On Proving that $e^x$ is continuous at $0$ utilizing a limit result.

Question

I was assigned the task to prove that $e^x$ is continuous in $x=0$ utilizing the fact that $$\lim_{x \rightarrow 0} \frac{e^x - 1} {x} = 1 $$

I think I am supposed to show that from the fact that for every $\epsilon >0$ I can choose a $\delta \in R$ s.t. $|x| < \delta \implies | \frac{e^x - 1} {x} - 1| < \epsilon $ that even $|e^x - 1|$ can be made less than $\epsilon$.

I wrote that $$ \frac{|e^x - 1| - |x|} {|x|} \le|\frac{e^x - 1} {x} - 1| < \epsilon$$

But I can't find the inequality I would like to be true, How could I do?

Answer 1

Given $\epsilon>0$, we are given that there eixts $\delta>0$ such that $0<|x|<\delta$ implies $\left|\frac{e^x-1}{x}-1\right|<\epsilon$. So $|e^x-(1+x)|<|x|\epsilon$ for such $x$. If we impose additionally that $|x|<\frac12$, this gives us $|e^x-(1+x)|<\frac\epsilon2$, so that if we (additionally again) impose the condition $|x|<\frac\epsilon2$ we obtain $$|e^x-1|\le |e^x-1|+|x|<\frac\epsilon2+|x|<\epsilon $$ As $e^0=1$ it follows that $x\mapsto e^x$ is continuous at $0$.

---

Actually, if we consider $f(x)=e^x$, note that that we want to show that $f$ is continuos at a point where we are given that $f'$ exists. This is always the case.

Answer 2

You know that $e^0=1$?

Using your notation, it follows that $$ |e^x-1|<(1+ϵ)·|x| $$ which gives you local Lipschitz continuity.