Exponentials and Logarithms
The tricky part of proving $\displaystyle \lim_{x \rightarrow a} e^x = e^a$ is to avoid circularity. Many people try to use the natural log in the proof, but often implicitly assume it is continuous, which is what we are trying to prove in the first place.
I can’t take full credit for this proof. This is heavily adapted from a community effort in this StackExchange post.
Some Lemmas
In order to prove this result, we need two lemmas. Lemma 1 is $e^{x + y} = e^{x} \cdot e^{y}$. This is a fundamental property of exponentials, so I leave this to real analysis textbooks for its proof. Lemma 2 is $e^{x} \leq 1 + x$, which I prove below.
For this series, I will use the following definition for $e^x$.
\[e^x = \lim_{n \rightarrow \infty} \left ( 1 + \frac{x}{n} \right )^n\]I chose this definition since the proof is trivial using the power series definition, and I didn’t want to invoke all of that machinery.
\[\begin{align} e^x &= \lim_{n \rightarrow \infty} \left ( 1 + \frac{x}{n} \right )^n \\[10pt] &= \lim_{n \rightarrow \infty} \sum_{k=0}^n \binom{n}{k} \left ( \frac{x}{n} \right )^k \\[10pt] &\leq \lim_{n \rightarrow \infty} \sum_{k=0}^1 \binom{n}{k} \left ( \frac{x}{n} \right )^k \\[10pt] &= \lim_{n \rightarrow \infty} \binom{n}{0} \left ( \frac{x}{n} \right )^0 + \binom{n}{1} \left ( \frac{x}{n} \right )^1 \\[10pt] &= \lim_{n \rightarrow \infty} 1 \cdot 1 + n \cdot \left ( \frac{x}{n} \right ) \\[10pt] &= \lim_{n \rightarrow \infty} 1 + x \\[10pt] &= 1 + x \end{align}\]A corollary to this (which will be used in the proof) is the following
\[e^x \leq 1 + x \quad\implies\quad e^{-x} \leq 1 - x \quad\implies\quad e^{x} \leq \frac{1}{1 - x} \quad\implies\quad e^{x} - 1 \leq \frac{x}{1 - x}\]Finally, Lemma 3 is that if $\abs{h} < \frac{1}{2}$, then $\abs{\frac{h}{1-h}} < 2 \abs{h}$, which I leave as an exercise to the reader.
Exponentials
Fix any $\epsilon > 0$. Let $\delta = \frac{1}{2} \cdot \min(1, \frac{\epsilon}{e^a})$. Therefore, we have
\[\begin{align} &\exists \delta > 0 \quad\text{s.t.}\quad \lvert x - a \rvert < \delta \quad \implies \quad \lvert x - a \rvert < \delta \\[10pt] &\exists \delta > 0 \quad\text{s.t.}\quad \lvert x - a \rvert < \delta \quad \implies \quad \lvert x - a \rvert < \frac{1}{2} \cdot \min \left ( 1, \frac{\epsilon}{e^a} \right ) \\[10pt] &\exists \delta > 0 \quad\text{s.t.}\quad \lvert x - a \rvert < \delta \quad \implies \quad 2 \lvert x - a \rvert < \min \left ( 1, \frac{\epsilon}{e^a} \right ) \\[10pt] &\exists \delta > 0 \quad\text{s.t.}\quad \lvert x - a \rvert < \delta \quad \implies \quad \left \lvert \frac{(x-a)}{1 - (x-a)} \right \rvert < \min \left ( 1, \frac{\epsilon}{e^a} \right ) \qquad \text{(Lemma 3)} \\[10pt] &\exists \delta > 0 \quad\text{s.t.}\quad \lvert x - a \rvert < \delta \quad \implies \quad \left \lvert \frac{(x-a)}{1 - (x-a)} \right \rvert < \frac{\epsilon}{e^a} \\[10pt] &\exists \delta > 0 \quad\text{s.t.}\quad \lvert x - a \rvert < \delta \quad \implies \quad \lvert e^{x-a} - 1 \rvert < \frac{\epsilon}{e^a} \qquad \text{(Lemma 2)} \\[10pt] &\exists \delta > 0 \quad\text{s.t.}\quad \lvert x - a \rvert < \delta \quad \implies \quad e^a \cdot \lvert e^{x-a} - 1 \rvert < \epsilon \\[10pt] &\exists \delta > 0 \quad\text{s.t.}\quad \lvert x - a \rvert < \delta \quad \implies \quad \lvert e^x - e^a \rvert < \epsilon \qquad \text{(Lemma 1)} \end{align}\]which completes the proof. Therefore, $e^x$ is a continuous function.
Logarithms
Since $e^x$ is a continuous function, its inverse function $\ln x$ is also continuous. Therefore
\[\lim_{x \rightarrow a} \ln x = \ln a\]Furthermore, using the composition law, we have
\[\lim_{x \rightarrow a} \ln f(x) = \ln \left ( \lim_{x \rightarrow a} f(x) \right )\]General Exponentiation Law
The previous statement combined with the logarithm identity
\[\ln (a^b) = b \cdot \ln a\]is a very powerful tool.
\[\lim_{x \rightarrow a} f(x)^{g(x)} = \left ( \lim_{x \rightarrow a} f(x) \right )^{\displaystyle \left ( \lim_{x \rightarrow a} g(x) \right )}\]We have to be careful here that we don’t get an indeterminant form. For example,
\[\lim_{n \rightarrow \infty} \left ( 1 + \frac{x}{n} \right )^n = \left ( \lim_{n \rightarrow \infty} \left ( 1 + \frac{x}{n} \right ) \right )^{ \lim_{n \rightarrow \infty} n} = 1^{\infty}\]Many naively would think that the solution is $1^{\infty} = 1$, but when dealing with limits it does not. As we know, this limit actually equals $e^x$.