Question

$f\left(x\right)=\frac{e^{1/x^2}}{e^{1/x^2}-1}$ , $x\ne 0$, $f\left(0\right)=1$, then $f$ at $x=0$ is:

• A. discontinuous
• B. left continuous
• C. right continuous
• D. both B and C

Answer 1

The correct option is D both B and C

Given: $f\left(x\right)=\frac{e^{1/x^2}}{e^{1/x^2}-1}$

Divide by $e^{1/x^2}$ to numerator and denominator

The function can be written in the form of,

$\underset{x\to 0}{lim}\frac{1}{1-e^{\frac{-1}{x^2}}}$

Applying the limit,

$f\left(0\right)=\underset{x\to 0}{lim}\frac{1}{1-e^{\frac{-1}{x^2}}}=1$

Hence, the function is continuous at $x=0$

So, the function is both left and right continuous.