Proving that $\lim_{x\rightarrow a}f(x) = \lim_{h\rightarrow 0} f(a+h) $ The Question: 

Verify that if $\lim_{x\to a}f(x)$ exists, then $\lim_{x\to a}f(x)=\lim_{h\to 0}f(a+h)$.

My plan is to use the theorem showing that the limit of a sum is the sum of the limits.  By that theorem, $\lim_{h\to 0}f(a+h)=\lim_{h\to 0}f(a)+\lim_{h\to 0}f(h)=f(a)+0=f(a)=\lim_{x\to a}f(x)$.

Does this proof hold?  Please let me know if there is anything that I need to clarify.
The limit $\lim_{x\to a}f(x)$ exists and is equal to $b$: $$\lim_{x\to a}f(x)=b$$ if and only if for each $\varepsilon\in\mathbb R^+$ exists $\delta\in\mathbb R^+$ such that $$\forall x,0&lt;|x-a|&lt;\delta\implies|f(x)-b|&lt;\varepsilon.$$

Hint:
Apply analogous definition for $\lim_{h\to0}f(a+h)$ and tell if you can derive one from the other.
Well first we have to prove that differentiability implies continuity. (We have to prove that $\lim_{x\to a}f(x)$ exists.)

Thus, suppose we have a differentiable function $f(x)$ such that $\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$. This equals $f'(a)$. 

Now, we multiply each side by $h$. Thus, we get $\lim_{h\to 0}hf'(a)$. (What is the answer here?)

Thus, we get $\lim_{h\to 0}f(a+h) \to f(a)$. (Why?)

Thus, we set $x = a + h$. Thus, $\lim_{x-a \to 0}f(x) \to f(a)$.

Thus, we have proven that such a $\lim_{x\to a}f(x) = f(a)$ exists.

The answer to your question:


   To answer your question, go backwards!