Question

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

Answer 1

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: -
$f^{\prime }(c)=\underset{x\to c}{lim}{\displaystyle \frac{f(x)-f(c)}{x-c}}$ Exists
Continuous functions: - A function is said to be continuous at point $x=c$ , if function exists at that point and is given by: -
$\underset{x\to c}{lim}f(x)$ Exists
$\underset{x\to c}{lim}f(x)=f(c)$

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