Chapter 5

CONTINUITY AND DIFFERENTIABILITY

5.1 Overview

5.1.1 Continuity of a function at a point

Let $f$ be a real function on a subset of the real numbers and let $c$ be a point in the domain of $f$. Then $f$ is continuous at $c$ if

$\underset{x\to c}{lim}f(x)=f(c)$

More elaborately, if the left hand limit, right hand limit and the value of the function at $x=c$ exist and are equal to each other, i.e.,

$\underset{x\to c^{-}}{lim}f(x)=f(c)=\underset{x\to c^{+}}{lim}f(x)$

then $f$ is said to be continuous at $x=c$.

5.1.2 Continuity in an interval

(i) $f$ is said to be continuous in an open interval $(a,b)$ if it is continuous at every point in this interval.

(ii) $f$ is said to be continuous in the closed interval $[a,b]$ if

• $f$ is continuous in $(a,b)$
• $\underset{x\to a^{+}}{lim}f(x)=f(a)$
• $\underset{x\to b^{-}}{lim}f(x)=f(b)$

5.1.3 Geometrical meaning of continuity

(i) Function $f$ will be continuous at $x=c$ if there is no break in the graph of the function at the point $(c,f(c))$.

(ii) In an interval, function is said to be continuous if there is no break in the graph of the function in the entire interval.

5.1.4 Discontinuity

The function $f$ will be discontinuous at $x=a$ in any of the following cases :

(i) $\underset{x\to a^{-}}{lim}f(x)$ and $\underset{x\to a^{+}}{lim}f(x)$ exist but are not equal.

(ii) $\underset{x\to a^{-}}{lim}f(x)$ and $\underset{x\to a^{+}}{lim}f(x)$ exist and are equal but not equal to $f(a)$.

(iii) $f(a)$ is not defined.

5.1.5 Continuity of some of the common functions

$$ Function f(x) $$ Interval in which f is continuous

$\begin{array}{cc}\text{1. The constant function,}i.e.f(x)=c& \mathbf{R}\\ \text{2. The identity function,}i.e.f(x)=x& \mathbf{R}\\ \text{3. The polynomial function,}i.e.\\ f(x)=a_0{x}^n+a_1{x}^{n-1}+\dots +a_{n-1}x+a_n& \mathbf{R}\\ \text{4. |x-a| }& (-\mathrm{\infty },\mathrm{\infty })\\ \text{5.}{x}^{-n},\text{n is a positive integer}& (-\mathrm{\infty },\mathrm{\infty })-\{0\}\\ \text{6. p(x) / q(x), where p(x) and q(x) are polynomials in x}& \mathbf{R}-\{x:q(x)=0\}\\ \text{7.}\sin x,\cos x& \mathbf{R}\\ \text{8.}\tan x,\sec x& \mathbf{R}-\{(2n+1)\frac{\pi }{2}:n\in \mathbf{Z}\}\\ \text{9.}\cot x,\mathrm{cosec}x& \mathbf{R}-\{n\pi :n\in \mathbf{Z}\}\\ \text{10.}{e}^x& \mathbf{R}\\ \text{11.}\log x& (0,\mathrm{\infty })\\ \text{12. The inverse trigonometric functions, i.e. }& \text{In their respective domains }\end{array}$

5.1.6 Continuity of composite functions

Let $f$ and $g$ be real valued functions such that ( $f\circ g$ ) is defined at $a$. If $g$ is continuous at $a$ and $f$ is continuous at $g(a)$, then $(f\circ g)$ is continuous at $a$.

5.1.7 Differentiability

The function defined by $f^{\prime }(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$, wherever the limit exists, is defined to be the derivative of $f$ at $x$. In other words, we say that a function $f$ is differentiable at a point $c$ in its domain if both $\underset{h\to 0^{-}}{lim}\frac{f(c+h)-f(c)}{h}$, called left hand derivative, denoted by $L{f}^{\prime }(c)$, and $\underset{h\to 0^{+}}{lim}\frac{f(c+h)-f(c)}{h}$, called right hand derivative, denoted by $R{f}^{\prime }(c)$, are finite and equal.

(i) The function $y=f(x)$ is said to be differentiable in an open interval $(a,b)$ if it is differentiable at every point of $(a,b)$

(ii) The function $y=f(x)$ is said to be differentiable in the closed interval $[a,b]$ if $R{f}^{\prime }(a)$ and $L{f}^{\prime }(b)$ exist and $f^{\prime }(x)$ exists for every point of $(a,b)$.

