Question

The value of the $\displaystyle\lim_{x\to1} \frac{\sin\left(e^{x-1}-1\right)}{\log \, x}$ is

• A. $0$
• B. $e$
• C. $\frac{1}{e}$
• D. $1$

Answer 1

Answer: (D)

$\displaystyle\lim_{x\to1} \frac{\sin\left(e^{x-1}-1\right)}{\log \, x}$

$=\displaystyle\lim_{h\to0} \frac{\sin \left(e^{h}-1\right)}{\log \left(1+h\right)}$

$=\displaystyle\lim_{h\to0} \frac{\sin\left(e^{h}-1\right)}{\log \left(1+h\right)}$

$=\displaystyle\lim_{h\to0} \frac{\sin \left(e^{h}-1\right)}{\left(e^{h}-1\right)}\times\frac{\left(e^{h}-1\right)}{\log \left(1+h\right)}$

$=1\times\displaystyle\lim_{h\to0} \frac{\left(h+\frac{h_{2}}{2!}+\ldots\right)}{\left(h-\frac{h^{2}}{2}+\ldots\infty\right)}$

$=1\times1=1$