Question

If $f\left(x\right)=\frac{\sin (e^{x-2}-1)}{ln(x-1)}$, then $\underset{x\to 2}{lim}f\left(x\right)$ is equal to

• A. -2
• B. -1
• C. 0
• D. 1

Answer 1

The correct option is D 1

Consider, $\underset{x\to 2}{lim}f\left(x\right)$

$=\underset{x\to 2}{lim}\frac{\sin (e^{x-2}-1)}{ln(x-1)}$

By applying L-Hospital's Rule

$=\underset{x\to 2}{lim}\frac{\cos (e^{x-2}-1).e^{x-2}}{\frac{1}{x-1}}$

$=\frac{1\times 1}{1}=1$

Hence, D is the correct option.