(iii) Every differentiable function is continuous, but the converse is not true

5.1.8 Algebra of derivatives

If $u,v$ are functions of $x$, then

(i) $\frac{d(u\pm v)}{dx}=\frac{du}{dx}\pm \frac{dv}{dx}$

(ii) $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$

(iii) $\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$

5.1.9 Chain rule is a rule to differentiate composition of functions. Let $f=vou$. If $t=u(x)$ and both $\frac{dt}{dx}$ and $\frac{dv}{dt}$ exist then $\frac{df}{dx}=\frac{dv}{dt}\cdot \frac{dt}{dx}$

5.1.10 Following are some of the standard derivatives (in appropriate domains)

1. $\frac{d}{dx}({\sin }^{-1}x)=\frac{1}{\sqrt{1-x^2}}$

2. $\frac{d}{dx}({\cos }^{-1}x)=\frac{-1}{\sqrt{1-x^2}}$

3. $\frac{d}{dx}({\tan }^{-1}x)=\frac{1}{1+x^2}$

4. $\frac{d}{dx}({\cot }^{-1}x)=\frac{-1}{1+x^2}$

5. $\frac{d}{dx}({\sec }^{-1}x)=\frac{1}{|x|\sqrt{x^2-1}},|x|>1$

6. $\frac{d}{dx}(cose{c}^{-1}x)=\frac{-1}{|x|\sqrt{x^2-1}},|x|>1$

5.1.11 Exponential and logarithmic functions

(i) The exponential function with positive base $b>1$ is the function $y=f(x)=b^x$. Its domain is $\mathbf{R}$, the set of all real numbers and range is the set of all positive real numbers. Exponential function with base 10 is called the common exponential function and with base $e$ is called the natural exponential function.

(ii) Let $b>1$ be a real number. Then we say logarithm of $a$ to base $b$ is $x$ if $b^x=a$, Logarithm of $a$ to the base $b$ is denoted by ${\log }_{b}a$. If the base $b=10$, we say it is common logarithm and if $b=e$, then we say it is natural logarithms. $\log x$ denotes the logarithm function to base $e$. The domain of logarithm function is ${\mathbf{R}}^{+}$, the set of all positive real numbers and the range is the set of all real numbers.

(iii) The properties of logarithmic function to any base $b>1$ are listed below:

1. ${\log }_b(xy)={\log }_{b}x+{\log }_{b}y$

2. ${\log }_b(\frac{x}{y})={\log }_{b}x-{\log }_{b}y$

3. ${\log }_b{x}^n=n{\log }_{b}x$

4. ${\log }_{b}x=\frac{{\log }_{c}x}{{\log }_{c}b}$, where $c>1$

5. ${\log }_{b}x=\frac{1}{{\log }_{x}b}$

6. ${\log }_{b}b=1$ and ${\log }_{b}1=0$

(iv) The derivative of $e^x$ w.r.t., $x$ is $e^x$, i.e. $\frac{d}{dx}(e^x)=e^x$. The derivative of $\log x$ w.r.t., $x$ is $\frac{1}{x}$; i.e. $\frac{d}{dx}(\log x)=\frac{1}{x}$.

5.1.12 Logarithmic differentiation is a powerful technique to differentiate functions of the form $f(x)=(u(x){)}^{v(x)}$, where both $f$ and $u$ need to be positive functions for this technique to make sense.

5.1.13 Differentiation of a function with respect to another function

Let $u=f(x)$ and $v=g(x)$ be two functions of $x$, then to find derivative of $f(x)$ w.r.t. to $g(x)$, i.e., to find $\frac{du}{dv}$, we use the formula

$\frac{du}{dv}=\frac{du}{\frac{dv}{dx}}$

5.1.14 Second order derivative

$\frac{d}{dx}\frac{dy}{dx}=\frac{d^{2}y}{d{x}^2}$ is called the second order derivative of $y$ w.r.t. $x$. It is denoted by $y^{\prime }$ or $y_2$, if $y=f(x)$.

5.1.15 Rolle’s Theorem

Let $f:[a,b]\to \mathbf{R}$ be continuous on $[a,b]$ and differentiable on $(a,b)$, such that $f(a)$ $=f(b)$, where $a$ and $b$ are some real numbers. Then there exists at least one point $c$ in $(a,b)$ such that $f^{\prime }(c)=0$.

Geometrically Rolle’s theorem ensures that there is at least one point on the curve $y=f(x)$ at which tangent is parallel to $x$-axis (abscissa of the point lying in $(a,b)$ ).

