Question

Show that the function f given by:

${\displaystyle f\left(x\right)=\{\begin{array}{lllll}\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}& \text{if}& x& \ne & 0\\ -1& \text{if}& x& =& 0\end{array}}$

is discontinuous at x = 0.

Answer 1

${\displaystyle f\left(x\right)=\{\begin{array}{lllll}\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}& \text{if}& x& \ne & 0\\ -1& \text{if}& x& =& 0\end{array}}$

LHL: ${\displaystyle \underset{x\to 0^{-}}{lim}\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}}$

= ${\displaystyle \underset{h\to 0}{lim}\frac{e^{-\frac{1}{h}}-1}{e^{-\frac{1}{h}}+1}=\frac{0-1}{0+1}=-1}$

RHL: ${\displaystyle \underset{h\to 0}{lim}\frac{e^{\frac{1}{h}}-1}{e^{\frac{1}{h}}+1}}$

= ${\displaystyle \underset{h\to 0}{lim}\frac{1-e^{-\frac{1}{h}}}{1+e^{-\frac{1}{h}}}=1}$

LHL ≠  RHL
∴ f(x) is discontinuous at x = 0.