Question

Solve: $\underset{x\to 2}{lim}\frac{\sin (e^{x-2}-1)}{\log (x-1)}$

Answer 1

$\underset{x\to 2}{lim}\frac{\sin (e^{x-2}-1)}{\log (x-1)}=\frac{0}{0}$ form

Using L'Hospital's Rule, we get

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

$=\frac{\cos (1-1)\times e^o}{1}$

$=\frac{\cos 0^o\times 1}{1}$

$=1$

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