Prove $\lim_{x \rightarrow \infty} (1-e^{-x})^{x} = 1$ without l'Hôpital

Question

A bit rusty on my calculus, trying to figure out whether this is correct.

I tried proving $\underset{x \rightarrow \infty}{\lim} (1-e^{-x})^{x} = 1$ without using L'hôpital.

$$\lim_{x \rightarrow \infty} \left(1-e^{-x}\right)^{x}=\lim_{x \rightarrow \infty}\left(1- \frac{x\cdot e^{-x}}{x}\right)^{x}\overset{(1)}{=}\lim_{x \rightarrow \infty} e^{e^{-x}\cdot x}=e^{\lim_{x \rightarrow \infty} e^{\frac{x}{e^{x}}}}=e^{0}=1$$

I'm a little unsure about the transition I marked as $(1)$.

Answer 1

We have that as $x \to \infty$

$$\left(1+\frac a{x}\right)^{x} \to e^a$$

is true only for $a\in \mathbb R$ fixed, here we can use that

$$\left(1-\frac1{e^{x}}\right)^{x}=e^{x\log\left(1-\frac1{e^{x}}\right) }=e^{\frac{x}{e^x}\frac{\log\left(1-\frac1{e^{x}}\right)}{\frac1{e^x}} }$$

and use standard limits to get the result indeed

• $\frac{x}{e^x} \to 0$
• $\frac{\log\left(1-\frac1{e^{x}}\right)}{\frac1{e^x}} \to -1$

Refer also to

• About $\lim \left(1+\frac {x}{n}\right)^n$
• How to prove that exponential grows faster than polynomial?
• Determine the following limit as x approaches 0: $\frac{\ln(1+x)}x$