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.

Another approach: define
$$x:=\frac1\delta\implies \delta\to 0\implies x\to\infty$$
and our limit is
$$\left[\left(1-\frac\lambda x\right)^x\right]^t\xrightarrow[x\to\infty]{}(e^{-\lambda})^t=e^{-\lambda t}$$
We used above the basic 
$$\lim_{x\to\infty}\left(1\pm\frac\lambda{f(x)}\right)^{f(x)}=e^{\pm\lambda}$$
for any function $\;f(x)\;$ s.t.
$$\lim_{x\to\infty}f(x)=\infty$$
