Show by series definition of exponential function that $\exp(-x) \rightarrow 0 $ for $x \rightarrow \infty.$

Question

There are many arguments I have seen using $\ln-$ arguments and other properties of the exponential function to show the existence of this limit $\exp(-x) \rightarrow 0 $ for $x \rightarrow \infty$. But does somebody have a good idea how to show the existence of this limit just by having the pure series definition of the exponential function?

Answer 1

We have for $x > 0$, $\exp(x) > 1 +x$ and $0 \leqslant \exp(-x) < (1+x)^{-1}$, since $\exp(x)\exp(-x)=1$.

Hence

$$\lim_{x \to \infty}\exp(-x) = 0.$$

All of this follows from the power series representation which converges for all $x \in \mathbb{R}$:

$$\exp(x) = \sum_{j=0}^\infty\frac{x^j}{j!}.$$

By Mertens', theorem the Cauchy product of the series for $\exp(x)$ and $\exp(y)$ converges to the product of the sums

$$\exp(x) \exp(y) = \sum_{j=0}^\infty\frac{x^j}{j!}\sum_{k=0}^\infty\frac{y^k}{k!} = \sum_{j=0}^\infty \sum_{k=0}^j \frac{x^{j-k}}{(j-k)!} \frac{y^{k}}{k!} \\ = \sum_{j=0}^\infty \frac{1}{j!}\sum_{k=0}^j \frac{j!}{(j-k)!k!} x^{j-k}y^k .$$

By the binomial theorem

$$\sum_{k=0}^j \frac{j!}{(j-k)!k!} x^{j-k}y^k = (x+y)^j.$$

Hence, $\exp(x)\exp(y) = \exp(x+y).$ With $y = -x$ we have $\exp(x)\exp(-x) = \exp(0) = 1.$