Answer
Verified
478.2k+ views
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE