Question

lf $f\left(x\right)=\frac{\sin (e^{x-2}-1)}{\log (x-1)}$ then $\underset{x\to 2}{lim}f\left(x\right)$ is given by

• A. $-2$
• B. $-1$
• C. $0$
• D. $1$

Answer 1

Answer: (D)

$1$

Substitute, $x=2+h$

when $x\to 2h\to 0$,
$\therefore \underset{x\to 2}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(h+2)=\underset{h\to 0}{lim}\frac{\sin (e^h-1)}{\log (1+h)}$
$=\underset{h\to 0}{lim}\frac{\sin (e^h-1)}{e^h-1}\cdot \frac{e^h-1}{h}\cdot \frac{h}{\log (1+h)}=1\cdot 1\cdot 1=1$