## Derivative implies Continuity but Converse NOT True:

In this section, we will prove that if a function is differentiable at a point, then the function is continuous at that point. Its converse statement is the following: if a function is continuous, then it is not necessarily differentiable. We prove the converse statement by providing examples.

**If $f(x)$ is differentiable at $x=a$, then it is continuous at $x=a$.**

__Theorem 1:____Proof:__Given that $f(x)$ is differentiable at $x=a$.

So the derivative of $f(x)$ at $x=a$ is finite.

In other words,

$f'(a)=\lim\limits_{h \to 0} \dfrac{f(a+h)-f(a)}{h}$ exists.

Now, we have

$\lim\limits_{h \to 0} [f(a+h)-f(a)]$

$=\lim\limits_{h \to 0} [\dfrac{f(a+h)-f(a)}{h} \times h]$

$=\lim\limits_{h \to 0} \dfrac{f(a+h)-f(a)}{h} \times \lim\limits_{h \to 0} h$

$=f'(a) \times 0$

$=0$.

Thus, we obtain that

$\lim\limits_{h \to 0} [f(a+h)-f(a)]=0$

$\Rightarrow \lim\limits_{h \to 0} f(a+h) – \lim\limits_{h \to 0} f(a)=0$

$\Rightarrow \lim\limits_{h \to 0} f(a+h) – f(a)=0$

$\Rightarrow \lim\limits_{h \to 0} f(a+h)=f(a)$

Therefore, $f(x)$ is continous at $x=a$.

**Supporting Example:**

Let $f(x)=x$.

We will show that $f(x)$is continuous and differentiable at $x=2$.

__Continuity Checking:__Note that $f(2)=2$

We have $\lim\limits_{x \to 2-} f(x)=\lim\limits_{x \to 2-} x=2$

Again, $\lim\limits_{x \to 2+} f(x) = \lim\limits_{x \to 2+} x=2$.

Thus $\lim\limits_{x \to 2-} f(x)=\lim\limits_{x \to 2+} f(x)=f(2)$

$\therefore f(x)$ is continuous at $x=2$.

__Differentiability Checking:__

By the definition of derivative, we have

$f'(2)=\lim\limits_{h \to 0} \dfrac{f(2+h)-f(2)}{h}$

$=\lim\limits_{h \to 0} \dfrac{2+h-2}{h}$

$=\lim\limits_{h \to 0} \dfrac{h}{h}$

$=\lim\limits_{h \to 0} \dfrac{1}{1}$

$=1$.

Thus, $f(x)=x$ is differentiable at $x=2$.

As $f(x)$ is continuous and differentiable at $x=2$, this example supports the above Theorem 1.

**But the converse is NOT True.**

Let $f(x)=|x|$ (the absolute value of $x$).

The function $f(x)=|x|$ is continuous

at $x=0$ but Not differentiable at $x=0$.