Question

Find the value of \[\displaystyle \lim_{x \to 0}{{\left| x \right|}^{\sin x}}\]
(a) 0
(b) 1
(c) -1
(d) None of the above

Answer 1

Hint: We solve this problem by using the left hand limit and right hand limit.
For a limit \[\displaystyle \lim_{x \to a}f\left( x \right)\] the left hand limit and the right hand limit are given as
\[\Rightarrow LHL=\displaystyle \lim_{x \to {{a}^{-}}}f\left( x \right)\]
\[\Rightarrow RHL=\displaystyle \lim_{x \to {{a}^{+}}}f\left( x \right)\]
We have the definition of limit that is
\[\displaystyle \lim_{x \to a}f\left( x \right)=f\left( a \right)\]
Then if LHL and RHL both exist and are equal then we can say that the original limit \[\displaystyle \lim_{x \to a}f\left( x \right)\] is defined and equal to LHL and RHL. If the LHL and RHL doesn’t exist then we can say that the limit \[\displaystyle \lim_{x \to a}f\left( x \right)\] is not defined. We also use the modulus function definition that is
\[\Rightarrow \left| x \right|=\left\{ \begin{align}
  & x,x>0 \\
 & -x,x<0 \\
\end{align} \right.\]
Here we can see that there is a function power of another function then we apply logarithm on both sides to find the limit if we get an indeterminate form. Then we can use the L - Hopital’s rule to find the limit. The L – hopital’s rule is given as
\[\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)}{g\left( x \right)}=\displaystyle \lim_{x \to a}\dfrac{{f}'\left( x \right)}{{g}'\left( x \right)}\]

Complete step-by-step answer:
We are given that the limit as \[\displaystyle \lim_{x \to 0}{{\left| x \right|}^{\sin x}}\]
Now, let us find the left hand limit and the right hand limit.
We know that for a limit \[\displaystyle \lim_{x \to a}f\left( x \right)\] the left hand limit and the right hand limit are given as
\[\Rightarrow LHL=\displaystyle \lim_{x \to {{a}^{-}}}f\left( x \right)\]
\[\Rightarrow RHL=\displaystyle \lim_{x \to {{a}^{+}}}f\left( x \right)\]
By using this formula to given limit then we get the left hand limit as
\[\Rightarrow LHL=\displaystyle \lim_{x \to {{0}^{-}}}{{\left| x \right|}^{\sin x}}\]
We know that the modulus function is defined as
\[\Rightarrow \left| x \right|=\left\{ \begin{align}
  & x,x>0 \\
 & -x,x<0 \\
\end{align} \right.\]
Here we can see that for the LHL the \['x'\] approaches 0 from the left side.
We know that the numbers that are left of 0 are negative numbers that are less than zero,
So, by using the definition of the modulus function in LHL we get
\[\Rightarrow LHL=\displaystyle \lim_{x \to {{0}^{-}}}{{\left( -x \right)}^{\sin x}}\]
We know that the definition of limit that is
\[\displaystyle \lim_{x \to a}f\left( x \right)=f\left( a \right)\]
By using the this formula to above equation we get
\[\begin{align}
  & \Rightarrow LHL={{\left( -0 \right)}^{\sin 0}} \\
 & \Rightarrow LHL={{0}^{0}} \\
\end{align}\]
We know that \[{{0}^{0}}\] is an indeterminate form
So, let us apply the logarithm function for LHL then we get
\[\begin{align}
  & \Rightarrow \ln \left( LHL \right)=\ln \left( \displaystyle \lim_{x \to {{0}^{-}}}{{\left( -x \right)}^{\sin x}} \right) \\
 & \Rightarrow \ln \left( LHL \right)=\displaystyle \lim_{x \to {{0}^{-}}}\left( \ln {{\left( -x \right)}^{\sin x}} \right) \\
\end{align}\]

