Question

$\underset{y\to 0}{lim}\frac{\sin \left(e^y-1\right)}{\log \left(y+1\right)}$

Answer 1

$\underset{y⟶0}{lim}\frac{\sin (e^y-1)}{\log (y+1)}\underset{y⟶0}{lim}\frac{\sin (e^y-1)}{e^y-1}\times \frac{e^y-1}{y}\times \frac{y}{\log (y+1)}$

So, $\underset{y⟶0}{lim}\frac{\sin (e^y-1)}{e^y-1}=\underset{y⟶0}{lim}\frac{e^y-1}{y}=\underset{y⟶0}{lim}\frac{y}{\log (y+1)}=1\underset{y⟶0}{lim}\frac{\sin (e^y-1)}{\log (y+1)}=1$