Question

If $f(x)=\frac{\sin \left(e^{x-2}-1\right)}{\ln (x-1)},x\ne 2$ is continuous at $x=2$, then $f(2)=$

• A. 0
• B. 2
• C. 1
• D. -2

Answer 1

Answer: (C)

Put $x=2+h$
$f\left(0^{+}\right)=\underset{h\to 0}{\mathrm{Lim}}\frac{\sin \left(e^h-1\right)}{\ln (1+h)}$
$=\underset{h\to 0}{\mathrm{Lim}}\frac{e^h-1}{h}\times \frac{h}{\ln (1+h)}=1$
Put $x=2-h$
$f\left(0^{-}\right)=\underset{h\to 0}{\mathrm{Lim}}\frac{\sin \left(e^{-h}-1\right)}{\ln (1-h)}$
$=\underset{h\to 0}{\mathrm{Lim}}\left(\frac{e^{-h}-1}{-h}\right)\times \left(\frac{-h}{\ln (1-h)}\right)=1$
$\therefore f(2)=1$ as $f(x)$ is continuous at $x=2$