# Differentiability implies continuity with alternate form

I understand graphically how a differentiability implies continuity. However how can I prove it using this form of the derivative definition:

$$f'(a)=\lim_{h\to 0}\frac{f(a+h) - f(a)}{h}$$

• The definition assume that $f(a+h)\to f(a)$. If is not the case then the limit of the derivative in $a$ doesnt exist. Commented Jun 10, 2016 at 15:27
• $f(a+h)-f(a)=h\cdot\left(\frac{f(a+h)-f(a)}{h}\right)$. As $h\to0$, this expression tends to 0, so $f(a+h)$ tends to $f(a)$, which means $f$ is continuous at $x=a$. Commented Jun 10, 2016 at 15:34

If that ratio tends to a finite limit when the denominator tends to zero, that means that also the numerator tends to zero. And that means continuity.

• Nice.
– BCLC
Commented May 24 at 10:09

$$f'(x)=\lim_{x \rightarrow a} \frac{f(x) - f(a)}{x-a}$$ Then $$\frac{f(x) - f(a)}{x-a}=f'(x)+\alpha (x),$$ where if $x \rightarrow a$ then $\alpha (x) \rightarrow0$.

Then $$f(x) - f(a)=\left( f'(x)+\alpha (x)\right)(x-a)$$

Consequently $f(x) \rightarrow f(a)$ at $x \rightarrow a$

First notice that the derivative exists implies that the function $f$ is defined at $x=a$, and furthermore on some little interval $(a-\epsilon,a+\epsilon)$ where $\epsilon>0$. Now

$$f'(a):=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$$

by definition. In particular for all $h$ with $0<|h|<\epsilon$ we have the following by basic algebra:

$$f(a+h)-f(a)=h\cdot\frac{f(a+h)-f(a)}{h}$$

This noting that the limits both exist we have the following by basic limit properties:

\begin{align*} \lim_{h\to0}(f(a+h)-f(a)) &=\lim_{h\to0}\left(h\cdot\frac{f(a+h)-f(a)}{h}\right) \\ &=\left(\lim_{h\to0}h\right)\cdot\left(\lim_{h\to0}\frac{f(a+h)-f(a)}{h}\right) \\ &=0\cdot f'(a)=0 \end{align*}

Now I'll leave it to you to show the following straightforward result (if you haven't already):

Lemma: For a real function $f$ defined at $a$ the following are equivalent:

1) Function $f$ is continuous at $x=a$.

2) The limit $\lim_{x\to a}f(x)=f(a)$.

3) The limit $\lim_{h\to0}(f(a+h)-f(a))=0$.

Using this Lemma we have right away that $f$ is continuous (as desired).