Question

If $f(x)=\frac{\sin \left(e^{x-2}-1\right)}{\log (x-1)}$, then $\displaystyle\lim _{x \rightarrow 2} f(x)$ is given by

• A. -2
• B. -1
• C. 0
• D. 1

Answer 1

Answer: (D)

We have,
$\displaystyle\lim _{x \rightarrow 2} f(x) =\lim _{x \rightarrow 2} \frac{\sin \left(e^{x-2}-1\right)}{\log (x-1)}$
$=\displaystyle\lim _{x \rightarrow 2}\left\{\frac{\sin \left(e^{x-2}-1\right)}{e^{x-2}-1} \cdot \frac{e^{x-2}-1}{x-2} \cdot \frac{x-2}{\log [1+(x-2)]}\right\} $
$=1 \times 1 \times 1=1$