Differentiability implies Continuity but Converse not True

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.
Theorem 1: If $f(x)$ is differentiable at $x=a$, then it is continuous at $x=a$.
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
 $\lim\limits_{h \to 0} [f(a+h)-f(a)]$
$=\lim\limits_{h \to 0} [\dfrac{f(a+h)-f(a)}{h} \times h]$
$=\lim\limits_{h \to 0} \dfrac{f(a+h)-f(a)}{h} \times \lim\limits_{h \to 0} h$
$=f'(a) \times 0$
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
$f'(2)=\lim\limits_{h \to 0} \dfrac{f(2+h)-f(2)}{h}$
$=\lim\limits_{h \to 0} \dfrac{2+h-2}{h}$
$=\lim\limits_{h \to 0} \dfrac{h}{h}$
$=\lim\limits_{h \to 0} \dfrac{1}{1}$
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 $x=0$.
Spread the love

Leave a Comment