Help with my proof that $\lim_{x\to\infty}(1+e^{-x})^{x^2}=1$

Question

$ (1+e^{-x})^{x^2} = \Big(\big(1+e^{-x}\big)^{e^{x}}\Big)^{\frac{x^2}{e^x}} $
Now I want to say, we know that $f(x)=(1+1/x)^x\xrightarrow[x\to\infty]{}e$, so we look at $e^x \xrightarrow[x\to\infty]{}\infty$ and by Heine we get $f(e^x)\xrightarrow[x\to\infty]{}e$. But how do I denote "$e^x$ is a sequence real numbers" in terms of
$x_n=e^n$ when $n\in\mathbb{R}$?

Answer 1

I have the following interpretation of the question. Define $f(x)=(1+\frac 1 x)^x$. You seem to know that $\lim_{x\to\infty}f(x) = e$ and you want to prove that $\lim_{x\to\infty} f(e^x) = e$ using: for every sequence $x_n\to\infty$ one has $\lim_{n\to\infty} f(x_n) = e$.

Your problem is now: what sequence should I choose to prove the claim with $f(e^x)$?

One can do it with sequences, but then one needs to do it in two steps. Let $y_n\to\infty$ be an arbitrary sequence. Choose $x_n = e^{y_n}$ and notice that it also tends to infinity as $n\to\infty$. One has $\lim_{n\to\infty} f(x_n)=e$ , or in other words: $\lim_{n\to\infty} f(e^{y_n}) = e$. Since $y_n$ was arbitrary we have proven that $\lim_{x\to\infty} f(e^x) = e$.

Answer 2

Yes we have that

$$\large{\left(1+\frac1{e^{x}}\right)^{x^2}=\left[\left(1+\frac1{e^{x}}\right)^{e^x}\right]^\frac{x^2}{e^x}}\to e^0=1$$

indeed since eventually $e^x>x^3$ we have

$$\frac{x^2}{e^x}<\frac{x^2}{x^3}=\frac1x\to 0$$