Prove that $\lim_{x\to a}e^x=e^a$

Question

Prove by epsilon-delta limit definition that: $\lim_{x\to a}e^x=e^a$

My definition of exponential function is $e^x=\lim_{n\to \infty}(1+x/n)^n$. My teacher said that we need to use it but when I use the epsilon-delta definition I have that $$|\lim_{n\to \infty}(1+x/n)^n-\lim_{n\to \infty}(1+a/n)^n|<\epsilon$$ then I don't know how to proceed from here

Answer 1

$|e^x - e^a| = e^a|e^{x - a} - 1|$, then it is enough to show that $\lim_{x \to 0}e^x = 1$.

Use that $$a^n - b^n = (a - b)(a^{n - 1} + ba^{n - 2} + \dots + b^{n - 1})$$ to rewrite the limit as follows: $$\lim_{x \to 0}\big|\lim_{n\to \infty}(1 + xn^{-1})^n - 1\big| = \lim_{x \to 0}\big|\lim_{n \to \infty}(1 + xn^{-1} - 1)((1 + xn^{-1})^{n - 1} + \dots)\big|$$ Now, without loss of generality, assume that $x \le 1$. Noticing that in the second expression between parenthesis we have $n$ terms, all of which are smaller than $e$, we get: $$\lim_{x \to 0}\big|\lim_{n \to \infty}(1 + xn^{-1} - 1)((1 + xn^{-1})^{n - 1} + \dots)\big| \le \lim_{x \to 0}\lim_{n \to \infty}\frac{x}{n}\cdot ne = \lim_{x \to 0}xe = 0$$

Answer 2

One of my math mantras is "always expand around zero".

In this case, that means we want to show that $\lim_{x \to 0} e^{x+a} = e^a $.

But $e^{x+a}-e^a =e^a(e^x-1) =e^a(e^x-e^0) $.

Suppose we can show that $e^x$ is continuous at $0$. This means that, for any $\delta > 0$ we can find an $\epsilon(\delta) > 0$ such that $|x| < \epsilon(\delta)$ implies that $|e^x-1| < \delta$. I'll leave this as an exercise.

To then show that $e^x$ is continuous at $a$, we need to show that, for any $\delta > 0$ we can find an $\epsilon > 0$ such that $|x| < \epsilon$ implies that $|e^{x+a}-e^a| < \delta$. But this is the same as $|e^{x}-1| < \delta e^{-a}$.

Under the assumption that $e^x$ is continuous at zero, consider $\epsilon(\delta e^{-a})$. By the definition of $\epsilon(\delta)$, if $|x| < \epsilon(\delta e^{-a})$ then $|e^x-1| < \delta e^{-a} $, which is the same as $|e^{x+a}-e^x| < \delta $.

Therefore, $e^x$ continuous at zero implies that $e^x$ is continuous everywhere.

A hint for showing $e^x$ is continuous at zero: $e^x-1 =\int_0^x e^t\, dt $.