Question

Let f(x)=\left\\{ \begin{matrix} \dfrac{x}{1+{{e}^{\dfrac{1}{x}}}},x\ne 0 \\\ 0,x=0 \\\ \end{matrix} \right.
Then
(a) f is continuous at $x=0$
(b) ${{f}^{'}}(0-0)=1$
(c) ${{f}^{'}}(0+0)=0$
(d) ${{f}^{'}}(0)=1$

Answer 1

Hint: First find the left hand limit and right hand limit to find out about the continuity. Then find the first order derivative at x = 0. Similarly find out the value of left hand derivative and right hand derivative at x = 0.

Complete step-by-step answer:
The given expression is, f(x)=\left\\{ \begin{matrix} \dfrac{x}{1+{{e}^{\dfrac{1}{x}}}},x\ne 0 \\\ 0,x=0 \\\ \end{matrix} \right.
For the given function to be continuous, the left hand limit and the right hand limit should be equal to f(0), i.e., $f(0)=LHL=RHL$.
First, we shall find the left hand limit, i.e., LHL.
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{x}{1+{{e}^{\dfrac{1}{x}}}}$
Here we take $x=0-h$, where $h\to 0$then,
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{0-h}{1+{{e}^{\dfrac{1}{0-h}}}}$
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{-h}{1+{{e}^{-\dfrac{1}{h}}}}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{-h}{1+\dfrac{1}{{{e}^{\dfrac{1}{h}}}}}$
Now applying the limits, we get
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{-0}{1+\dfrac{1}{{{e}^{\dfrac{1}{0}}}}}$
But we know, $\dfrac{1}{0}=\infty$ , so above equation becomes,
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{0}{1+\dfrac{1}{{{e}^{\infty }}}}$
We also know, $\dfrac{1}{{{e}^{\infty }}}=\dfrac{1}{\infty }\to 0$ , so above equation becomes,
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{0}{1+0}=0$
So, $LHL=0..........(i)$
Now, we shall find the right hand limit, i.e., RHL.
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{x}{1+{{e}^{\dfrac{1}{x}}}}$
Here we take $x=0+h$, where $h\to 0$then,
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{0+h}{1+{{e}^{\dfrac{1}{0+h}}}}$
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{h}{1+{{e}^{\dfrac{1}{h}}}}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{h}{1+{{e}^{\dfrac{1}{h}}}}$
Now applying the limits, we get
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{0}{1+{{e}^{\dfrac{1}{0}}}}$
But we know, $\dfrac{1}{0}=\infty$ , so above equation becomes,
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{0}{1+{{e}^{\infty }}}$
We also know, ${{e}^{\infty }}=\infty$ , so above equation becomes,
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{0}{1+\infty }=0$
So, $RHL=0..........(ii)$
From equation (i), (ii) and given value, we get
$f(0)=LHL=RHL$
Therefore, the given function is continuous at $x=0$.
So, option (a) is correct.
But as the given options consists of the values of Left hand derivative (LHD) & right hand derivative (RHD) let us check them too.
Now, let us consider the LHD.
