Show that $\lim_{\delta \to 0}(1-\lambda \delta)^{1/\delta} = e^{-\lambda}$ My professor said that 
$$\lim_{\delta \to 0}(1-\lambda \delta)^{t/\delta}=e^{-\lambda t}$$
can be shown with L'Hospital's rule. I don't know what he meant. What is the best way to show this (or, more simply, $\lim_{\delta \to 0}(1-\lambda \delta)^{1/\delta} = e^{-\lambda}$)?
If I try as follows 
$$\lim_{\delta \to 0}\left(1-\lambda\delta \right)^{1/\delta} = \lim_{\eta \to \infty} \frac{(\eta-\lambda)^\eta}{\eta^\eta},$$
then I'm getting led into confusion trying LHR on the last one.

Let $$L = \lim_{\delta \to 0}(1-\lambda \delta)^{t/\delta}.$$ 
Then $$\ln{L} = \lim_{\delta \to 0} \frac{t}{\delta} \ln{(1-\lambda \delta)}$$
Apply LHR once to this expression and get that $\ln L = -\lambda t$, therefore $L = e^{-\lambda  t}$.
