Continuity proof for exponential

Question

Show that $f(x) = e^x$ is continuous using the epsilon-delta definition.

I can't seem to write down anything meaningful...

Answer 1

Let $a$ be a positive real number. Then the function $f: \mathbb{R}\to \mathbb{R}$ defined by $x\mapsto a^x$ is continuous.

Proof:

1) First we prove the continuous at $0$.

We may assume that $a>1$. Let $\varepsilon>0$ be arbitrary. Since $a^{1/k}\to 1$ and $a^{-1/k}\to 1$ as $k\to \infty$, we choose $K$ such that both are $\varepsilon$-close to $1$. Let $\delta=1/K$, so for $|x|<1/K$ we have

\begin{align}-1/K<x<1/K\\a^{-1/K}<a^{x}<a^{1/K} \end{align}

which proves that $a^{x}$ is $\varepsilon$-close to $1$ as desired (the case $a\le1$ is handled similarly just with the inequality in the other direction).

2) To conclude we prove the continuity in the general case.

Let $x_0\in \mathbb{R}$, we have to show that $\lim_{x\to x_0}a^x=a^{x_0}$. Since $x-x_0\to 0$ as $x\to x_0$, then $a^{x-x_0}\to1$ by ($1$) and thus

$$\lim_{x\to x_0}a^x=\lim_{x\to x_0}a^{x_0}a^{x-x_0}=a^{x_0}\lim_{x\to x_0}a^{x-x_0}=a^{x_0}$$

as was to be shown.

Since $e\in \mathbb{R}^{>0}$, thus $e^x$ is continuous.

Answer 2

Start with $|x-c|<\delta$.

Let's take a look at the case that $x\ge c$. Then: $$x-c <\delta$$ $$x < c+ \delta$$ $$e^x < e^{c+\delta}$$ $$0 \le e^x-e^c < e^c(e^\delta - 1)$$ Let's pick $\delta$ such that $\varepsilon = e^c(e^\delta - 1)$. Then: $$\delta = \ln(1+\varepsilon e^{-c})$$ Repeat for $x<c$.

In other words, you are able to find a $\delta$ for any $\varepsilon>0$.

Qed.

Answer 3

Almost identical questions have been answered $n + 1$ times on this site. The proof is trivial. I will present a slight variation of the standard proof.

Recall that, by definition, $f(x)$ is continuous on $\mathbb{R}$ if and only if $$(\forall x_0 \in \mathbb{R})(\forall \varepsilon > 0)(\exists \delta > 0)(\forall x \in \mathbb{R})(0 < |x - x_0| < \delta \implies |f(x) - f(x_0)| < \varepsilon).$$

Combining a well-worn result $$e^x = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^n,$$ and the fact that, by definition, $\lim_{x \to x_0} f(x) = L$ if and only if $$(\forall \varepsilon > 0)(\exists \delta > 0)(0 < |x - x_0| < \delta \implies |f(x) - L| < \varepsilon),$$ proves that $e^x$ is continuous.

Edit: To summarize, we know that the sequence $$s_n = \left(1 + \frac{x}{n}\right)^n$$ converges to $e^x$ as $n \to \infty$. We see that $$\lim_{x \to x_0} \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n = \lim_{x \to x_0} e^x = e^{x_0} = \lim_{n \to \infty} \left(1 + \frac{x_0}{n}\right)^n = \lim_{n \to \infty} \lim_{x \to x_0} \left(1 + \frac{x}{n}\right)^n.$$ Moreover, by definition, $\lim_{x \to x_0} f(x) = L$ if and only if $$(\forall \varepsilon > 0)(\exists \delta > 0)(0 < |x - x_0| < \delta \implies |f(x) - L| < \varepsilon).$$ Yet, since $x_0$ is arbitrary, we have $$(\forall x_0 \in \mathbb{R})(\forall \varepsilon > 0)(\exists \delta > 0)(\forall x \in \mathbb{R})(0 < |x - x_0| < \delta \implies |f(x) - f(x_0)| < \varepsilon).$$ Thus, $e^x$ is continuous for all $x$.

Answer 4

As discussed in this forum, it suffices to show continuity at zero. Using the fact that $E(0)=1,$ we see that for any $\delta<1,$ every $x\in\mathbb{R}$ (or $x\in\mathbb{C}$) such that $|x|<\delta$ satisfies \begin{align*} \left|E(x) - E(0)\right| \hspace{1mm}=\hspace{1mm} \left|\sum_{n=1}^\infty \frac{x^n}{n!}\right| \hspace{1mm}\leq\hspace{1mm} \sum_{n=1}^\infty \frac{|x|^n}{n!} \hspace{1mm}<\hspace{1mm} \sum_{n=1}^\infty \frac{\delta^n}{n!} \hspace{1mm}<\hspace{1mm} \sum_{n=1}^\infty \delta^n \hspace{1mm}=\hspace{0.5mm} \frac{1}{1-\delta} - 1, \end{align*} at which point the reader is invited to choose $\delta$ in terms of a given $\varepsilon$.