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{d}{dx}\left[ \dfrac{x}{1+{{e}^{\dfrac{1}{x}}}} \right]$
By applying Quotient rule for differentiation, i.e., $\left( \dfrac{u\left( x \right)}{v\left( x \right)} \right)'=\dfrac{{u}'\left( x \right).v\left( x \right)-u\left( x \right).{v}'\left( x \right)}{v{{\left( x \right)}^{2}}}$
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{\left( 1+{{e}^{\dfrac{1}{x}}} \right)\dfrac{d}{dx}(x)-x\dfrac{d}{dx}\left( 1+{{e}^{\dfrac{1}{x}}} \right)}{{{\left( 1+{{e}^{\dfrac{1}{x}}} \right)}^{2}}}$
Now we will apply the exponential function rule, i.e., ${{\left( {{e}^{u(x)}} \right)}^{'}}={{e}^{u(x)}}.u'(x)$ , so we get
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{\left( 1+{{e}^{\dfrac{1}{x}}} \right)(1)-x{{e}^{\dfrac{1}{x}}}\dfrac{d}{dx}\left( \dfrac{1}{x} \right)}{{{\left( 1+{{e}^{\dfrac{1}{x}}} \right)}^{2}}}$
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{\left( 1+{{e}^{\dfrac{1}{x}}} \right)(1)-x{{e}^{\dfrac{1}{x}}}\left( (-1){{x}^{-1-1}} \right)}{{{\left( 1+{{e}^{\dfrac{1}{x}}} \right)}^{2}}}$
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{\left( 1+{{e}^{\dfrac{1}{x}}} \right)(1)+x{{e}^{\dfrac{1}{x}}}\left( {{x}^{-2}} \right)}{{{\left( 1+{{e}^{\dfrac{1}{x}}} \right)}^{2}}}$
This can be written as,
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{\left( 1+{{e}^{\dfrac{1}{x}}} \right)+{{x}^{1-2}}{{e}^{\dfrac{1}{x}}}}{{{\left( 1+{{e}^{\dfrac{1}{x}}} \right)}^{2}}}$
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{\left( 1+{{e}^{\dfrac{1}{x}}} \right)+{{x}^{-1}}{{e}^{\dfrac{1}{x}}}}{{{\left( 1+{{e}^{\dfrac{1}{x}}} \right)}^{2}}}$
Now, let us take $x=0-h,h\to 0$, so the above equation becomes,
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{f}^{'}}(0-h)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{\left( 1+{{e}^{\dfrac{1}{0-h}}} \right)+{{(0-h)}^{-1}}{{e}^{\dfrac{1}{0-h}}}}{{{\left( 1+{{e}^{\dfrac{1}{0-h}}} \right)}^{2}}}$
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{f}^{'}}(0-h)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{\left( 1+{{e}^{-\dfrac{1}{h}}} \right)+{{(-h)}^{-1}}{{e}^{-\dfrac{1}{h}}}}{{{\left( 1+{{e}^{-\dfrac{1}{h}}} \right)}^{2}}}$
Now by applying the limits, we get
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{f}^{'}}(0-0)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{\left( 1+{{e}^{-\dfrac{1}{0}}} \right)+{{(0)}^{-1}}{{e}^{-\dfrac{1}{0}}}}{{{\left( 1+{{e}^{-\dfrac{1}{0}}} \right)}^{2}}}$
As ${{e}^{-\dfrac{1}{0}}}={{e}^{-\infty }}=0$, so above equation becomes,
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{f}^{'}}(0-0)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{\left( 1+0 \right)+{{(0)}^{-1}}(0)}{{{\left( 1+(0) \right)}^{2}}}$
On solving, we get
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{f}^{'}}(0-0)=1.........(iii)$
So, option (b) is also correct.
Now, let us consider the RHD.
