# How to prove the function $e^x$ is continous for all real numbers

We have been asked to prove the above statement through the two sub-proofs:

Firstly we must prove that $$e^x$$ is increasing and then prove that $$\log(x)$$ is also an increasing function.

After we are asked to prove that using the delta-epsilion definition of a limit that $$\lim_{x \to 0} e^x = 1$$.

I have done all of this but still don't see where the instructor is going with this.

• What's your definition of $e^x$? There are a couple of different (but equivalent) ones, and the proof is going to depend on which one you're working with. – anomaly Nov 11 '18 at 7:55
• Do you have already that $\exp (\log x) = \log (\exp x) = x$? If so, you can prove in general that an increasing surjective function is continuous. – Patrick Stevens Nov 11 '18 at 9:07

The function $$f$$ is continuous at some point $$c$$ of its domain if the limit of $$f(x)$$, as $$x$$ approaches $$c$$ through the domain of $$f$$, exists and is equal to $$f(c)$$. In mathematical notation, this is written as $$\lim_{x \to c}=f(c)$$
Back to your question, to prove that the limit of $$e^x$$ is $$e^{x_0}$$ as $$x$$ approaches $$x_0$$ with $$\varepsilon - \delta$$ definition
$$\begin{gather} |e^x-e^{x_0}|<\varepsilon \\ -\varepsilon < e^x - e^{x_0} < \varepsilon \\ e^{x_0} - \varepsilon < e^x < e^{x_0} + \varepsilon \\ \ln(e^{x_0} - \varepsilon) < x < \ln(e^{x_0} + \varepsilon) \quad (*)\\ \ln(e^{x_0} - \varepsilon) - x_0 < x - x_0 < \ln(e^{x_0} + \varepsilon) - x_0 \end{gather}$$ Hence there exist a $$\delta(\varepsilon)$$ s.t. $$\delta=\min \big\{x_0 - \ln(e^{x_0} - \varepsilon), \ln(e^{x_0} + \varepsilon) - x_0 \big\}$$ The point $$x_0$$ is arbitrary, so $$e^x$$ is continuous $$\forall \, x \in \mathbb{R}$$.
$$(*)\,\, f^{-1}(x)=\ln x$$ is an increasing function.