Question

$f\left(x\right)=1/(1-e^{-1/x}),x\ne 0,f\left(0\right)=0$ at $x=0$ Is function continuous at x=0?

Answer 1

Here $f\left(x\right)=\frac{1}{1-e^{-1/x}}$
$\therefore f(0+0)=\underset{h\to 0}{lim}\frac{1}{1-e^{-1(0+h)}}=\underset{h\to 0}{lim}\frac{1}{1-e^{-1/h}}=\frac{1}{1-e^{-\infty }}=\frac{1}{1-0}=1.$ and $\therefore f(0-0)=\underset{h\to 0}{lim}\frac{1}{1-e^{-1(0+h)}}=\underset{h\to 0}{lim}\frac{1}{1-e^{-1/h}}=\underset{h\to 0}{lim}\frac{1}{1-e^{\infty }}=\frac{1}{1-\infty }=0.$ Also $f\left(0\right)=0.$
Hence f is discontinuous at $x=0$.The function has a jump of one unit at 0 Since $f(0+0)-f(0-0)=1.$