Limit of a sequence, power of one minus an exponential

Question

What is the limit of the following quantity $L \rightarrow \infty$, $$ (1 - \exp(-cL))^{\delta L} $$ for any $c$ and $\delta$ positive constants?

Answer 1

With $f(L) = (1 - \exp(-cL))^{\delta L}$, apply L'Hospital's rule to $\ln f(L)$.

$$\lim_{L \rightarrow \infty}\ln f(L)=\lim_{L \rightarrow \infty}\frac{\delta \ln(1 - \exp(-cL))}{1/L}=\lim_{L \rightarrow \infty}\frac{-c\delta L^2\exp(-cL)}{(1 - \exp(-cL))}=0,$$

using $L^2\exp(-cL) \rightarrow 0$ as $L \rightarrow \infty$ for $c>0.$

Then $\lim_{L \rightarrow \infty} f(L) =\lim_{L \rightarrow \infty} \exp[\ln f(L)] = \exp(0) = 1,$ since the exponential function is continuous.