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{d}{dx}\left[ \dfrac{x}{1+{{e}^{\dfrac{1}{x}}}} \right]$
By applying Quotient rule for differentiation, i.e., $\left( \dfrac{u\left( x \right)}{v\left( x \right)} \right)'=\dfrac{{u}'\left( x \right).v\left( x \right)-u\left( x \right).{v}'\left( x \right)}{v{{\left( x \right)}^{2}}}$
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left( 1+{{e}^{\dfrac{1}{x}}} \right)\dfrac{d}{dx}(x)-x\dfrac{d}{dx}\left( 1+{{e}^{\dfrac{1}{x}}} \right)}{{{\left( 1+{{e}^{\dfrac{1}{x}}} \right)}^{2}}}$
Now we will apply the exponential function rule, i.e., ${{\left( {{e}^{u(x)}} \right)}^{'}}={{e}^{u(x)}}.u'(x)$ , so we get
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left( 1+{{e}^{\dfrac{1}{x}}} \right)(1)-x{{e}^{\dfrac{1}{x}}}\dfrac{d}{dx}\left( \dfrac{1}{x} \right)}{{{\left( 1+{{e}^{\dfrac{1}{x}}} \right)}^{2}}}$
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left( 1+{{e}^{\dfrac{1}{x}}} \right)(1)-x{{e}^{\dfrac{1}{x}}}\left( (-1){{x}^{-1-1}} \right)}{{{\left( 1+{{e}^{\dfrac{1}{x}}} \right)}^{2}}}$
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left( 1+{{e}^{\dfrac{1}{x}}} \right)(1)+x{{e}^{\dfrac{1}{x}}}\left( {{x}^{-2}} \right)}{{{\left( 1+{{e}^{\dfrac{1}{x}}} \right)}^{2}}}$
This can be written as,
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left( 1+{{e}^{\dfrac{1}{x}}} \right)+{{x}^{1-2}}{{e}^{\dfrac{1}{x}}}}{{{\left( 1+{{e}^{\dfrac{1}{x}}} \right)}^{2}}}$
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left( 1+{{e}^{\dfrac{1}{x}}} \right)+{{x}^{-1}}{{e}^{\dfrac{1}{x}}}}{{{\left( 1+{{e}^{\dfrac{1}{x}}} \right)}^{2}}}$
Now, let us take $x=0+h,h\to 0$, so the above equation becomes,
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(0+h)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left( 1+{{e}^{\dfrac{1}{0+h}}} \right)+{{(0+h)}^{-1}}{{e}^{\dfrac{1}{0+h}}}}{{{\left( 1+{{e}^{\dfrac{1}{0+h}}} \right)}^{2}}}$
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(0+h)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left( 1+{{e}^{\dfrac{1}{h}}} \right)+{{(h)}^{-1}}{{e}^{\dfrac{1}{h}}}}{{{\left( 1+{{e}^{\dfrac{1}{h}}} \right)}^{2}}}$
Now by applying the limits, we get
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(0+h)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left( 1+{{e}^{\dfrac{1}{0}}} \right)+{{(0)}^{1}}{{e}^{\dfrac{1}{0}}}}{{{\left( 1+{{e}^{\dfrac{1}{0}}} \right)}^{2}}}$
As ${{e}^{\dfrac{1}{0}}}={{e}^{\infty }}=\infty$, so above equation becomes,
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(0+h)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left( 1+\infty \right)+{{(0)}^{-1}}(\infty )}{{{\left( 1+\infty \right)}^{2}}}=\dfrac{\infty }{\infty }$
So this is of indeterminate form, now we will apply L’Hopsital rule and as per L’hospital rule we have to differentiate numerator and denominator separately as shown below:
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}\left( \left( 1+{{e}^{\dfrac{1}{x}}} \right)+{{x}^{-1}}{{e}^{\dfrac{1}{x}}} \right)}{\dfrac{d}{dx}\left( {{\left( 1+{{e}^{\dfrac{1}{x}}} \right)}^{2}} \right)}$
Now we will apply the exponential function rule, i.e., ${{\left( {{e}^{u(x)}} \right)}^{'}}={{e}^{u(x)}}.u'(x)$ , so we get

& \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{{{e}^{\dfrac{1}{x}}}\dfrac{d}{dx}\left( \dfrac{1}{x} \right)+{{x}^{-1}}\dfrac{d}{dx}\left( {{e}^{\dfrac{1}{x}}} \right)+{{e}^{\dfrac{1}{x}}}\dfrac{d}{dx}\left( {{x}^{-1}} \right)}{2\left( 1+{{e}^{\dfrac{1}{x}}} \right)\dfrac{d}{dx}\left( {{e}^{\dfrac{1}{x}}} \right)} \\\ & \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{{{e}^{\dfrac{1}{x}}}\left( \dfrac{{{x}^{-1-1}}}{-1-1} \right)+{{x}^{-1}}{{e}^{\dfrac{1}{x}}}\dfrac{d}{dx}\left( \dfrac{1}{x} \right)+{{e}^{\dfrac{1}{x}}}\left( \dfrac{{{x}^{-1-1}}}{-1-1} \right)}{2\left( 1+{{e}^{\dfrac{1}{x}}} \right){{e}^{\dfrac{1}{x}}}\dfrac{d}{dx}\left( \dfrac{1}{x} \right)} \\\ & \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{{{e}^{\dfrac{1}{x}}}\left( \dfrac{{{x}^{-2}}}{-2} \right)+{{x}^{-1}}{{e}^{\dfrac{1}{x}}}\left( \dfrac{{{x}^{-1-1}}}{-1-1} \right)+{{e}^{\dfrac{1}{x}}}\left( \dfrac{{{x}^{-2}}}{-2} \right)}{2\left( 1+{{e}^{\dfrac{1}{x}}} \right){{e}^{\dfrac{1}{x}}}\left( \dfrac{{{x}^{-1-1}}}{-1-1} \right)} \\\ & \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{2{{e}^{\dfrac{1}{x}}}\left( \dfrac{{{x}^{-2}}}{-2} \right)+{{x}^{-1}}{{e}^{\dfrac{1}{x}}}\left( \dfrac{{{x}^{-2}}}{-2} \right)}{2\left( 1+{{e}^{\dfrac{1}{x}}} \right){{e}^{\dfrac{1}{x}}}\left( \dfrac{{{x}^{-2}}}{-2} \right)} \\\ & \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{-\left[ {{e}^{\dfrac{1}{x}}}{{x}^{-2}}+{{e}^{\dfrac{1}{x}}}\left( \dfrac{{{x}^{-3}}}{2} \right) \right]}{-\left( 1+{{e}^{\dfrac{1}{x}}} \right){{e}^{\dfrac{1}{x}}}{{x}^{-2}}} \\\ & \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left[ {{e}^{\dfrac{1}{x}}}{{x}^{-2}}+{{e}^{\dfrac{1}{x}}}\left( \dfrac{{{x}^{-3}}}{2} \right) \right]}{\left( 1+{{e}^{\dfrac{1}{x}}} \right){{e}^{\dfrac{1}{x}}}{{x}^{-2}}} \\\ \end{aligned}$$ Taking out the common term and cancelling, we get $$\begin{aligned} & \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{{{e}^{\dfrac{1}{x}}}{{x}^{-2}}\left[ 1+\left( \dfrac{{{x}^{-1}}}{2} \right) \right]}{\left( 1+{{e}^{\dfrac{1}{x}}} \right){{e}^{\dfrac{1}{x}}}{{x}^{-2}}} \\\ & \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left[ 1+\left( \dfrac{{{x}^{-1}}}{2} \right) \right]}{\left( 1+{{e}^{\dfrac{1}{x}}} \right)} \\\ \end{aligned}$$ Now, let us take $$x=0+h,h\to 0$$, so the above equation becomes, $$\Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left[ 1+\left( \dfrac{{{(0+h)}^{-1}}}{2} \right) \right]}{\left( 1+{{e}^{\dfrac{1}{0+h}}} \right)}$$ Now by applying the limits, we get $$\Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\dfrac{\left[ 1+\left( \dfrac{{{(0+0)}^{-1}}}{2} \right) \right]}{\left( 1+{{e}^{\dfrac{1}{0+0}}} \right)}$$ As $${{e}^{\dfrac{1}{0}}}={{e}^{\infty }}=\infty $$, so above equation becomes, $$\Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\dfrac{1}{\infty }=\infty $$ This is undefined, so option (c) is not correct. From equation (iii) and (iv), we can see that $$LHD\ne RHD$$ Therefore, $f'(0)$ is undefined, so option (d) is also not correct. So, we can conclude that the correct answer is option (a), (b), Note: One common mistake is once you get the left hand limit or left hand derivative, we can write similarly the right hand limit or right hand derivative is this, instead of solving. So don’t conclude before solving anything.