Question

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

Answer 1

We have, $\underset{x\to 2}{lim}\frac{sin(e^{x-2}-1)}{log(x-1)}$

At x=2, the value of the given expression

$\frac{sin(e^{2-2}-1)}{log(2-1)}=\frac{sin0}{log1}=\frac{0}{0}form$

Applying L'Hospital Rule

$\underset{x\to 2}{lim}\frac{\frac{d}{dx}\left(sin\right(e^{x-2}-1\left)\right)}{\frac{d}{dx}log(x-1)}$

$\underset{x\to 2}{lim}\frac{cos\left(e^{(x-2)-1}\right).\frac{d}{dx}\left\{e^{(x-2)-1}\right\}}{\frac{1}{x-1}\frac{d}{dx}(x-1)}$

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

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

Putting x=2

(2-1) $cos(e^{2-2}-1)\times e^{(2-2)}$

$=1\times cos0\times e^0$

$=1\times 1\times 1=1$