Differentiability and continuity

An intuitive but extremely important fact underlying Mathematical Analysis is that differentiable functions are always continuous:

Let’s see the proof of this result together. Let’s first recall the mathematical definition of continuity: $f$ is said to be continuous at $x_0$ if

$\displaystyle \lim_{x \to x_0} f(x) = f(x_0).$

Proof.

Since $f$ is differentiable at $x_0$. by definition this means that the limit of the difference quotient of $f$ at $x_0$ exists finitely:

$\displaystyle \lim_{h\to 0}\frac{f(x_0 + h) - f(x_0)}{h}=f'(x_0).$

1. Let’s show that $\Delta f$ tends to 0.
   
   Recall the writing
   
   

   $\Delta f=f(x_0+h)-f(x_0).$

   
   To show that $\Delta f$ tends to 0 when $h$ tends to 0 we reason like this: we rewrite $\Delta f$ for $h\neq 0$ multiplying and dividing by $h$:
   
   

   $\Delta f= h \cdot \frac{f(x_0 + h) - f(x_0)}{h},$

   
   therefore the two terms will have the same limit:
   
   

   $\displaystyle \lim_{h\to 0}\Delta f= \displaystyle \lim_{h\to 0} h \cdot \frac{f(x_0 + h) - f(x_0)}{h}.$

   
   Since the limit of a product is the product of the limits when these exist finitely, the limit of the second term is
   
   

   $\displaystyle \lim_{h\to 0} h \cdot \frac{f(x_0 + h) - f(x_0)}{h}= \displaystyle \lim_{h\to 0} h \cdot \displaystyle \lim_{h\to 0} \frac{f(x_0 + h) - f(x_0)}{h}=0\cdot f'(x_0),$

   
   Since $f'(x_0)$ is finite, let $0\cdot f'(x_0)=0$, so the limit is zero.
   
   We have thus shown that if $f$ is differentiable at$x_0$ then
   $\displaystyle \lim_{h\to 0} \Delta f=0.$
   
   

   (1)

2. Let’s recognize continuity.
   
   Let’s write $f(x_0+h)$ by adding and subtracting $f(x_0)$:
   
   

   $f(x_0+h)=f(x_0)+[f(x_0+h)-f(x_0)]=f(x_0)+\Delta f.$

   
   Now let’s remember that the limit of a sum is the sum of the limits, when these exist and are finite: possiamo we can therefore move to the limit by exploiting (1):
   
   

   $\displaystyle \lim_{h \to 0} f(x_0 + h) = \displaystyle \lim_{h\to 0} f(x_0) + \displaystyle \lim_{h \to 0} \Delta f=f(x_0)+0.$

   
   We have thus obtained that
   
   

   $\displaystyle \lim_{h \to 0} f(x_0 + h) =f(x_0).$

   
   
   

3. Conclusion.
   
   Setting $x=x_0+h$ and observing that $x$ tends to $0$ if and only if h tends to $h$ tends to 0, the limit just obtained can be rewritten as
   
   

   $\displaystyle \lim_{x\to x_0} f(x) =f(x_0).$

   
   Remembering the definition of continuity, this shows that $f$ is continuous at $x_0$. $\checkmark$