5.1.16 Mean Value Theorem (Lagrange)

Let $f:[a,b]\to \mathbf{R}$ be a continuous function on $[a,b]$ and differentiable on $(a,b)$. Then there exists at least one point $c$ in $(a,b)$ such that $f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$.

Geometrically, Mean Value Theorem states that there exists at least one point $c$ in $(a,b)$ such that the tangent at the point $(c,f(c))$ is parallel to the secant joining the points $(a,f(a)$ and $(b,f(b))$.

5.2 Solved Examples

Short Answer (S.A.)

Example 1 Find the value of the constant $k$ so that the function $f$ defined below is

continuous at $x=0$, where f(x)= $\{\begin{array}{ll}\frac{1-\cos 4x}{8x^2},& x\ne 0.\\ k,& x=0\end{array}$

Solution It is given that the function $f$ is continuous at $x=0$. Therefore, $\underset{x\to 0}{lim}f(x)=f(0)$

$\begin{array}{cc}\Rightarrow & \underset{x\to 0}{lim}\frac{1-\cos 4x}{8x^2}=k\\ \Rightarrow & \underset{x\to 0}{lim}\frac{2{\sin }^{2}2x}{8x^2}=k\\ \Rightarrow & \underset{x\to 0}{lim}(\frac{\sin 2x^2}{2x})=k\\ \Rightarrow & k=1\end{array}$

Thus, $f$ is continuous at $x=0$ if $k=1$.

Example 2 Discuss the continuity of the function $f(x)=\sin x\cdot \cos x$.

Solution Since $\sin x$ and $\cos x$ are continuous functions and product of two continuous function is a continuous function, therefore $f(x)=\sin x\cdot \cos x$ is a continuous function.

Example 3 If f(x)= $\{\begin{array}{ll}\frac{x^3+x^2-16x+20}{(x-2{)}^2},& x\ne 2\\ k,& x=2\end{array}$ is continuous at $x=2$, find

the value of $k$.

Solution Given $f(2)=k$.

Now, $\underset{x\to 2^{-}}{lim}f(x)=\underset{x\to 2^{+}}{lim}f(x)=\underset{x\to 2}{lim}\frac{x^3+x^2-16x+20}{(x-2{)}^2}$

$=\underset{x\to 2}{lim}\frac{(x+5)(x-2{)}^2}{(x-2{)}^2}=\underset{x\to 2}{lim}(x+5)=7$

As $f$ is continuous at $x=2$, we have

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

$\Rightarrow k=7$

Example 4 Show that the function $f$ defined by

f(x)= $\{\begin{array}{l}x\sin \frac{1}{x},x\ne 0\\ 0,x=0\end{array}$

is continuous at $x=0$.

Solution Left hand limit at $x=0$ is given by

$\underset{x\to 0^{-}}{lim}f(x)=\underset{x\to 0^{-}}{lim}x\sin \frac{1}{x}=0[\text{ since, }-1<\sin \frac{1}{x}<1]$

Similarly $\underset{x\to 0^{+}}{lim}f(x)=\underset{x\to 0^{+}}{lim}x\sin \frac{1}{x}=0$. Moreover $f(0)=0$.

Thus $\underset{x\to 0^{-}}{lim}f(x)=\underset{x\to 0^{+}}{lim}f(x)=f(0)$. Hence $f$ is continuous at $x=0$

Example 5 Given $f(x)=\frac{1}{x-1}$. Find the points of discontinuity of the composite function $y=f[f(x)]$.

Solution We know that $f(x)=\frac{1}{x-1}$ is discontinuous at $x=1$

Now, for $x\ne 1$,

$f(f(x))=f(\frac{1}{x-1})=\frac{1}{\frac{1}{x-1}-1}=\frac{x-1}{2-x},$

which is discontinuous at $x=2$.

Hence, the points of discontinuity are $x=1$ and $x=2$.

Example 6 Let $f(x)=x|x|$, for all $x\in \mathbf{R}$. Discuss the derivability of $f(x)$ at $x=0$

Solution We may rewrite $f$ as f(x)= $\{\begin{array}{ll}{x}^2& \text{, if }x\ge 0\\ -x^2& \text{, if }x<0\end{array}$

Now $L{f}^{\prime }(0)=\underset{h\to 0^{-}}{lim}\frac{f(0+h)-f(0)}{h}=\underset{h\to 0^{-}}{lim}\frac{-h^2-0}{h}=\underset{h\to 0^{-}}{lim}-h=0$

Now Rf $(0)=\underset{h\to 0^{+}}{lim}\frac{f(0+h)-f(0)}{h}=\underset{h\to 0^{+}}{lim}\frac{h^2-0}{h}=\underset{h\to 0^{-}}{lim}h=0$

Since the left hand derivative and right hand derivative both are equal, hence $f$ is differentiable at $x=0$.

