Q84 / 170 · Manipal (MET) 2016 · Mathematics

If $f(x)=\frac{e^{1 / x}-1}{e^{1 / x}+1}, x \neq 0$ and $f(0)=0$, then $f(x)$ is

  • A
    left continuous at 0
  • B
    right continuous at 0
  • Correct answer
    C
    discontinuous at 0
  • D
    continuous at 0

Solution

$\because f^{\prime}\left(0^{-}\right)=\lim _{h \rightarrow 0} f(0-h)$ $\Rightarrow \quad f^{\prime}\left(0^{-}\right)=\lim _{h \rightarrow 0} \frac{e^{-1 / h}-1}{e^{1 /-h}+1}$ $\begin{array}{ll}\Rightarrow & f^{\prime}\left(0^{-}\right)=\frac{0-1}{0+1} \\ \therefore & f^{\prime}\left(0^{-}\right)=-1\end{array}$ and $\quad f^{\prime}\left(0^{+}\right)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} \frac{e^{1 / h}-1}{e^{1 / h}+1}$ $\Rightarrow \quad f^{\prime}\left(0^{+}\right)=\frac{1-0}{1+0}=1$

Hence, the correct answer is option: 'C'.

Relevant Questions