Question

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

• A. $1$
• B. $2$
• C. $0$
• D. $3$

Answer 1

The correct option is C $0$

Given $f\left(x\right)$ is continuous at $x=0$
$LHL=RHL=f\left(0\right)$
$f\left(x\right)=\frac{x(e^{1/x}-e^{-1/x})}{e^{1/x}+e^{-1/x}}$
$f\left(x\right)=\frac{x(1-e^{-2/x})}{1+e^{-2/x}}$
So, $\underset{x\to 0}{lim}\frac{x(1-e^{-2/x})}{1+e^{-2/x}}=0$
$\Rightarrow f\left(0\right)=0$