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# Let f\left( x \right)=\left\{ \begin{align} & 1-\sqrt{1-{{x}^{2}}}\text{ for }-1\le x\le 1 \\ & 1+\log \dfrac{1}{x}\text{ for }x>1 \\ \end{align} \right.Then(a) $f$is continuous at $x=1$(b) $f$is not differentiable at $x=1$(c) $f$is continuous and differentiable at $x=1$(d) ${{f}^{'}}\left( x \right)$exists for all $x\in \left( 0,1 \right)$

Last updated date: 12th Sep 2024
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Hint: If the value of limit of the function at a point $x=a$ is equal to the value of the function at $x=a$ , the function is said to be continuous at $x=a$. A function is differentiable at $x=a$ , if the left-hand derivative of the function is equal to the right hand derivative of the function at $x=a$.

The given function is f\left( x \right)=\left\{ \begin{align} & 1-\sqrt{1-{{x}^{2}}}\text{ for }-1\le x\le 1 \\ & 1+\log \dfrac{1}{x}\text{ for }x>1 \\ \end{align} \right.
We will check if the function is continuous or differentiable at critical points of the function .
A function is differentiable at $x=a$ , if the left-hand derivative of the function is equal to the right hand derivative of the function at $x=a$.
We know , the left hand derivative of $f\left( x \right)$ at $x=a$ is given as ${{L}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( a-h \right)-f\left( a \right)}{-h}$ and the right hand derivative of $f\left( x \right)$ at $x=a$ is given as ${{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( a+h \right)-f\left( a \right)}{h}$.
First , we will check the differentiability of the function $f\left( x \right)$ at $x=1$.
The left-hand derivative of $f\left( x \right)$at $x=1$is given by
${{L}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( 1-h \right)-f\left( 1 \right)}{-h}$
$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left( 1-\sqrt{1-{{\left( 1-h \right)}^{2}}} \right)-\left( 1-\sqrt{1-{{1}^{2}}} \right)}{-h}$
$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{1-\sqrt{1-1+2h-{{h}^{2}}}-1}{-h}$
$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{-\sqrt{2h-{{h}^{2}}}}{-h}$
Now , on substituting $h=0$ in the limit , we can see that it gives an indeterminate value $\dfrac{0}{0}$. In such conditions, we apply L’ Hopital’s rule to evaluate the limit.
L’ Hopital’s Rule states that “ if $L=\underset{x\to c}{\mathop{\lim }}\,\dfrac{f(x)}{g(x)}$ and $f(x)=g(x)=0$ or $\infty$, then $L=\underset{x\to c}{\mathop{\lim }}\,\dfrac{f'(x)}{g'(x)}$ .”
So , to find the value of the limit , we must differentiate the numerator and the denominator with respect to $x$.
${{L}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\dfrac{-1}{2\sqrt{2h-{{h}^{2}}}}.2-2h}{-1}$
$=\dfrac{2}{0}=\infty$
The right-hand derivative of $f\left( x \right)$at $x=1$is given by
${{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( 1+h \right)-f\left( 1 \right)}{h}$
${{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{1+\log \dfrac{1}{\left( 1+h \right)}-\left( 1+\log 1 \right)}{h}$
${{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{1+\log \dfrac{1}{\left( 1+h \right)}-1-0}{h}$
${{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\log \dfrac{1}{\left( 1+h \right)}}{h}$
Applying L’ Hopital’s Rule , we get
\begin{align} & {{R}^{'}}=\dfrac{\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{1}{\dfrac{1}{1+h}}\times \dfrac{-1}{{{(1+h)}^{2}}}}{1} \\ & =-1 \\ \end{align}
Since the left hand derivative is not equal to the right hand derivative, hence , the function is not differentiable at $x=1$.
Now, we will check the continuity of the function at $x=1$.
A function is said to be continuous at $x=a$ if the value of limit of the function at a point $x=a$ is equal to the value of the function at $x=a$ .
The right-hand limit of $f\left( x \right)$at $x=1$ is given by
$\underset{h\to 0}{\mathop{\lim }}\,f\left( 1+h \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( 1+\log \dfrac{1}{1+h} \right)$
\begin{align} & =1+0 \\ & =1 \\ \end{align}
The left-hand limit of $f\left( x \right)$at $x=1$is given by
\begin{align} & \underset{h\to 0}{\mathop{\lim }}\,f\left( 1-h \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( 1-\sqrt{1-{{\left( 1-h \right)}^{2}}} \right) \\ & =1 \\ \end{align}
Value of function at $x=1$is given as
\begin{align} & f\left( 1 \right)=1-\sqrt{1-{{1}^{2}}} \\ & =1 \\ \end{align}
Clearly , left hand limit $=$right hand limit$=f\left( 1 \right)$
So , the function is continuous at $x=1$.
Now , in the interval $\left( 0,1 \right)$, $f\left( x \right)=1-\sqrt{1-{{x}^{2}}}$. On differentiating $f(x)$ with respect to $x$ , we get
$\dfrac{d}{dx}(f(x))=\dfrac{-1}{2\sqrt{1-{{x}^{2}}}}\times (-2x)$
$=\dfrac{x}{\sqrt{1-{{x}^{2}}}}$
Clearly , $f'(x)$ exists $\forall x\in (-1,1)$. Hence , ${{f}^{'}}\left( x \right)$exists for all $x\in \left( 0,1 \right)$.
Note: If a function is differentiable at a, it should necessarily be continuous. But if a function is continuous, it is not necessary that it is differentiable at $x=a$.