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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.

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