Example 7 Differentiate $\sqrt{\tan \sqrt{x}}$ w.r.t. $x$

Solution Let $y=\sqrt{\tan \sqrt{x}}$. Using chain rule, we have

$\begin{array}{rl}& \frac{dy}{dx}=\frac{1}{2\sqrt{\tan \sqrt{x}}}\cdot \frac{d}{dx}(\tan \sqrt{x})\\ & =\frac{1}{2\sqrt{\tan \sqrt{x}}}\cdot {\sec }^2\sqrt{x}\frac{d}{dx}(\sqrt{x})\\ & =\frac{1}{2\sqrt{\tan \sqrt{x}}}({\sec }^2\sqrt{x})(\frac{1}{2\sqrt{x}})\\ & =\frac{({\sec }^2\sqrt{x})}{4\sqrt{x}\sqrt{\tan \sqrt{x}}}.\end{array}$

Example 8 If $y=\tan (x+y)$, find $\frac{dy}{dx}$.

Solution Given $y=\tan (x+y)$. differentiating both sides w.r.t. $x$, we have

$\begin{array}{rl}\frac{dy}{dx}=& {\sec }^2(x+y)\frac{d}{dx}(x+y)\\ & ={\sec }^2(x+y)(1+\frac{dy}{dx})\end{array}$

or $[1-{\sec }^2(x+y]\frac{dy}{dx}={\sec }^2(x+y).$

Therefore, $\frac{dy}{dx}=\frac{{\sec }^2(x+y)}{1-{\sec }^2(x+y)}=-cose{c}^2(x+y)$.

Example 9 If $e^x+e^y=e^{x+y}$, prove that

$\frac{dy}{dx}=-e^{y-x}.$

Solution Given that $e^x+e^y=e^{x+y}$. Differentiating both sides w.r.t. $x$, we have

$e^x+e^y\frac{dy}{dx}=e^{x+y}(1+\frac{dy}{dx})$

or $(e^y-e^x+y)\frac{dy}{dx}=e^x+y-e^x$

which implies that $\frac{dy}{dx}=\frac{e^{x+y}-e^x}{e^y-e^{x+y}}=\frac{e^x+e^y-e^x}{e^y-e^x-e^y}=-e^{y-x}$.

Example 10 Find $\frac{dy}{dx}$, if $y={\tan }^{-1}(\frac{3x-x^3}{1-3x^2}),-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}$.

Solution Put $x=\tan \theta$, where $\frac{-\pi }{6}<\theta <\frac{\pi }{6}$.

Therefore, $y={\tan }^{-1}(\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta })$

$={\tan }^{-1}(\tan 3\theta )$

$\begin{array}{rl}& =3\theta \text{ (because }\frac{-\pi }{2}<3\theta <\frac{\pi }{2}\text{ ) }\\ & =3{\tan }^{-1}x\end{array}$

Hence, $\frac{dy}{dx}=\frac{3}{1+x^2}$.

Example 11 If $y={\sin }^{-1}\{x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2}\}$ and $0<x<1$, then find $\frac{dy}{dx}$.

Solution We have $y={\sin }^{-1}\{x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2}\}$, where $0<x<1$.

Put

$x=\sin A\text{ and }\sqrt{x}=\sin B$

Therefore, $y={\sin }^{-1}\{\sin A\sqrt{1-{\sin }^{2}B}-\sin B\sqrt{1-{\sin }^{2}A}\}$

$\begin{array}{rl}& ={\sin }^{-1}\{\sin A\cos B-\sin B\cos A\}\\ & ={\sin }^{-1}\{\sin (A-B)\}=A-B\end{array}$

Thus

$y={\sin }^{-1}x-{\sin }^{-1}\sqrt{x}$

Differentiating w.r.t. $x$, we get

$\begin{array}{c}\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-{\sqrt{(x)}}^2}}\cdot \frac{d}{dx}(\sqrt{x})\\ =\frac{1}{\sqrt{1-x^2}}-\frac{1}{2\sqrt{x}\sqrt{1-x}}\end{array}$

Example 12 If $x=a{\sec }^3\theta$ and $y=a{\tan }^3\theta$, find $\frac{dy}{dx}$ at $\theta =\frac{\pi }{3}$.

Solution We have $x=a{\sec }^3\theta$ and $y=a{\tan }^3\theta$.

Differentiating w.r.t. $\theta$, we get

$\frac{dx}{d\theta }=3a{\sec }^2\theta \frac{d}{d\theta }(\sec \theta )=3a{\sec }^3\theta \tan \theta$

and $\frac{dy}{d\theta }=3a{\tan }^2\theta \frac{d}{d\theta }(\tan \theta )=3a{\tan }^2\theta {\sec }^2\theta$.

Thus $\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{3a{\tan }^2\theta {\sec }^2\theta }{3a{\sec }^3\theta \tan \theta }=\frac{\tan \theta }{\sec \theta }=\sin \theta$.

Hence, $(\frac{dy}{dx}{)}_{at\theta =\frac{\pi }{3}}=\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$.

Example 13 If $x^y=e^{x-y}$, prove that $\frac{dy}{dx}=\frac{\log x}{(1+\log x{)}^2}$.

Solution We have $x^y=e^{x-y}$. Taking logarithm on both sides, we get

$y\log x=x-y$

$\Rightarrow y(1+\log x)=x$

$\text{ i.e. }y=\frac{x}{1+\log x}$

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

$\frac{dy}{dx}=\frac{(1+\log x)\cdot 1-x(\frac{1}{x})}{(1+\log x{)}^2}=\frac{\log x}{(1+\log x{)}^2}.$

Example 14 If $y=\tan x+\sec x$, prove that $\frac{d^{2}y}{d{x}^2}=\frac{\cos x}{(1-\sin x{)}^2}$.

Solution We have $y=\tan x+\sec x$. Differentiating w.r.t. $x$, we get

$\begin{array}{c}\frac{dy}{dx}={\sec }^{2}x+\sec x\tan x\\ =\frac{1}{{\cos }^{2}x}+\frac{\sin x}{{\cos }^{2}x}=\frac{1+\sin x}{{\cos }^{2}x}=\frac{1+\sin x}{(1+\sin x)(1-\sin x)}.\end{array}$

thus $\frac{dy}{dx}=\frac{1}{1-\sin x}$.

Now, differentiating again w.r.t. $x$, we get

$\frac{d^{2}y}{d{x}^2}=\frac{-(-\cos x)}{(1-\sin x{)}^2}=\frac{\cos x}{(1-\sin x{)}^2}$

Example 15 If $f(x)=|\cos x|$, find $f^{\prime }(\frac{3\pi }{4})$.

Solution When $\frac{\pi }{2}<x<\pi ,\cos x<0$ so that $|\cos x|=-\cos x$, i.e., $f(x)=-\cos x$

$\Rightarrow f^{\prime }(x)=\sin x$.

Hence, $f^{\prime }(\frac{3\pi }{4})=\sin (\frac{3\pi }{4})=\frac{1}{\sqrt{2}}$

Example 16 If $f(x)=|\cos x-\sin x|$, find $f^{\prime }(\frac{\pi }{6})$.

Solution When $0<x<\frac{\pi }{4},\cos x>\sin x$, so that $\cos x-\sin x>0$, i.e.,

$f(x)=\cos x-\sin x$

$\Rightarrow f^{\prime }(x)=-\sin x-\cos x$

Hence $f^{\prime }(\frac{\pi }{6})=-\sin \frac{\pi }{6}-\cos \frac{\pi }{6}=-\frac{1}{2}(1+\sqrt{3})$.

Example 17 Verify Rolle’s theorem for the function, $f(x)=\sin 2x$ in $[0,\frac{\pi }{2}]$.

Solution Consider $f(x)=\sin 2x$ in $[0,\frac{\pi }{2}]$. Note that:

(i) The function $f$ is continuous in $[0,\frac{\pi }{2}]$, as $f$ is a sine function, which is always continuous.

(ii) $f^{\prime }(x)=2\cos 2x$, exists in $(0,\frac{\pi }{2})$, hence $f$ is derivable in $0,(\frac{\pi }{2})$.

(iii) $f(0)=\sin 0=0$ and $f(\frac{\pi }{2})=\sin \pi =0\Rightarrow f(0)=f(\frac{\pi }{2})$.

Conditions of Rolle’s theorem are satisfied. Hence there exists at least one $c\in (0,\frac{\pi }{2})$ such that $f^{\prime }(c)=0$. Thus

$2\cos 2c=0\Rightarrow 2c=\frac{\pi }{2}\Rightarrow c=\frac{\pi }{4}.$

Example 18 Verify mean value theorem for the function $f(x)=(x-3)(x-6)(x-9)$ in $[3,5]$.

Solution (i) Function $f$ is continuous in $[3,5]$ as product of polynomial functions is a polynomial, which is continuous.

(ii) $f^{\prime }(x)=3x^2-36x+99$ exists in $(3,5)$ and hence derivable in $(3,5)$.

Thus conditions of mean value theorem are satisfied. Hence, there exists at least one $c\in (3,5)$ such that

$\begin{array}{rl}& f^{\prime }(c)=\frac{f(5)-f(3)}{5-3}\\ & \Rightarrow 3c^2-36c+99=\frac{8-0}{2}=4\\ & \Rightarrow c=6\pm \sqrt{\frac{13}{3}}.\end{array}$

Hence $c=6-\sqrt{\frac{13}{3}}$ (since other value is not permissible).

Long Answer (L.A.)

Example 19 If $f(x)=\frac{\sqrt{2}\cos x-1}{\cot x-1},x\ne \frac{\pi }{4}$

find the value of $f(\frac{\pi }{4})$ so that $f(x)$ becomes continuous at $x=\frac{\pi }{4}$.

Solution Given, $f(x)=\frac{\sqrt{2}\cos x-1}{\cot x-1},x\ne \frac{\pi }{4}$

Therefore, $\underset{x\to \frac{\pi }{4}}{lim}f(x)=\underset{x\to (\frac{\pi }{4})}{lim}\frac{\sqrt{2}\cos x-1}{\cot x-1}$

$\begin{array}{rl}& =\underset{x\to \frac{\pi }{4}}{lim}\frac{(\sqrt{2}\cos x-1)\sin x}{\cos x-\sin x}\\ & =\underset{x\to \frac{\pi }{4}}{lim}\frac{(\sqrt{2}\cos x-1)}{(\sqrt{2}\cos x+1)}\cdot \frac{(\sqrt{2}\cos x+1)}{(\cos x-\sin x)}\cdot \frac{(\cos x+\sin x)}{(\cos x+\sin x)}\cdot \sin x\end{array}$

$\begin{array}{rl}& =\underset{x\to \frac{\pi }{4}}{lim}\frac{2{\cos }^{2}x-1}{{\cos }^{2}x-{\sin }^{2}x}\cdot \frac{\cos x+\sin x}{\sqrt{2}\cos x+1}\cdot (\sin x)\\ & =\underset{x\to \frac{\pi }{4}}{lim}\frac{\cos 2x}{\cos 2x}\cdot (\frac{\cos x+\sin x}{\sqrt{2}\cos x+1})\cdot (\sin x)\\ & =\underset{x\to \frac{\pi }{4}}{lim}\frac{(\cos x+\sin x)}{\sqrt{2}\cos x+1}\sin x\\ & =\frac{\frac{1}{\sqrt{2}}(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})}{\sqrt{2}\cdot \frac{1}{\sqrt{2}}+1}=\frac{1}{2}\end{array}$

Thus, $\underset{x\to \frac{\pi }{4}}{lim}f(x)=\frac{1}{2}$

If we define $f(\frac{\pi }{4})=\frac{1}{2}$, then $f(x)$ will become continuous at $x=\frac{\pi }{4}$. Hence for $f$ to be continuous at $x=\frac{\pi }{4},f(\frac{\pi }{4})=\frac{1}{2}$.

Example 20 Show that the function $f$ given by f(x)= $\{\begin{array}{ll}\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1},& ifx\ne 0\\ 0,& ifx=0\end{array}$

is discontinuous at $x=0$.

Solution The left hand limit of $f$ at $x=0$ is given by

$\underset{x\to 0^{-}}{lim}f(x)=\underset{x\to 0^{-}}{lim}\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}=\frac{0-1}{0+1}=-1$

Similarly, $\underset{x\to 0^{+}}{lim}f(x)=\underset{x\to 0^{+}}{lim}\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}$

$=\underset{x\to 0^{+}}{lim}\frac{1-\frac{1}{e^{\frac{1}{x}}}}{1+\frac{1}{e^{\frac{1}{x}}}}=\underset{x\to 0^{+}}{lim}\frac{1-e^{\frac{-1}{x}}}{1+e^{\frac{-1}{x}}}=\frac{1-0}{1+0}=1$

Thus $\underset{x\to 0^{-}}{lim}f(x)\ne limf(x)$, therefore, $\underset{x\to 0^{+}}{lim}f(x)$ does not exist. Hence $f$ is discontinuous at $x=0$.

Example 21 Let f(x)= $\{\begin{array}{ll}\frac{1-\cos 4x}{x^2}& \text{, if }x<0\\ a& ,ifx=0\\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}& \text{, if }x>0\end{array}$

For what value of $a,f$ is continuous at $x=0$ ?

Solution Here $f(0)=a$ Left hand limit of $f$ at 0 is

$\begin{array}{rl}& \underset{x\to 0^{-}}{lim}f(x)=\underset{x\to 0^{-}}{lim}\frac{1-\cos 4x}{x^2}=\underset{x\to 0^{-}}{lim}\frac{2{\sin }^{2}2x}{x^2}\\ & =\underset{2x\to 0^{-}}{lim}8(\frac{\sin 2x}{2x}{)}^2=8(1{)}^2=8.\end{array}$

and right hand limit of $f$ at 0 is

$\begin{array}{rl}& \underset{x\to 0^{+}}{lim}f(x)=\underset{x\to 0^{+}}{lim}\frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}\\ =& \underset{x\to 0^{+}}{lim}\frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{(\sqrt{16+\sqrt{x}}+4)(\sqrt{16+\sqrt{x}}-4)}\end{array}$

$=\underset{x\to 0^{+}}{lim}\frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{16+\sqrt{x}-16}=\underset{x\to 0^{+}}{lim}(\sqrt{16+\sqrt{x}}+4)=8$

Thus, $\underset{x\to 0^{+}}{lim}f(x)=\underset{x\to 0^{-}}{lim}f(x)=8$. Hence $f$ is continuous at $x=0$ only if $a=8$.

Example 22 Examine the differentiability of the function $f$ defined by

f(x)= $\{\begin{array}{l}2x+3\text{, if }-3\le x<-2\\ x+1,\text{ if }-2\le x<0\\ x+2,\text{ if }0\le x\le 1\end{array}$

Solution The only doubtful points for differentiability of $f(x)$ are $x=-2$ and $x=0$. Differentiability at $x=-2$.

Now L $f^{\prime }(-2)=\underset{h\to 0^{-}}{lim}\frac{f(-2+h)-f(-2)}{h}$

$=\underset{h\to 0^{-}}{lim}\frac{2(-2+h)+3-(-2+1)}{h}=\underset{h\to 0^{-}}{lim}\frac{2h}{h}=\underset{h\to 0^{-}}{lim}2=2.$

and $R{f}^{\prime }(-2)=\underset{h\to 0^{+}}{lim}\frac{f(-2+h)-f(-2)}{h}$

$\begin{array}{rl}& =\underset{h\to 0^{-}}{lim}\frac{-2+h+1-(-2+1)}{h}\\ & =\underset{h\to 0^{-}}{lim}\frac{h-1-(-1)}{h}=\underset{h\to 0^{+}}{lim}\frac{h}{h}=1\end{array}$

Thus $R{f}^{\prime }(-2)\ne L{f}^{\prime }(-2)$. Therefore $f$ is not differentiable at $x=-2$. Similarly, for differentiability at $x=0$, we have

$\begin{array}{rl}L(f^{\prime }(0).& =\underset{h\to 0^{-}}{lim}\frac{f(0+h)-f(0)}{h}\\ & =\underset{h\to 0^{-}}{lim}\frac{0+h+1-(0+2)}{h}\\ & =\underset{h\to 0^{-}}{lim}\frac{h-1}{h}=\underset{h\to 0^{-}}{lim}(1-\frac{1}{h})\end{array}$

which does not exist. Hence $f$ is not differentiable at $x=0$.

Example 23 Differentiate ${\tan }^{-1}(\frac{\sqrt{1-x^2}}{x})$ with respect to ${\cos }^{-1}(2x\sqrt{1-x^2})$, where $x\in (\frac{1}{\sqrt{2}},1)$.

Solution Let $u={\tan }^{-1}(\frac{\sqrt{1-x^2}}{x})$ and $v={\cos }^{-1}(2x\sqrt{1-x^2})$.

We want to find $\frac{du}{dv}=\frac{\frac{du}{dx}}{\frac{dv}{dx}}$

Now $u={\tan }^{-1}(\frac{\sqrt{1-x^2}}{x})$. Put $x=\sin \theta .(\frac{\pi }{4}<\theta <\frac{\pi }{2})$.

Then $u={\tan }^{-1}(\frac{\sqrt{1-{\sin }^2\theta }}{\sin \theta })={\tan }^{-1}(\cot \theta )$

$={\tan }^{-1}\{\tan (\frac{\pi }{2}-\theta )\}=\frac{\pi }{2}-\theta =\frac{\pi }{2}-{\sin }^{-1}x$

Hence

$\frac{du}{dx}=\frac{-1}{\sqrt{1-x^2}}$.

Now

$\begin{array}{rl}v& ={\cos }^{-1}(2x\sqrt{1-x^2})\\ & =\frac{\pi }{2}-{\sin }^{-1}(2x\sqrt{1-x^2})\\ & =\frac{\pi }{2}-{\sin }^{-1}(2\sin \theta \sqrt{1-{\sin }^2\theta })=\frac{\pi }{2}-{\sin }^{-1}(\sin 2\theta )\\ & =\frac{\pi }{2}-{\sin }^{-1}\{\sin (\pi -2\theta )\}[\text{ since }\frac{\pi }{2}<2\theta <\pi ]\end{array}$

$\frac{\pi }{2}-(\pi -2\theta )=\frac{-\pi }{2}+2\theta$

$\Rightarrow v=\frac{-\pi }{2}+2{\sin }^{-1}x$

$\Rightarrow \frac{dv}{dx}=\frac{2}{\sqrt{1-x^2}}.$

$\text{ Hence }\frac{du}{dv}=\frac{\frac{du}{dx}}{\frac{dv}{dx}}=\frac{\frac{-1}{\sqrt{1-x^2}}}{\frac{2}{\sqrt{1-x^2}}}=\frac{-1}{2}$ .

Objective Type Questions

Choose the correct answer from the given four options in each of the Examples 24 to 35.

Example 24 The function f(x)= $\{\begin{array}{ll}\frac{\sin x}{x}+\cos x& \text{, if }x\ne 0\\ k& \text{, if }x=0\end{array}$

is continuous at $x=0$, then the value of $k$ is

(A) 3

(B) 2

(C) 1

(D) 1.5

Solution (B) is the Correct answer.

Example 25 The function $f(x)=[x]$, where $[x]$ denotes the greatest integer function, is continuous at

(A) 4

(B) -2

(C) 1

(D) $1.5$

Solution (D) is the correct answer. The greatest integer function $[x]$ is discontinuous at all integral values of $x$. Thus $D$ is the correct answer.

Example 26 The number of points at which the function $f(x)=\frac{1}{x-[x]}$ is not continuous is

(A) 1

(B) 2

(C) 3

(D) none of these

Solution (D) is the correct answer. As $x-[x]=0$, when $x$ is an integer so $f(x)$ is discontinuous for all $x\in \mathbf{Z}$.

Example 27 The function given by $f(x)=\tan x$ is discontinuous on the set

(A) $\{n\pi :n\in Z\}$

(B) $\{2n\pi :n\in Z\}$

(C) $\{2n+1)\frac{\pi }{2}:n\in Z\}$

(D) $\{\frac{n\pi }{2}:n\in \mathbb{Z}\}$

Solution C is the correct answer.

Example 28 Let $f(x)=|\cos x|$. Then,

(A) $f$ is everywhere differentiable.

(B) $f$ is everywhere continuous but not differentiable at $n=n\pi ,n\in Z$.

(C) $f$ is everywhere continuous but not differentiable at $x=(2n+1)\frac{\pi }{2}$, $n\in \mathbf{Z}$.

(D) none of these.

Solution $C$ is the correct answer.

Example 29 The function $f(x)=|x|+|x-1|$ is

(A) continuous at $x=0$ as well as at $x=1$.

(B) continuous at $x=1$ but not at $x=0$.

(C) discontinuous at $x=0$ as well as at $x=1$.

(D) continuous at $x=0$ but not at $x=1$.

Solution Correct answer is A.

Example 30 The value of $k$ which makes the function defined by

f(x)= $\{\begin{array}{ll}\sin \frac{1}{x},& \text{ if }x\ne 0\\ k,& \text{ if }x=0\end{array}$, continuous at $x=0$ is

(A) 8

(B) 1

(C) $-1$

(D) none of these

Solution (D) is the correct answer. Indeed $\underset{x\to 0}{lim}\sin \frac{1}{x}$ does not exist.