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# Prove that if the function is differentiable at a point c, then it is continuous at that point.

Last updated date: 13th Jun 2024
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Hint: Here in this question firstly we should know the basic definition of differentiable and continuous functions which is mentioned below: -
Differentiable functions: - A function is said to be differentiable if it has derivatives there. The derivative of a function at a point x=c (c is a point in its domain) is given by: -
$f'(c) = \mathop {\lim }\limits_{x \to c} \dfrac{{f(x) - f(c)}}{{x - c}}$ Exists
Continuous functions: - A function is said to be continuous at point $x = c$ , if function exists at that point and is given by: -
$\mathop {\lim }\limits_{x \to c} f(x)$ Exists
$\mathop {\lim }\limits_{x \to c} f(x) = f(c)$

Complete step-by-step solution:
Let f(x) is function which is differentiable at point x=c, so according to differentiability definition $f'(c) = \mathop {\lim }\limits_{x \to c} \dfrac{{f(x) - f(c)}}{{x - c}}$ Exists.
Also when a function is continuous at point x=c then $\mathop {\lim }\limits_{x \to a} f(x) = f(c)$ exists.
We can also write equation of continuity as $\mathop {\lim }\limits_{x \to c} f(x) - f(c) = 0$ (rearranging the terms)
So to prove a function to be continuous we have to make $\mathop {\lim }\limits_{x \to c} f(x) - f(c) = 0$ equation satisfied.
Now we will multiply and divide $\mathop {\lim }\limits_{x \to c} (x - c)$in $\mathop {\lim }\limits_{x \to c} f(x) - f(c)$ so that we can prove a function continuous. Basically we are trying to obtain $\mathop {\lim }\limits_{x \to c} f(x) - f(c) = 0$ because it is the relation for continuous function, if this exists then only function is said to be continuous.
$\Rightarrow \mathop {\lim }\limits_{x \to c} f(x) - f(c) = \mathop {\lim }\limits_{x \to c} (f(x) - f(c))(\dfrac{{x - c}}{{x - c}})$
(Taking $\mathop {\lim }\limits_{x \to c} (x - c)$ in the denominator as well as in the multiplication) $\Rightarrow \mathop {\lim }\limits_{x \to c} f(x) - f(c) = \dfrac{{\mathop {\lim }\limits_{x \to c} (f(x) - f(c))}}{{\mathop {\lim }\limits_{x \to c} x - c}}\mathop {\lim }\limits_{x \to c} (x - c)$ ..........................equation(1)
(We know that $f'(c) = \mathop {\lim }\limits_{x \to c} \dfrac{{f(x) - f(c)}}{{x - c}}$ as function is differentiable given in question so we have used this relation in equation1)
$\Rightarrow \mathop {\lim }\limits_{x \to c} f(x) - f(c) = f('c)\mathop {\lim }\limits_{x \to c} (x - c)$
(Now we will put the limit in $\mathop {\lim }\limits_{x \to c} (x - c)$ which will result that term to zero)
$\Rightarrow \mathop {\lim }\limits_{x \to c} f(x) - f(c) = f('c)(0)$
$\therefore \mathop {\lim }\limits_{x \to c} f(x) - f(c) = (0)$
Hence $\mathop {\lim }\limits_{x \to c} f(x) - f(c) = (0)$ and by rearranging we also write this as $\mathop {\lim }\limits_{x \to c} f(x) = f(c)$ making function f(x) continuous at x=c.
Therefore we have used the function differentiability to prove its continuity. And hence if the function is differentiable at a point c, then it is continuous at that point.

Note: It should be noted that the converse is definitely not true for example, f(x) = |x| is continuous at x = 0, but not differentiable there. Solving limits can sometimes become confusing so make sure you are doing it cautiously. Also definition of differentiability and continuous function must be well known to a student.