Question

Solve

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

Answer 1

${lim}_{x⟶2}\frac{\sin (e^{x-2}-1)}{\log (x-1)}{lim}_{x⟶2}\frac{\sin (e^{x-2}-1)}{\log (1+(x-2))}{lim}_{x⟶2}\frac{\sin (e^{x-2}-1)}{\log (x-2)}\frac{(x-2)}{\log (1+(x-2))}$

So, ${lim}_{x⟶2}\frac{\sin (e^{x-2}-1)}{\log (x-1)}=1$