( $z-1$ ),

Sol. $\lim _{x \rightarrow 1} \frac{a x^2+b x+c}{c x^2+b x+a}=\frac{a(1)^2+b(1)+c}{c(1)^2+b(1)+a}=\frac{a+b+c}{c+b+a}=1$.

$\lim _{x \rightarrow-2} \frac{\frac{1}{x}+\frac{1}{2}}{x+2} \quad\left(\right.$ Putting $x=-2$, we get the Form $\left.\frac{0}{0}\right)$

Using L.C.M. to write as a single fraction:

$\lim _{x \rightarrow 0} \frac{\sin a x}{b x}=\lim _{x \rightarrow 0}\left(\frac{a}{b} \cdot \frac{\sin a x}{a x}\right)=\frac{a}{b} \lim _{a x \rightarrow 0}\left(\frac{\sin a x}{a x}\right)$

$\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x}=\lim _{x \rightarrow 0}\left(\frac{a}{b} \cdot \frac{\sin a x}{a x} \cdot \frac{b x}{\sin b x}\right)$

Put $\pi-x=t$ so that $t \rightarrow 0$ as $x \rightarrow \pi$.

$\lim _{x \rightarrow 0} \frac{\cos x}{\pi-x}=\frac{\cos 0}{\pi-0}=\frac{1}{\pi}$.

$\lim _{x \rightarrow 0} \frac{\cos 2 x-1}{\cos x-1}=\lim _{x \rightarrow 0} \frac{2 \cos ^2 x-1-1}{\cos x-1}$

Factorising the numerator:

Dividing numerator and denominator by $x$

$\lim _{x \rightarrow 0} x \sec x=\lim _{x \rightarrow 0} \frac{x}{\cos x}=\frac{0}{\cos 0}=\frac{0}{1}=0$.

Dividing numerator and denominator by $x$

$\lim _{x \rightarrow 0}(\operatorname{cosec} x-\cot x)=\lim _{x \rightarrow 0}\left(\frac{1}{\sin x}-\frac{\cos x}{\sin x}\right)$

(on changing all T-ratios in terms of $\sin x$ and $\cos x$ )

Put $x=\frac{\pi}{2}+t$ so that $t=x-\frac{\pi}{2}$ and as $x \rightarrow \frac{\pi}{2}, t \rightarrow 0$

In the neighbourhood of $0, f(x)$ is defined differently.

Therefore, we shall find both left hand limit and right hand limit.

In the neighbourhood of $1, f(x)=3(x+1)$

Here $f(x)$ is defined differently in the neighbourhood of 1 . Therefore, we shall find both left hand limit and right hand limit.

$\left[\because \quad\right.$ when $x \rightarrow 1^{+}, \quad x>1$ and $\left.f(x)=-x^2-1\right]$

Since $\lim _{x \rightarrow 1^{-}} f(x) \neq \lim _{x \rightarrow 1^{+}} f(x)$, therefore, $\lim _{x \rightarrow 1} f(x)$ does not exist.

We know that $|x|=\left\{\begin{array}{rll}x, & \text { if } & x \geq 0 \\ -x, & \text { if } & x<0\end{array}\right.$

We shall find both left hand limit and right hand limit.

Again $\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{|x|}{x}=\lim _{x \rightarrow 0^{+}} \frac{x}{x}=\lim _{x \rightarrow 0^{+}}(1)=1$

Since $\lim _{x \rightarrow 0^{-}} f(x) \neq \lim _{x \rightarrow 0^{+}} f(x)$, therefore, $\lim _{x \rightarrow 0} f(x)$ does not exist.

L.H.L. $=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{x}{|x|}$

Since L.H.L. $\neq$ R.H.L., therefore, $\lim _{x \rightarrow 0} f(x)$ does not exist.

Here $f(x)$ is defined differently in the neighbourhood of 1 . Therefore, we shall find both left hand limit and right hand limit.

Again $\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(b-a x)=b-a \times 1=b-a$

Also $\quad f(1)=4$

$\Rightarrow \quad a+b=4$ and $b-a=4$

Adding, $\quad 2 b=8 \quad \therefore \quad b=4$

Putting $b=4$ in $a+b=4$, we have $a+4=4$ or $a=0$.

$f(x)=\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_n\right)$

Let us find $\lim _{x \rightarrow a_i} f(x)$ for $i=1$

i.e., $\lim _{x \rightarrow a_1} f(x)=\lim _{x \rightarrow a_1}\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_n\right)$

Putting $x=a_1$

Again $\lim _{x \rightarrow a_i} f(x)=\lim _{x \rightarrow a_i}\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_i\right)$

Putting $x=a_i$

for all $i=1,2,3, \ldots, n$.

Again $\lim _{x \rightarrow a} f(x)$ for some $a \neq a_1, a_2, \ldots, a_n$

Putting $x=a$

We know that $|x|=\left\{\begin{array}{cc}x, & x>0 \\ -x, & x<0\end{array}\right.$

Since $a \in \mathrm{R}$, three cases arise.

