Chapter 5: Continuity and Differentiability

Ans: The given function is $\text{f(x)=cos(}{{\text{x}}^{\text{2}}}\text{)}$.

Note that, $\text{f}$ is defined for all real numbers and so $\text{f}$ can be expressed as the composition of two functions as, $\text{f=g}\circ \text{h}$, where $\text{g(x)=cosx}$ and $\text{h(x)=}{{\text{x}}^{\text{2}}}$.

$\text{ }\!\![\!\!\text{ }\therefore \text{(goh)(x)=g(h(x))=g(}{{\text{x}}^{\text{2}}}\text{)=cos(}{{\text{x}}^{\text{2}}}\text{)=f(x) }\!\!]\!\!\text{ }$

Now, it is to be Proven that, the functions $\text{g(x)=cosx}$ and $\text{h(x)=}{{\text{x}}^{\text{2}}}$ are continuous.

Since the function $\text{g}$ is defined for all the real numbers, let 's consider $\text{c}$ be a real number.

Then, $\text{g(c)=cosc}$.

Substitute $\text{x=c+h}$ into the function $\text{g}$.

When, $\text{x}\to \text{c}$, then \[\text{h}\to 0\].

Then we have,

$\begin{align} & \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{cosx} \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\text{cos}\left( \text{c+h} \right) \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\left[ \text{cosc}\cdot \text{cosh-sinc}\cdot \text{sinh} \right] \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{cosc}\cdot \text{cosh} \right)-\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{sinc}\cdot \text{sinh} \right) \\ & \text{=cosc}\cdot \text{cos0}-\text{sinc}\cdot \text{sin0} \\ & \text{=cosc }\!\!\times\!\!\text{ 1}-\text{sinc }\!\!\times\!\!\text{ 0} \\ & \text{=cosc} \\ \end{align}$

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{g(x)=g(c)}$.

Hence, the function $\text{g(x)=cosx}$ is continuous.

Again, $\text{h(x)=}{{\text{x}}^{\text{2}}}$ is defined for every real point.

So, let consider $\text{k}$ be a real number, then $\text{h(k)=}{{\text{k}}^{\text{2}}}$ and

$\underset{\text{x}\to \text{k}}{\mathop{\lim }}\,\text{h(x)=}\underset{\text{x}\to \text{k}}{\mathop{\lim }}\,{{\text{x}}^{\text{2}}}\text{=}{{\text{k}}^{\text{2}}}$.

Therefore, $\underset{\text{x}\to \text{k}}{\mathop{\lim }}\,\text{h(x)=h(k)}$.

Hence, the function $\text{h}$ is continuous.

Now, remember that for real valued functions $\text{g}$ and $\text{h}$ , such that $\text{(g }\circ \text{ h)}$ is defined at $\text{c}$ , if $\text{g}$ is continuous at $\text{c}$ and $\text{f}$ is continuous at $\text{g(c)}$, then $\text{(f }\circ \text{ h)}$is continuous at $\text{c}$.

Hence, the function $\text{f(x)=(g }\circ \text{ h)(x)=cos(}{{\text{x}}^{2}}\text{)}$ is continuous.

Ans: The given function is $\text{f(x)=}\left| \text{cosx} \right|$.

Note that, the function $\text{f}$ is defined for all real numbers. So, the function $\text{f}$ can be expressed as the composition of two functions as, $\text{f=g}\circ \text{h}$, where $\text{g(x)=}\left| \text{x} \right|$ and $\text{h(x)=cosx}$.

$[\because (\text{goh)(x)=g(h(x))=g(cosx)=}\left| \text{cosx} \right|\text{=f(x) }\!\!]\!\!\text{ }$

Now, it is to be proved that the functions $\text{g(x)=}\left| \text{x} \right|$ and $\text{h(x)=cosx}$ are continuous.

Remember that, $\text{g(x)=}\left| \text{x} \right|$, can be written as

$\text{g(x)=}\left\{ \begin{align} & \text{-x, if x}\,<\,\text{0} \\ & \text{ x, if x}\ge \text{0} \\ \end{align} \right.$

Now, since the function $\text{g}$ is defined for every real number, let consider $\text{c}$ be a real number.

Then there may arise three cases, either $\text{c}<\text{0}$, or $\text{c}>\text{0}$, or $\text{c=0}$.

Let's discuss the cases one after another.

Case I: When $\text{c}<\text{0}$.

Then, $\text{g(c)=-c}$.

Also, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{(-x)=-c}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{g(x)=g(c)}$.

Hence, the function $\text{g}$ is continuous at every point $\text{x}$, for $\text{x}<\text{0}$.

Case II: When $\text{c}>\text{0}$.

Then, $\text{g(c)=c}$.

Also, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{x=c}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{g(x)=g(c)}$.

Hence, the function $\text{g}$ is continuous at every point $\text{x}$ for $\text{x}>\text{0}$.

Case III: When \[\text{c=0}\].

Then, $\text{g(c)=g(0)=0}$.

Now, the left-hand-limit of the function $\text{g}$ at $\text{x=0}$ is

$\underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to {{0}^{\text{-}}}}{\mathop{\lim }}\,\text{(-x)=0}$ and the right-hand-limit is

$\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\text{(x)=0}$.

Therefore, $\underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\text{g(x)=g(0)}$.

Hence, the function $\text{g}$ is continuous at \[\text{x=0}\].

By observing the above three discussions, we can conclude that the function $\text{g}$ is continuous at every real point.

Now, since the function $\text{h(x)=cosx}$ is defined for all real numbers, let 's consider $c$ be a real number. Then, substitute $\text{x=c+h}$ into the function $\text{h}$.

So, when $\text{x}\to \text{c}$, then $\text{h}\to 0$.

Then, we have

\[\text{h}\left( \text{c} \right)=\text{cosc}\] and

\[\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{h(x)=}\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{cosx}\]

\[\text{=}\underset{\text{h}\to \text{0}}{\mathop{\text{lim}}}\,\text{cos}\left( \text{c+h} \right)\]

\[\text{=}\underset{\text{h}\to \text{0}}{\mathop{\text{lim}}}\,\left[ \text{cosc}\cdot \text{cosh-sinc}\cdot \text{sinh} \right]\]

\[\text{=}\underset{\text{h}\to \text{0}}{\mathop{\text{lim}}}\,\left( \text{cosc}\cdot \text{cosh} \right)-\underset{\text{h}\to \text{0}}{\mathop{\text{lim}}}\,\left( \text{sinc}\cdot \text{sinh} \right)\]

\[\begin{align} & \text{=cosc}\cdot \text{cos0}-\text{sinc}\cdot \text{sin0} \\ & \text{=cosc }\!\!\times\!\!\text{ 1-sinc }\!\!\times\!\!\text{ 0} \\ & \text{=cosc} \\ \end{align}\]

Therefore, \[\underset{x\to \text{c}}{\mathop{\lim }}\,\text{h}\left( \text{x} \right)=\text{h}\left( \text{c} \right)\].

Hence, the function \[\text{h(x)=cosx}\] is continuous.

Now remember that, for real valued functions $\text{g}$ and \[\text{h}\], such that $\text{(g}\circ \text{h)}$ is defined at $\text{x=c}$ only if $\text{g}$ is continuous at $\text{c}$ and $\text{f}$ is continuous at $\text{g(c)}$, then the composition functions $\text{(f }\circ \text{ g)}$ is continuous at $\text{x=c}$.

Thus, the function $\text{f(x)=(goh)(x)=g(h(x))=g(cosx)=}\left| \text{cosx} \right|$ is continuous.

Ans: First suppose that, $f\left( \text{x} \right)=\sin \left| \text{x} \right|$.

Now, note that the function $\text{f}$ is defined for all real numbers and so \[\text{f}\] can be expressed as the composition of  functions as, $\text{f=g}\circ \text{h,}$ where $\text{g(x)=sinx}$ and $\text{h(x)=}\left| \text{x} \right|$.

$\left[ \text{(g}\circ \text{h)(x)=g(h(x))=g(}\left| \text{x} \right|\text{)=sin}\left| \text{x} \right|\text{=f(x)} \right]$

So, it is to be proved that the functions $\text{g(x)=sinx}$ and \[\text{h(x)=}\left| \text{x} \right|\] are continuous.

Now, remember that, the function $\text{h(x)=}\left| \text{x} \right|$ can be written as 

$\text{h(x)=}\left\{ \begin{align} & \text{-x, if x}<\text{0} \\ & \text{x, if x}\ge \text{0} \\ \end{align} \right.$

Note that, the function $\text{h}$ is defined for every real number, and so let consider $\text{c}$ be a real number. 

Then, there may arise three cases, either $\text{c}<\text{0}$, or $\text{c}>\text{0}$, or $\text{c=0}$.

Let us discuss the cases one after another.

Case I: When $\text{c}<\text{0}$.

Then \[\text{h(c)=-c}\].

Also, \[\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{(-x)=}\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{x=-c}\].

Therefore, \[\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{h(x)=h(c)}\].

Hence, the function $\text{h}$ is continuous at every point $\text{x}$  for $\text{x}<\text{0}$.

Case II: When $\text{c}>\text{0}$.

Then, \[\text{h(c)=c}\]

Also, $\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{(-x)=}\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{x=c}$.

Therefore, \[\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{h(x)=h(c)}\].

Thus, the function $\text{h}$ is continuous at every point $\text{x}$  for $\text{x}>\text{0}$.

Case III: When \[\text{c = 0}\].

 Then, $\text{h(c)=h(0)=0}$.

Also, the left-hand-limit of the function $\text{h}$ at $\text{x=0}$ is

$\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\text{h(x)=}\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\text{(-x)=0}$ and the right-hand -limit is 

$\underset{\text{x}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{h(x)=}\underset{\text{x}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{(x)=0}$.

Therefore, $\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\text{h(x)=}\underset{\text{x}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{(x)=h(0)}$.

Thus, the function $\text{h}$ is continuous at \[\text{x = 0}\].

By observing the above three discussions, we can conclude that the function $\text{h}$ is continuous at every point. 

Again, since the function $\text{g(x)=sinx}$ is defined for all real numbers, so let consider $\text{c}$ be a real number and substitute $\text{x=c+k}$ into the function.

Now, when $\text{x}\to \text{c}$ then \[\text{k }\to \text{ 0}\].

Then, we have

$\text{g(c)=sinc}$.

Also,

$\begin{align} & \underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{g(x)=}\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{sinx} \\ & \text{=}\underset{\text{k}\to \text{0}}{\mathop{\text{lim}}}\,\text{sin}\left( \text{c+k} \right) \\ & \text{=}\underset{\text{k}\to \text{0}}{\mathop{\text{lim}}}\,\left[ \text{sinc}\cdot \text{cosk+cosc}\cdot \text{sink} \right] \\ & \text{=}\underset{\text{k}\to \text{0}}{\mathop{\text{lim}}}\,\left( \text{sinc}\cdot \text{cosk} \right)\text{+}\underset{\text{h}\to \text{0}}{\mathop{\text{lim}}}\,\left( \text{cosc}\cdot \text{sink} \right) \\ & \text{=sinc}\cdot \text{cos0+cosc}\cdot \text{sin0} \\ & \text{=sinc+0} \\ & \text{=sinc} \\ \end{align}$

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{g(x)=g(c)}$.

Hence, the function $\text{g}$ is continuous.

Now, remember that, for any two real valued functions $\text{g}$ and $\text{h}$, such that the composition of functions $\text{g}\circ \text{h}$ is defined at $\text{c}$, if $\text{g}$ is continuous at $\text{c}$ and $\text{f}$ is continuous at $\text{g(c)}$, then the composition function $\text{g}\circ \text{h}$ is continuous at $\text{c}$.

Thus, the function $\text{(g}\circ \text{h)(x)=g(h(x))=g(}\left| \text{x} \right|\text{)=sin}\left| \text{x} \right|\text{=f(x)}$ is continuous.

Ans: The given function is $\text{f(x)=}\left| \text{x} \right|\text{-}\left| \text{x+1} \right|$.

Let consider two functions

$\text{g(x)=}\left| \text{x} \right|$ and $\text{h(x)=}\left| \text{x+1} \right|$.

Then we get,  $\text{f=g-h}$.

Now, the function \[\text{g(x)=}\left| \text{x} \right|\] can be written as

$\text{g(x)=}\left\{ \begin{align} & \text{-x, if x}<\text{0} \\ & \text{x, if x}\ge \text{0} \\ \end{align} \right.$

Note that, the function $\text{g}$ is defined for every real number and so let consider $\text{c}$ be a real number.

Then there may arise three cases, either $\text{c}<\text{0}$, or $\text{c}>\text{0}$, or $\text{c=0}$.

Let us discuss the cases one after another.

Case I: When $\text{c}<\text{0}$.

Then, $\text{g(c)=g(0)=-c}$.

Also, $\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{g(x)=}\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{(-x)=-c}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{g(x)=g(c)}$.

Hence, the function $\text{g}$ is continuous at every point $\text{x}$ for \[\text{x}<\text{0}\].

Case II: When \[\text{c}>\text{0}\].

 Then $\text{g(c)=c}$.

Also, $\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{g(x)=}\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{x=c}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{g(x)=g(c)}$.

Hence, the function $\text{g}$ is continuous at every point $\text{x}$, where \[\text{x}>\text{0}\].

Case III: When \[\text{c = 0}\].

Then $\text{g(c)=g(0)=0}$.

Also, the left-hand-limit of the function $\text{g}$ at $\text{x=0}$ is

$\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\text{g(x)=}\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\text{(-x)=0}$ and the right-hand-limit is

$\underset{\text{x}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{g(x)=}\underset{\text{x}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{(x)=0}$.

Therefore, $\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\text{g(x)=}\underset{\text{x}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{(x)=g(0)}$.

Hence, the function $\text{g}$ is continuous at \[\text{x = 0}\].

Thus, we can conclude by observing the above three discussions that $\text{g}$ is continuous at every real point.

Now, remember that, the function $\text{h(x)=}\left| \text{x+1} \right|$ can be written as

$\text{h(x)=}\left\{ \begin{align} & \text{-x(x+1), if, x}<\text{-1} \\ & \text{x+1, }\,\,\,\,\,\,\,\,\text{if, x}\ge \text{-1} \\ \end{align} \right.$

Note that, the function $\text{h}$ is defined for all real numbers, and so let consider $\text{c}$ be a real number.

Case I: When \[\text{c}<\text{-1}\].

Then $\text{h(c)=-(c+1)}$.

Also, $\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\left[ \text{-(x+1)} \right]\text{=-(c+1)}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{h(x)=h(c)}$.

Hence, the function $\text{h}$ attains continuity at every real point $\text{x}$, where \[\text{x}<\text{-1}\].

Case II: When \[\text{c}>\text{-1}\].

Then, $\text{h(c)=c+1}$.

Also, $\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{h(x)=}\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{(x+1)=(c+1)}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{h(x)=h(c)}$.

Hence, the function $\text{h}$ satisfies continuity at every real point $\text{x}$ for \[\text{x}>\text{-1}\].

Case III: When \[\text{c =-1}\].

Then, $\text{h(c)=h(-1)=-1+1=0}$.

Also, the left-hand-limit of the function $\text{h}$ at $\text{x=1}$ is

$\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\text{lim}}}\,\text{h(x)=}\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\text{lim}}}\,\left[ \text{-(x+1)} \right]\text{=-(-1+1)=0}$ and the right-hand-limit is

$\underset{\text{x}\to {{\text{1}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{h(x)=}\underset{\text{x}\to {{\text{1}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{(x+1)=(-1+1)=0}$.

Therefore, $\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\text{lim}}}\,\text{h=}\underset{\text{x}\to {{\text{1}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{h(x)=h(-1)}$.

Thus, the function $\text{h}$ satisfies continuity at \[\text{x=-1}\].

Hence, by observing the above three discussions, we can conclude that the function $\text{h}$ is continuous for every real point.

Now, since the functions $\text{g}$ and $\text{h}$ are both continuous, so the function $\text{f=g-h}$ is also continuous.

Hence, the function $\text{f}$ does not have any discontinuity points.