Question

Let f\left( x \right)=\left\\{ \begin{aligned} & 1-\sqrt{1-{{x}^{2}}}\text{ for }-1\le x\le 1 \\\ & 1+\log \dfrac{1}{x}\text{ for }x>1 \\\ \end{aligned} \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 $x\in \left( 0,1 \right)$

Answer 1

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 $x=a$.

Complete step-by-step answer:
The given function is f\left( x \right)=\left\\{ \begin{aligned} & 1-\sqrt{1-{{x}^{2}}}\text{ for }-1\le x\le 1 \\\ & 1+\log \dfrac{1}{x}\text{ for }x>1 \\\ \end{aligned} \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

& {{R}^{'}}=\dfrac{\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{1}{\dfrac{1}{1+h}}\times \dfrac{-1}{{{(1+h)}^{2}}}}{1} \\\ & =-1 \\\ \end{aligned}$$ Since the left hand derivative is not equal to the right hand derivative, hence , the function is not differentiable at $$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{aligned} & =1+0 \\\ & =1 \\\ \end{aligned}$$ The left-hand limit of $$f\left( x \right)$$at $$x=1$$is given by $$\begin{aligned} & \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{aligned}$$ Value of function at $$x=1$$is given as $$\begin{aligned} & f\left( 1 \right)=1-\sqrt{1-{{1}^{2}}} \\\ & =1 \\\ \end{aligned}$$ Clearly , left hand limit $$=$$right hand limit$$=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)$$. Answer is (a),(b),(d) Note: If a function is differentiable at a, it should necessarily be continuous. But if a function is continuous, it is not necessary that it is differentiable at $$x=a$$.