Show that $\lim_{\delta \to 0}(1-\lambda \delta)^{1/\delta} = e^{-\lambda}$

Question

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.

Answer 1

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.

$$\lim_{x\to\infty}f(x)=\infty$$

Answer 2

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 $L = e^{-\lambda t}$.