# Let \[f(x)=\left\{ \begin{matrix}

\dfrac{x}{\sin x},x>0 \\

2-x,x\le 0 \\

\end{matrix} \right.\] and $g(x)=\left\{ \begin{matrix}

x+3,x<1 \\

{{x}^{2}}-2x-2,1\le x<2 \\

x-5,x\ge 2 \\

\end{matrix} \right.$. Find LHL and RHL of g(f(x)) at x=0 and hence find \[\underset{x\to 0}{\mathop{\lim }}\,g(f(x)).\].

Answer

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Hint: Find Left hand limit and Right Hand Limit of f(x) as well as g(f(x)). For finding out whether the limit exists, we should find the left hand limit and right hand limit. If they are equal then the limit exists.

In the question the functions given are:

\[f(x)=\left\{ \begin{matrix}

\dfrac{x}{\sin x},x>0 \\

2-x,x\le 0 \\

\end{matrix} \right.\]

$g(x)=\left\{ \begin{matrix}

x+3,x<1 \\

{{x}^{2}}-2x-2,1\le x<2 \\

x-5,x\ge 2 \\

\end{matrix} \right.$

Here we have to find the left hand limit.

For finding the left hand limit, we have to consider $x\to {{0}^{-}}$.

So, when x tends to ‘0’ from left hand side then,

\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,2-x\]

Applying the limit, we get

$\Rightarrow $\[f\left( {{0}^{-}} \right)=2\]

So, now considering the function g(f(x), here g(f(x) becomes g(2).

So, here we have to consider the function g(x) then x tends to${{2}^{+}}$ because the value of ‘x’ is greater than ‘2’.

Then,

\[\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,g\left( x \right)=\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,x-5\]

Applying the limit, we get

\[g\left( {{2}^{+}} \right)=-3\]

Then the left hand limit is ‘-3’ at x=0.

So the left hand limit of g(f(x)) is ‘-3’ at x=0.

For finding the right hand limit, we have to consider $x\to {{0}^{+}}$.

So referring to the given function, when x tends to ‘0’ from right hand side then,

\[\begin{align}

& \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{x}{\sin x} \\

& \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{1}{\dfrac{\sin x}{x}} \\

\end{align}\]

But we know $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1$ , so the above equation becomes,

\[\Rightarrow f({{0}^{+}})=1\]

So, now considering the function g(f(x), here g(f(x) becomes g(1).

So, here we have to consider the function g(x) then x tends to${{1}^{+}}$ because the value of ‘x’ is greater than ‘1’.

Then,

\[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,g\left( x \right)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,{{x}^{2}}-2x-2\]

Applying the limit, we get

\[g\left( {{1}^{+}} \right)=1-2-2=-3\]

Then the right hand limit is ‘-3’ at x=0.

So the right hand limit of g(f(x)) is ‘-3’ at x=0.

Hence LHL=RHL.

Hence, the correct answer is \[\underset{x\to 0}{\mathop{\lim }}\,g(f(x))=-3\]

Note: A possible mistake is when finding \[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,g\left( x \right)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,{{x}^{2}}-2x-2\], then student think this is the right hand limit of g(x) at x=1.

Students get confused when finding the left hand limit of g(f(x)).

When they find f(x) at x=0, they take the limit of g(x), i.e., \[\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,g\left( x \right)\]as ${{2}^{-1}}$ thinking this is left hand limit. But this will lead to the wrong answer.

In these types of questions students have confusions while finding the left hand and right hand limit so be careful.

In the question the functions given are:

\[f(x)=\left\{ \begin{matrix}

\dfrac{x}{\sin x},x>0 \\

2-x,x\le 0 \\

\end{matrix} \right.\]

$g(x)=\left\{ \begin{matrix}

x+3,x<1 \\

{{x}^{2}}-2x-2,1\le x<2 \\

x-5,x\ge 2 \\

\end{matrix} \right.$

Here we have to find the left hand limit.

For finding the left hand limit, we have to consider $x\to {{0}^{-}}$.

So, when x tends to ‘0’ from left hand side then,

\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,2-x\]

Applying the limit, we get

$\Rightarrow $\[f\left( {{0}^{-}} \right)=2\]

So, now considering the function g(f(x), here g(f(x) becomes g(2).

So, here we have to consider the function g(x) then x tends to${{2}^{+}}$ because the value of ‘x’ is greater than ‘2’.

Then,

\[\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,g\left( x \right)=\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,x-5\]

Applying the limit, we get

\[g\left( {{2}^{+}} \right)=-3\]

Then the left hand limit is ‘-3’ at x=0.

So the left hand limit of g(f(x)) is ‘-3’ at x=0.

For finding the right hand limit, we have to consider $x\to {{0}^{+}}$.

So referring to the given function, when x tends to ‘0’ from right hand side then,

\[\begin{align}

& \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{x}{\sin x} \\

& \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{1}{\dfrac{\sin x}{x}} \\

\end{align}\]

But we know $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1$ , so the above equation becomes,

\[\Rightarrow f({{0}^{+}})=1\]

So, now considering the function g(f(x), here g(f(x) becomes g(1).

So, here we have to consider the function g(x) then x tends to${{1}^{+}}$ because the value of ‘x’ is greater than ‘1’.

Then,

\[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,g\left( x \right)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,{{x}^{2}}-2x-2\]

Applying the limit, we get

\[g\left( {{1}^{+}} \right)=1-2-2=-3\]

Then the right hand limit is ‘-3’ at x=0.

So the right hand limit of g(f(x)) is ‘-3’ at x=0.

Hence LHL=RHL.

Hence, the correct answer is \[\underset{x\to 0}{\mathop{\lim }}\,g(f(x))=-3\]

Note: A possible mistake is when finding \[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,g\left( x \right)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,{{x}^{2}}-2x-2\], then student think this is the right hand limit of g(x) at x=1.

Students get confused when finding the left hand limit of g(f(x)).

When they find f(x) at x=0, they take the limit of g(x), i.e., \[\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,g\left( x \right)\]as ${{2}^{-1}}$ thinking this is left hand limit. But this will lead to the wrong answer.

In these types of questions students have confusions while finding the left hand and right hand limit so be careful.

Last updated date: 29th May 2023

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Total views: 328.2k

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