Question

Solve: $\underset{x\to 0}{lim}\frac{e^{2+x}-e^2}{x}$.

Answer 1

$li{m}_{x\to 0}\frac{e^{2+x}-e^2}{x}$

$=li{m}_{x\to 0}\frac{e^2(e^x-1)}{x}$

$=li{m}_{x\to 0}\frac{e^2.e^x}{1}$

$=e^2.e^0=e^2$