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.

  • 2
    $\begingroup$ 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. $\endgroup$ – anomaly Nov 11 '18 at 7:55
  • $\begingroup$ 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. $\endgroup$ – 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.


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