\displaystyle \lim_{y\rightarrow 0}\frac{e^{y+1}-e}{\ln (y+1)}=\lim_{y\rightarrow 0}\frac{\frac{d(e^{y+1}-e)}{dy}}{\frac{d(ln (y+1))}{dy}}=\lim_{y\rightarrow 0}\frac{e^{y+1}}{\frac{1}{y+1}}=e