
Find the value of $\underset{x\to 0}{\mathop{\lim }}\,{{\left( \sin x \right)}^{2\tan x}}$.
A. 2 B. 1 C. 0 D. doesn’t exist
Answer
584.7k+ views
Hint: We try to put the points on the graph to find out the limit of the function. From the graph, we find out if both sides limit for the function exists or not. If they exist and are equal then we place the value as a solution otherwise limit doesn’t exist.
Complete step by step answer:
The given is $y=\underset{x\to 0}{\mathop{\lim }}\,{{\left( \sin x \right)}^{2\tan x}}$.
We draw the graph of the function and get
From the figure, we can see the graph is discontinuous at x = 0.
Although the right-hand side limit exists at point $\left( 0,1 \right)$, the left-hand side doesn’t. so, the limit doesn’t exist altogether.
We now find the right-hand side limit.
We take logarithm both side of the function $y=\underset{x\to 0}{\mathop{\lim }}\,{{\left( \sin x \right)}^{2\tan x}}$.
So, \[\log y=\log \left( \underset{x\to 0}{\mathop{\lim }}\,{{\left( \sin x \right)}^{2\tan x}} \right)=\underset{x\to 0}{\mathop{\lim }}\,\log \left[ {{\left( \sin x \right)}^{2\tan x}} \right]=\underset{x\to 0}{\mathop{\lim }}\,\left[ 2\tan x\log \left( \sin x \right) \right]\]
Now, we apply the theorem of limit to break two multiplication.
We know $\underset{x\to a}{\mathop{\lim }}\,\left[ \left( f\left( x \right) \right)\left( g\left( x \right) \right) \right]=\underset{x\to a}{\mathop{\lim }}\,\left[ \left( f\left( x \right) \right) \right]\times \underset{x\to a}{\mathop{\lim }}\,\left[ \left( g\left( x \right) \right) \right]$
We get \[\underset{x\to 0}{\mathop{\lim }}\,\left[ 2\tan x\log \left( \sin x \right) \right]=\underset{x\to 0}{\mathop{\lim }}\,\left[ 2\tan x \right]\times \underset{x\to 0}{\mathop{\lim }}\,\left[ \log \left( \sin x \right) \right]\].
Now we find the value of the limit as
\[\begin{align}
& \log y=\underset{x\to 0}{\mathop{\lim }}\,\left[ 2\tan x \right]\times \underset{x\to 0}{\mathop{\lim }}\,\left[ \log \left( \sin x \right) \right] \\
& \Rightarrow \log y=0\times \underset{x\to 0}{\mathop{\lim }}\,\left[ \log \left( \sin x \right) \right]=0 \\
& \Rightarrow y={{e}^{0}}=1 \\
\end{align}\]
So, the value of the limit(right-hand) is 1. But the left-hand side doesn’t exist as for \[\log \left( \sin x \right)\] the value of $\sin x$ is negative when $x\to {{0}^{-}}$. So, in that case we get logarithm of negative value which is not possible.
So, the correct option is (D).
Note:
Although we find the graph value as a limit non-existent, we still need to show the right-hand limit. In case of left-hand limit, although the \[\underset{x\to 0}{\mathop{\lim }}\,\left[ 2\tan x \right]\] omit the all possible options for rest, the rest still has to be viable for existence. That’s why that limit can’t exist.
Complete step by step answer:
The given is $y=\underset{x\to 0}{\mathop{\lim }}\,{{\left( \sin x \right)}^{2\tan x}}$.
We draw the graph of the function and get
From the figure, we can see the graph is discontinuous at x = 0.
Although the right-hand side limit exists at point $\left( 0,1 \right)$, the left-hand side doesn’t. so, the limit doesn’t exist altogether.
We now find the right-hand side limit.
We take logarithm both side of the function $y=\underset{x\to 0}{\mathop{\lim }}\,{{\left( \sin x \right)}^{2\tan x}}$.
So, \[\log y=\log \left( \underset{x\to 0}{\mathop{\lim }}\,{{\left( \sin x \right)}^{2\tan x}} \right)=\underset{x\to 0}{\mathop{\lim }}\,\log \left[ {{\left( \sin x \right)}^{2\tan x}} \right]=\underset{x\to 0}{\mathop{\lim }}\,\left[ 2\tan x\log \left( \sin x \right) \right]\]
Now, we apply the theorem of limit to break two multiplication.
We know $\underset{x\to a}{\mathop{\lim }}\,\left[ \left( f\left( x \right) \right)\left( g\left( x \right) \right) \right]=\underset{x\to a}{\mathop{\lim }}\,\left[ \left( f\left( x \right) \right) \right]\times \underset{x\to a}{\mathop{\lim }}\,\left[ \left( g\left( x \right) \right) \right]$
We get \[\underset{x\to 0}{\mathop{\lim }}\,\left[ 2\tan x\log \left( \sin x \right) \right]=\underset{x\to 0}{\mathop{\lim }}\,\left[ 2\tan x \right]\times \underset{x\to 0}{\mathop{\lim }}\,\left[ \log \left( \sin x \right) \right]\].
Now we find the value of the limit as
\[\begin{align}
& \log y=\underset{x\to 0}{\mathop{\lim }}\,\left[ 2\tan x \right]\times \underset{x\to 0}{\mathop{\lim }}\,\left[ \log \left( \sin x \right) \right] \\
& \Rightarrow \log y=0\times \underset{x\to 0}{\mathop{\lim }}\,\left[ \log \left( \sin x \right) \right]=0 \\
& \Rightarrow y={{e}^{0}}=1 \\
\end{align}\]
So, the value of the limit(right-hand) is 1. But the left-hand side doesn’t exist as for \[\log \left( \sin x \right)\] the value of $\sin x$ is negative when $x\to {{0}^{-}}$. So, in that case we get logarithm of negative value which is not possible.
So, the correct option is (D).
Note:
Although we find the graph value as a limit non-existent, we still need to show the right-hand limit. In case of left-hand limit, although the \[\underset{x\to 0}{\mathop{\lim }}\,\left[ 2\tan x \right]\] omit the all possible options for rest, the rest still has to be viable for existence. That’s why that limit can’t exist.
Recently Updated Pages
Master Class 10 Computer Science: Engaging Questions & Answers for Success

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 English: Engaging Questions & Answers for Success

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

Trending doubts
What is the median of the first 10 natural numbers class 10 maths CBSE

Which women's tennis player has 24 Grand Slam singles titles?

Who is the Brand Ambassador of Incredible India?

Why is there a time difference of about 5 hours between class 10 social science CBSE

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

State and prove converse of BPT Basic Proportionality class 10 maths CBSE

