Take the function $f(x) = e^{x}$ and I want to prove this is continuous at $x=0$ using epsilon-delta definition.
ie. $\forall \epsilon>0, \exists\delta>0 \text{ s.t. } |x - 0| < \delta \implies |e^{x}-1| < \epsilon$
So $|e^{x}-1| < |e^{x}| + |1| = e^{x}+1$
Now $|x - 0| < \delta$ or $|x|<\delta$ so $x<\delta$
Hence $e^{x}+1 < e^{\delta}+1$
Take $\epsilon = e^{\delta}+1$ hence $\delta = \ln{(\epsilon -1)}$. But this doesn't show it because if we choose an $\epsilon \le 2$ we can't find a $\delta > 0$ such that the implication holds.
You could use the following $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots$$
Therefore,
$$|e^x -1| = \left|x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots\right|$$
Taking $|x-0| < \delta$, where $0<\delta<1$ and $\frac{\delta}{1-\delta}<\epsilon$, we have that $$|e^x -1| = \left|x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \right| < \delta + \frac{\delta^2}{2!} + \frac{\delta^3}{3!} + \ldots<$$ $$< \delta + \delta^2 + \delta^3 + \ldots = \frac{\delta}{1-\delta} < \epsilon.$$
PS: if you don't know this definition for $e^x$, disregard my answer.