Question

Is function $f\left( x \right)=\left\{ \begin{align}
  & 1+{{x}^{2}}\, if\,0\le x\le 1 \\
 & 2-x\, if\,x>1 \\
\end{align} \right\}$ is discontinuous at $x=1$

Answer 1

Hint: Here we have to find whether the function given is discontinuous at the point given. Firstly, as we know, a function is continuous only if its Left-Hand limit is equal to the Right-hand limit at the point given. So we will start by finding the both limits at the point given and check whether they are equal or not. If they are equal, that means the function is continuous, otherwise the function is not continuous.

Complete step-by-step solution:
The function is given as below,
$f\left( x \right)=\left\{ \begin{align}
  & 1+{{x}^{2}}\,\,,if\,0\le x\le 1 \\
 & 2-x\,\,\,,if\,x>1 \\
\end{align} \right\}$
We have to find whether it is continuous at $x=1$ or not.
For the function to be continuous, LHL should be equal to RHL.
Formula for LHL and RHL is as follows,
LHL at $x=1$ is,
$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,f\left( 1-h \right)$….$\left( 1 \right)$
RHL at $x=1$ is,
$\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,f\left( 1+h \right)$…..$\left( 2 \right)$
So the function value for LHL is $f\left( x \right)=1+{{x}^{2}}$ substitute it in equation (1) as follows,
$\underset{h\to 0}{\mathop{\lim }}\,f\left( 1-h \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( 1+{{\left( 1-h \right)}^{2}} \right)$
$\Rightarrow \underset{h\to 0}{\mathop{\lim }}\,f\left( 1-h \right)=\underset{h\to 0}{\mathop{\lim }}\,1+1+{{h}^{2}}-2h$
Put the limit,
$\underset{h\to 0}{\mathop{\lim }}\,f\left( 1-h \right)=1+1+0-0$
$\Rightarrow \underset{h\to 0}{\mathop{\lim }}\,f\left( 1-h \right)=2$
So we got the LHL for $2$ .
Next the function for RHL is $f\left( x \right)=2-x$ substitute it in equation (2) as follows,
$\underset{h\to 0}{\mathop{\lim }}\,f\left( 1+h \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( 2-\left( 1+h \right) \right)$
$\Rightarrow \underset{h\to 0}{\mathop{\lim }}\,f\left( 1+h \right)=\underset{h\to 0}{\mathop{\lim }}\,1+h$
Put the limit,
$\underset{h\to 0}{\mathop{\lim }}\,f\left( 1+h \right)=1+0$
$\Rightarrow \underset{h\to 0}{\mathop{\lim }}\,f\left( 1+h \right)=1$
So we get the RHL as $1$
As we can see that LHL $\ne $ RHL
Hence the function is discontinuous at $x=1$

Note: A function is said to be continuous if a small change in the output can be assured by restricting a small change in the input of the function. If a function is not continuous it is known as discontinuous function. In simple words we can say that a function is continuous if we can draw its curved surface on a graph without lifting our pen even once. If the function is continuous, its LHL (Left hand limit) and RHL (Right hand limit) are always the same on the given point and the value of function exists and is equal to them.