Just for fun, I decided I'd try to prove that if a function is continuous, then $\displaystyle \displaystyle \begin{align*} \lim_{x \to \alpha} f(x) = f(\alpha) \end{align*}$. I haven't cheated by looking for a proof on the web.

Anyway, my idea is to make use of the Mean Value Theorem, namely $\displaystyle \displaystyle \begin{align*} \frac{f(b) - f(a)}{b - a} = f'(c) \end{align*}$ for some $\displaystyle \displaystyle \begin{align*} c \in (a, b) \end{align*}$. It therefore follows that

$\displaystyle \displaystyle \begin{align*} f(b) - f(a) &= (b - a)f'(c) \\ |f(b) - f(a)| &= |(b - a)f'(c)| \\ |f(b) - f(a)| &= |b - a||f'(c)| \end{align*}$

Now, in order to prove $\displaystyle \displaystyle \begin{align*} \lim_{ x \to \alpha } f(x) = f(\alpha) \end{align*}$, we need to show that $\displaystyle \displaystyle \begin{align*} 0 < |x - \alpha| < \delta \implies |f(x) - f(\alpha)| < \epsilon \end{align*}$.

Now by the Mean Value Theorem we have

$\displaystyle \displaystyle \begin{align*} |f(x) - f(\alpha)| &< \epsilon \\ |x - \alpha| |f'(c)| &< \epsilon \textrm{ where } c \in (x, \alpha) \\ |x - \alpha| &< \frac{\epsilon}{|f'(c)|} \end{align*}$

So we can let $\displaystyle \displaystyle \begin{align*} \delta = \frac{\epsilon}{|f'(c)|} \end{align*}$.

Proof: Let $\displaystyle \displaystyle \begin{align*} \epsilon > 0 \end{align*}$ and $\displaystyle \displaystyle \begin{align*} \delta = \frac{\epsilon}{|f'(c)|} \end{align*}$, where $\displaystyle \displaystyle \begin{align*} c \in (x, \alpha) \end{align*}$. Then

$\displaystyle \displaystyle \begin{align*} 0 < |x - \alpha| &< \delta \\ |x - \alpha| &< \frac{\epsilon}{|f'(c)|} \\ |f'(c)||x - \alpha| &< \epsilon \\ |f(x) - f(\alpha)| &< \epsilon \textrm{ by the Mean Value Theorem} \end{align*}$

Q. E. D.

My only worry is the application of the Mean Value Theorem assumes evaluating derivatives, which assumes evaluating limits. Would we have already needed to prove the result that if a function is continuous at a point then the limit of that function is equal to the function value at that point in order to make use of the Mean Value Theorem?