
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
523.8k+ 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 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

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

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Write examples of herbivores carnivores and omnivo class 10 biology CBSE

10 examples of evaporation in daily life with explanations

Choose the feminine form of the given noun Fox AFoxess class 10 english CBSE
