In every proof I have seen of this, one manipulates the definition of continuity (a function $f$ is continuous at a point $a$ if the value of the limit of $f$ as $x$ approaches $a$ is $f(a)$) to prove the result.
I am wondering if somebody can help me reach this same conclusion by specifying a delta value using the full definition of limits.
For any $x\in\mathbb R$ there is some $L=f'(x)\in\mathbb R$ with
$$\forall \epsilon >0,\,\exists\delta>0,\,\forall h \text{ with } 0<|h|<\delta ,\, \left|\frac{f(x+h)-f(x)}{h}-L\right|<\epsilon$$
Now, the last inequality can be manipulated to yield
\begin{align} |f(x+h)-f(x)|-|Lh| &\leq\Big||f(x+h)-f(x)|-|Lh|\Big|&\\ &\leq|f(x+h)-f(x)-Lh|&<\epsilon|h| \end{align}
which implies that
$$|f(x+h)-f(x)|<|h|\,(\epsilon + |L|)<\delta\,(|L|+\epsilon)$$
We need to choose $\delta$ suitable so that $\delta\, (|L|+\epsilon)\leq \epsilon$. It suffices to take $\delta\leq \frac{\epsilon}{|L|+\epsilon}$, which concludes the proof. $\square$