Case 1. When $a<0, f(x)=-x+1$

Case 2. When $a>0, f(x)=x-1$

Case 3. When $a=0$.

Since $\lim _{x \rightarrow a^{-}} f(x) \neq \lim _{x \rightarrow a^{+}} f(x)$ for $a=0, \lim _{x \rightarrow a} f(x)$ does not exist.

Hence $\lim _{x \rightarrow a} f(x)$ exists for all $a \neq 0$.

Given: $\quad \lim _{x \rightarrow 1} \frac{f(x)-2}{x^2-1}=\pi$

But $\lim _{x \rightarrow 1}\left(x^2-1\right)=1^2-1=1-1=0$, therefore we must have $\lim _{x \rightarrow 1}(f(x)-2)=0$,

because if $\lim _{x \rightarrow 1}(f(x)-2)$ is non-zero, then the given limit becomes $\frac{\text { non-zero }}{0}=\infty$ which does not exist and hence can’t be $\pi$ (given).

$\left[\begin{array}{r}\text { Remark : If } \lim _{x \rightarrow a} \frac{g(x)}{h(x)}=a \text { real number } l \text { and } \lim _{x \rightarrow a} h(x)=h(a)=0, \\ \text { then } \lim _{x \rightarrow a} g(x) \text { must be } 0 .\end{array}\right]$

Here $f(x)$ is defined differently in the neighbourhood of 0 as well as 1 .

Therefore, we shall find both left hand limit and right hand limit both for $x=0$ and $x=1$.

Now $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}\left(m x^2+n\right)=m \times 0^2+n=n$

Again $\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}(n x+m)=n \times 0+m=m$

$\because \quad \lim _{x \rightarrow 0} f(x)$ exist (given),

$\therefore \quad \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x) \quad \therefore \quad n=m$

Also $\quad \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(n x+m)=n \times 1+m=n+m$

Again $\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}\left(n x^3+m\right)=n \times 1^3+m=n+m$

Here $\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=n+m$, therefore, $\lim _{x \rightarrow 1} f(x)$ exists for all values of $m$ and $n$ and $\lim _{x \rightarrow 1} f(x)=m+n$

From (i), we conclude that $m$ and $n$ must be equal integers.

Here $f(x)=x^2-2,(x=) a=10$.

We know that $f^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$

Here, $\quad f(x)=99 x, \quad(x=) a=100$

Hence the derivative of $99 x$ at $x=100$ is 99 .

Here $f(x)=x, a=1$.

(i) Let $f(x)=x^3-27$.

Changing $x$ to $x+h, f(x+h)=(x+h)^3-27$.

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

(ii) Let $f(x)=(x-1)(x-2)=x^2-3 x+2$.

Changing $x$ to $x+h, f(x+h)=(x+h)^2-3(x+h)+2$.

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

(iii) Let $f(x)=\frac{1}{x^2}$.

Changing $x$ to $x+h, f(x+h)=\frac{1}{(x+h)^2}$.

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

Taking L.C.M.

(iv) Here $f(x)=\frac{x+1}{x-1}$

Changing $x$ to $x+h, f(x+h)=\frac{x+h+1}{x+h-1}$

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

Using L.C.M.,

Given $f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\ldots+\frac{x^2}{2}+x+1$

Putting $x=0$ and $x=1$ in ( $i$ ), we have

[By (ii)]

$f(x)=x^n+a x^{n-1}+a^2 x^{n-2}+\ldots+a^{n-1} x+a^n$

(i) Let $f(x)=(x-a)(x-b)$; | uv form By the product rule:

(ii) Let $f(x)=\left(a x^2+b\right)^2=a^2 x^4+2 a b x^2+b^2$, then

$\left[\because b\right.$ is constant (given) $\Rightarrow b^2$ is also constant. For example $3^2=9=$ constant]

(iii) Here $f(x)=\frac{x-a}{x-b}$

Using the quotient rule:

Let $f(x)=\frac{x^n-a^n}{x-a}$, then using the quotient rule

(i) Let $f(x)=2 x-\frac{3}{4}$, then

(ii) Here $f(x)=\left(5 x^3+3 x-1\right)(x-1)$

Using the product rule

(iii) Here $f(x)=x^{-3}(5+3 x)$

(iv) Let $f(x)=x^5\left(3-6 x^{-9}\right)=3 x^5-6 x^{-4}$, then

(v) Let $f(x)=x^{-4}\left(3-4 x^{-5}\right)=3 x^{-4}-4 x^{-9}$, then

(vi) Let $f(x)=\frac{2}{x+1}-\frac{x^2}{3 x-1}$

Using the quotient rule:

Let $f(x)=\cos x$.

Changing $x$ to $x+h, f(x+h)=\cos (x+h)$

We know that

(i) Let $\quad f(x)=\sin x \cos x$; | uv form

Applying Product rule,

(ii) Here $f(x)=\sec x=\frac{1}{\cos x}$

Applying the quotient rule:

(iii) Here $f(x)=5 \sec x+4 \cos x$

(iv) Here $f(x)=\operatorname{cosec} x=\frac{1}{\sin x}$

Applying the quotient rule:

(v) Here $f(x)=3 \cot x+5 \operatorname{cosec} x$

(vi) Here $f(x)=5 \sin x-6 \cos x+7$

(vii) Here $f(x)=2 \tan x-7 \sec x$

(i) Here $f(x)=-x$

Changing $x$ to $x+h, f(x+h)=-(x+h)$

(ii) Let $f(x)=(-x)^{-1}=\frac{1}{-x}=-\frac{1}{x}$.

Changing $x$ to $x+h, f(x+h)=\frac{-1}{x+h}$.

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

Taking L.C.M.

(iii) Here $f(x)=\sin (x+1)$

changing $x$ to $x+h, f(x+h)=\sin (x+h+1)$

(iv) Let $f(x)=\cos \left(x-\frac{\pi}{8}\right)$

Changing $x$ to $x+h ; f(x+h)=\cos \left(x+h-\frac{\pi}{8}\right)$.

We know that $\boldsymbol{f}^{\prime}(\boldsymbol{x})=\boldsymbol{\operatorname { l i m }}_{\boldsymbol{h} \rightarrow \mathbf{0}} \frac{\boldsymbol{f}(\boldsymbol{x}+\boldsymbol{h})-\boldsymbol{f}(\boldsymbol{x})}{\boldsymbol{h}}$

Find the derivative of the following functions (it is to be understood that $a, b, c, d, p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers):

Here $f(x)=x+a$

Let $f(x)=(p x+q)\left(\frac{r}{x}+s\right)=p r+p s x+\frac{q r}{x}+q s$, then

Here $f(x)=(a x+b)(c x+d)^2$.

Using the product rule

[By the chain rule: here $\frac{d}{d x} u^n=n u^{n-1} \frac{d}{d x} u$ ]

Here $f(x)=\frac{a x+b}{c x+d}$.

Applying the quotient rule:

Let $f(x)=\frac{1+\frac{1}{x}}{1-\frac{1}{x}}=\frac{\frac{x+1}{x}}{\frac{x-1}{x}}=\frac{x+1}{x-1}$

Applying the quotient rule:

Let $f(x)=\frac{1}{a x^2+b x+c}$

By the quotient rule:

Here $f(x)=\frac{a x+b}{p x^2+q x+r}$.

Applying the quotient rule:

Here $f(x)=\frac{p x^2+q x+r}{a x+b}$.

Using the quotient rule

Let $f(x)=\frac{a}{x^4}-\frac{b}{x^2}+\cos x=a x^{-4}-b x^{-2}+\cos x$, then

Let $f(x)=4 \sqrt{x}-2$, then

Let $f(x)=(a x+b)^n$

Differentiating both sides w.r.t $x$,

[By the chain rule: Here $\frac{d}{d x} u^n=n u^{n-1} \frac{d}{d x} u$ where $u$ is a function of $x$ ]

Let $f(x)=(a x+b)^n(c x+d)^m$, then

By the product rule:

[By the chain rule: here $\frac{d}{d x} u^n=n u^{n-1} \frac{d}{d x} u$ ]

Here $f(x)=\sin (x+a)=\sin x \cos a+\cos x \sin a$.

Alternatively:

Let $\quad f(x)=\operatorname{cosec} x \cot x$; I uv form

By the product rule:

Here $f(x)=\frac{\cos x}{1+\sin x}$.

Applying the quotient rule:

Let $f(x)=\frac{\sin x+\cos x}{\sin x-\cos x}$, using the quotient rule:

Here $f(x)=\frac{\sec x-1}{\sec x+1}$

Applying the quotient rule:

Let $f(x)=\sin ^n x=(\sin x)^n$

Applying the chain rule:

Here $f(x)=\frac{a+b \sin x}{c+d \cos x}$

Applying the quotient rule:

Here $f(x)=\frac{\sin (x+a)}{\cos x}$.

Applying the quotient rule:

Using $\cos \mathrm{A} \cos \mathrm{B}+\sin \mathrm{A} \sin \mathrm{B}=\cos (\mathrm{A}-\mathrm{B})$,

Here $f(x)=x^4(5 \sin x-3 \cos x)$

Applying the product rule:

Let $f(x)=\left(x^2+1\right) \cos x$,

Applying the product rule:

Here $f(x)=\left(a x^2+\sin x\right)(p+q \cos x)$

Applying the product rule:

Let $\quad f(x)=(x+\cos x)(x-\tan x) ; \quad$ I u form Applying the product rule:

Let $f(x)=\frac{4 x+5 \sin x}{3 x+7 \cos x}$, using the quotient rule:

Sol. Let

Using the quotient rule:

Here $f(x)=\frac{x}{1+\tan x}$.

Applying the quotient rule:

Here $f(x)=(x+\sec x)(x-\tan x)$

Applying the product rule:

Let $f(x)=\frac{x}{\sin ^n x}$, using the quotient rule: