Question

If $f\left(x\right)=\left\{\begin{array}{ll}\frac{1}{1+e^{1/x}}& ,x\ne 0\\ 0& ,x=0\end{array}\right.$ then f (x) is

(a) continuous as well as differentiable at x = 0
(b) continuous but not differentiable at x = 0
(c) differentiable but not continuous at x = 0
(d) none of these

Answer 1

(d) none of these

we have,

$$\begin{gathered} (\mathrm{LHL}atx=0) \\ =\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(0-h)=\underset{h\to 0}{lim}f(-h) \\ =\underset{h\to 0}{lim}\frac{1}{1+e^{1/-h}} \\ =\underset{h\to 0}{lim}\frac{1}{1+{\displaystyle \frac{1}{e^{1/h}}}}[\underset{h\to 0}{lim}\frac{1}{e^{1/h}}=0] \\ =\frac{1}{1+0} \\ =1 \\ (RHLatx=0) \\ =\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(0+h) \\ =\underset{h\to 0}{lim}\frac{1}{1+e^{1/h}} \\ =\frac{1}{1+e^{1/0}}=\frac{1}{1+e^{\infty }}=\frac{1}{1+\infty } \end{gathered}$$

So, f(x) is not continuous at x = 0

Differentiability at x = 0

$$\begin{gathered} (LHDatx=0) \\ =\underset{x\to 0^{-}}{lim}\frac{f\left(x\right)-f\left(0\right)}{x-0} \\ =\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{0-h-0} \\ =\underset{h\to 0}{lim}\frac{f(-h)-0}{-h} \\ =\underset{h\to 0}{lim}\frac{\frac{1}{1+e^{1/-h}}}{-h} \\ =\underset{h\to 0}{lim}\frac{\frac{1}{1+{\displaystyle \frac{1}{e^{1/h}}}}}{-h} \\ =\underset{h\to 0}{lim}\frac{\frac{1}{1+{\displaystyle 0}}}{-h}=\underset{h\to 0}{lim}\frac{{\displaystyle 1}}{-h}=-\infty \\ (RHDatx=0) \\ =\underset{x\to 0^{+}}{lim}\frac{f\left(x\right)-f\left(0\right)}{x-0} \\ =\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{0+h-0} \\ =\underset{h\to 0}{lim}\frac{f\left(h\right)-0}{h} \\ =\underset{h\to 0}{lim}\frac{\frac{1}{1+e^{1/h}}}{h}=\infty \\ So,f\left(x\right)isalsonotdifferentiableatx=0. \end{gathered}$$