# If $f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}

{\dfrac{{{x^2} - 1}}{{x - 1}};{\text{ for }}x \ne 1} \\

{2;{\text{ for }}x = 1}

\end{array}} \right.$. Find whether $f\left( x \right)$ continuous at $x = 1$.

Last updated date: 18th Mar 2023

•

Total views: 306.9k

•

Views today: 2.85k

Answer

Verified

306.9k+ views

Hint- Calculate LHL and RHL of the given function.

LHL $\mathop { = \lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} f\left( {1 - h} \right)$, RHL$ = \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} f\left( {1 + h} \right)$

Given function

$f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}

{\dfrac{{{x^2} - 1}}{{x - 1}};{\text{ for }}x \ne 1} \\

{2;{\text{ for }}x = 1}

\end{array}} \right.$

We have to check its continuity at $x = 1$.

So, consider LHL

$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right)$where ‘-’ sign indicates LHL.

Now convert the limit into $h \to 0$ by substituting $\left( {1 - h} \right)$ in place of x.

For LHL the function is left sided to 1. i.e. for $x \ne 1$.

LHL$ = \mathop {\lim }\limits_{h \to 0} f\left( {1 - h} \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\left( {1 - h} \right)}^2} - 1}}{{\left( {1 - h} \right) - 1}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{1 + {h^2} - 2h - 1}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \left( {\dfrac{{h - 2}}{{ - 1}}} \right)$

Now substitute h = 0, we have

LHL $ = \dfrac{{0 - 2}}{{ - 1}} = 2$

Now consider RHL

$\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)$where ‘+’ sign indicates RHL.

Now convert the limit into $h \to 0$ by substituting $\left( {1 + h} \right)$ in place of x.

For RHL the function is right sided to 1. i.e. for $x \ne 1$.

RHL$ = \mathop {\lim }\limits_{h \to 0} f\left( {1 + h} \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\left( {1 + h} \right)}^2} - 1}}{{\left( {1 + h} \right) - 1}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{1 + {h^2} + 2h - 1}}{h} = \mathop {\lim }\limits_{h \to 0} \left( {\dfrac{{h + 2}}{1}} \right)$

Now substitute h = 0, we have

RHL $ = \dfrac{{0 + 2}}{1} = 2$.

Also $f\left( 1 \right) = 2$ (given).

Now, since

$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} f\left( x \right) = 2$

Hence $f\left( x \right)$ is continuous at $x = 1$.

Note- In such types of questions the key concept we have to remember is that if the left hand limit, right hand limit and the value of function at a given point are equal then the function is continuous at a given point. So calculate LHL, RHL at a given point and check whether they are equal and also check they are equal to the value of the function at that point if yes then the function is continuous if not then the function is not continuous.

LHL $\mathop { = \lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} f\left( {1 - h} \right)$, RHL$ = \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} f\left( {1 + h} \right)$

Given function

$f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}

{\dfrac{{{x^2} - 1}}{{x - 1}};{\text{ for }}x \ne 1} \\

{2;{\text{ for }}x = 1}

\end{array}} \right.$

We have to check its continuity at $x = 1$.

So, consider LHL

$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right)$where ‘-’ sign indicates LHL.

Now convert the limit into $h \to 0$ by substituting $\left( {1 - h} \right)$ in place of x.

For LHL the function is left sided to 1. i.e. for $x \ne 1$.

LHL$ = \mathop {\lim }\limits_{h \to 0} f\left( {1 - h} \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\left( {1 - h} \right)}^2} - 1}}{{\left( {1 - h} \right) - 1}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{1 + {h^2} - 2h - 1}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \left( {\dfrac{{h - 2}}{{ - 1}}} \right)$

Now substitute h = 0, we have

LHL $ = \dfrac{{0 - 2}}{{ - 1}} = 2$

Now consider RHL

$\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)$where ‘+’ sign indicates RHL.

Now convert the limit into $h \to 0$ by substituting $\left( {1 + h} \right)$ in place of x.

For RHL the function is right sided to 1. i.e. for $x \ne 1$.

RHL$ = \mathop {\lim }\limits_{h \to 0} f\left( {1 + h} \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\left( {1 + h} \right)}^2} - 1}}{{\left( {1 + h} \right) - 1}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{1 + {h^2} + 2h - 1}}{h} = \mathop {\lim }\limits_{h \to 0} \left( {\dfrac{{h + 2}}{1}} \right)$

Now substitute h = 0, we have

RHL $ = \dfrac{{0 + 2}}{1} = 2$.

Also $f\left( 1 \right) = 2$ (given).

Now, since

$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} f\left( x \right) = 2$

Hence $f\left( x \right)$ is continuous at $x = 1$.

Note- In such types of questions the key concept we have to remember is that if the left hand limit, right hand limit and the value of function at a given point are equal then the function is continuous at a given point. So calculate LHL, RHL at a given point and check whether they are equal and also check they are equal to the value of the function at that point if yes then the function is continuous if not then the function is not continuous.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE