Question

The value of $\underset{x\to 0}{lim}\frac{e-1}{e^2+e^{1/x}-1}$ is

• A. $0$
• B. $\frac{1}{e+1}$
• C. $\frac{e-1}{e^2}$
• D. Limit does not exist

Answer 1

The correct option is D Limit does not exist

$L=\underset{x\to 0}{lim}\frac{e-1}{e^2+e^{1/x}-1}$
$\text{RHL}=\underset{x\to 0^{+}}{lim}\frac{e-1}{e^2+e^{1/x}-1}$
Since, $x\to 0^{+}\Rightarrow \frac{1}{x}\to \infty \Rightarrow e^{1/x}\to \infty$
$\Rightarrow \text{RHL}=0$

$\text{LHL}=\underset{x\to 0^{-}}{lim}\frac{e-1}{e^2+e^{1/x}-1}$
Since, $x\to 0^{-}\Rightarrow \frac{1}{x}\to -\infty \Rightarrow e^{1/x}\to 0$
$=\underset{x\to 0^{-}}{lim}\frac{e-1}{e^2-1}=\frac{1}{e+1}$

$\Rightarrow \text{LHL}\ne \text{RHL}$
$\therefore$ Limit does not exist.