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# Show that every differentiable function is continuous (converse is not true i.e., a function may be continuous but not differentiable).

Last updated date: 13th Jun 2024
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Hint: Here we will use the basic definition of the differential function. Then we will form the condition of a continuous function. We will then show that the function is continuous to prove that every differentiable function is a continuous function.

Let $f$ be the differentiable function at $x = a$.
Then according to the basic definition of the differentiation, Differentiation of a function is equals to
$f'\left( c \right) = \mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right) - f\left( a \right)}}{{x - a}}$……………………….$\left( 1 \right)$
We know that the for a function to be continuous at a point it must satisfy the equation
$\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right)$
We can write the above equation as
$\Rightarrow \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = 0$……………………$\left( 2 \right)$
So, for a function to be continuous it must satisfy the equation $\left( 2 \right)$.
Now we will find the value of $\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right)$ for the given differentiable function.
Therefore we can write $\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right)$ as,
$\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = \mathop {\lim }\limits_{x \to a} \left( {\dfrac{{f\left( x \right) - f\left( a \right)}}{{x - a}}\left( {x - a} \right)} \right)$
$\Rightarrow \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = \mathop {\lim }\limits_{x \to a} \left( {\dfrac{{f\left( x \right) - f\left( a \right)}}{{x - a}}} \right) \times \mathop {\lim }\limits_{x \to a} \left( {x - a} \right)$
By using the equation $\left( 1 \right)$ in the above equation, we get
$\Rightarrow \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = f'\left( c \right) \times \mathop {\lim }\limits_{x \to a} \left( {x - a} \right)$
By putting the limit on the RHS of the equation, we get
$\Rightarrow \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = f'\left( c \right) \times \left( {a - a} \right)$
$\Rightarrow \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = f'\left( c \right) \times 0$
$\Rightarrow \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = 0$
Hence as per the condition of the equation $\left( 2 \right)$ we can say that the given function $f$ is a continuous function.
Hence, every differentiable function is continuous.

Note: Here we have to note that continuous function is the function whose value does not change or value remains constant. When the function is continuous at a point then the left hand limit of the function and the right hand limit of the function is equal to the value of the function at that point.
$\mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{x \to {a^ + }} f(x) = f(a)$
Also, a differentiable function is always continuous but the converse is not true which means a function may be continuous but not always differentiable. A differentiable function may be defined as is a function whose derivative exists at every point in its range of domain.