Prove that $\lim_{x\to 0-}e^x=1,$ using $\epsilon-\delta$ definition.

Question

Prove that $\lim_{x\to 0-}e^x=1,$ using $\epsilon-\delta$ definition.

This means we need to show that, $\forall\epsilon\gt 0,\exists \delta \gt 0$ such that, if $0\lt -x\lt\delta$ then, $|e^x-1|\lt\epsilon.$

Now, $|e^x-1|\lt \epsilon\iff 1-\epsilon\lt e^x\lt 1+\epsilon.$ But $e^x\gt 1-\epsilon$ is always true. So, the task is to find a $\delta $ such that if $0\lt -x\lt\delta\iff -\delta\lt x\lt 0$ then, $e^x\lt 1+\epsilon.$

This is the part, where I am stuck. I don't know how to proceed further. If we take log on both sides of the inequality, $e^x\lt 1+\epsilon,$ then we get, $x\lt\ln (1+\epsilon),$ but this is not what we desire.

Answer 1

For $\forall \epsilon>0$ sufficiently small, there exists $\delta = -\ln(1-\epsilon)>0$ such that for $-\delta<x<0$ we have $$|e^x-1|<\epsilon.$$