We know that the formula of logarithms that is
\[\Rightarrow \ln \left( {{z}^{b}} \right)=b\ln z\]
By using this formula to above equation we get
\[\begin{align}
  & \Rightarrow \ln \left( LHL \right)=\displaystyle \lim_{x \to {{0}^{-}}}\left( \sin x.\ln \left( -x \right) \right) \\
 & \Rightarrow \ln \left( LHL \right)=\displaystyle \lim_{x \to {{0}^{-}}}\left( \dfrac{\ln \left( -x \right)}{\csc x} \right) \\
\end{align}\]
We know that the L – hopital’s rule is given as
\[\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)}{g\left( x \right)}=\displaystyle \lim_{x \to a}\dfrac{{f}'\left( x \right)}{{g}'\left( x \right)}\]
By using this rule to above equation we get
\[\Rightarrow \ln \left( LHL \right)=\displaystyle \lim_{x \to {{0}^{-}}}\left( \dfrac{\dfrac{d}{dx}\left( \ln \left( -x \right) \right)}{\dfrac{d}{dx}\left( \csc x \right)} \right)\]
We know that the formulas of derivatives that are
\[\begin{align}
  & \dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x} \\
 & \dfrac{d}{dx}\left( \csc x \right)=-\csc x.\cot x \\
\end{align}\]
By using the above formulas in LHL we get
\[\begin{align}
  & \Rightarrow \ln \left( LHL \right)=\displaystyle \lim_{x \to {{0}^{-}}}\left( \dfrac{\dfrac{1}{-x}}{-\csc x.\cot x} \right) \\
 & \Rightarrow \ln \left( LHL \right)=\displaystyle \lim_{x \to {{0}^{-}}}\left( \dfrac{\sin x.\tan x}{x} \right) \\
\end{align}\]
Here we can see that the RHS is getting an indeterminate form.
So, by applying the L – hopital’s rule again we get
\[\Rightarrow \ln \left( LHL \right)=\displaystyle \lim_{x \to {{0}^{-}}}\left( \dfrac{\dfrac{d}{dx}\left( \sin x.\tan x \right)}{\dfrac{d}{dx}\left( x \right)} \right)\]
We know that the chain rule of derivatives that is
\[\dfrac{d}{dx}\left( f\left( x \right)\times g\left( x \right) \right)={f}'\left( x \right)\times g\left( x \right)+f\left( x \right)\times {g}'\left( x \right)\]
By using the chain rule in above equation we get
\[\Rightarrow \ln \left( LHL \right)=\displaystyle \lim_{x \to {{0}^{-}}}\left( \dfrac{\cos x.\tan x+\sin x.{{\sec }^{2}}x}{1} \right)\]
Now, by expanding the limit we get
\[\begin{align}
  & \Rightarrow \ln \left( LHL \right)=\cos 0.\tan 0+\sin 0.{{\sec }^{2}}0 \\
 & \Rightarrow \ln \left( LHL \right)=0 \\
 & \Rightarrow LHL={{e}^{0}}=1 \\
\end{align}\]
Therefore, we can see that the value of LHL is 1
Now let us find the value of RHL then we get
\[\Rightarrow RHL=\displaystyle \lim_{x \to {{0}^{+}}}{{\left| x \right|}^{\sin x}}\]
Now, by using the definition of modulus function we get
\[\Rightarrow RHL=\displaystyle \lim_{x \to {{0}^{+}}}{{\left( x \right)}^{\sin x}}\]
We know that the definition of limit that is
\[\displaystyle \lim_{x \to a}f\left( x \right)=f\left( a \right)\]
By using the this formula to above equation we get
\[\begin{align}
  & \Rightarrow RHL={{\left( 0 \right)}^{\sin 0}} \\
 & \Rightarrow RHL={{0}^{0}} \\
\end{align}\]
We know that \[{{0}^{0}}\] is an indeterminate form
So, let us apply the logarithm function for RHL then we get
\[\begin{align}
  & \Rightarrow \ln \left( RHL \right)=\ln \left( \displaystyle \lim_{x \to {{0}^{+}}}{{\left( x \right)}^{\sin x}} \right) \\
 & \Rightarrow \ln \left( RHL \right)=\displaystyle \lim_{x \to {{0}^{+}}}\left( \ln {{\left( x \right)}^{\sin x}} \right) \\
\end{align}\]
We know that the formula of logarithms that is
\[\Rightarrow \ln \left( {{z}^{b}} \right)=b\ln z\]
By using this formula to above equation we get
\[\begin{align}
  & \Rightarrow \ln \left( RHL \right)=\displaystyle \lim_{x \to {{0}^{+}}}\left( \sin x.\ln \left( x \right) \right) \\
 & \Rightarrow \ln \left( RHL \right)=\displaystyle \lim_{x \to {{0}^{+}}}\left( \dfrac{\ln \left( x \right)}{\csc x} \right) \\
\end{align}\]
We know that the L – hopital’s rule is given as
\[\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)}{g\left( x \right)}=\displaystyle \lim_{x \to a}\dfrac{{f}'\left( x \right)}{{g}'\left( x \right)}\]
By using this rule to above equation we get
\[\Rightarrow \ln \left( RHL \right)=\displaystyle \lim_{x \to {{0}^{-}}}\left( \dfrac{\dfrac{d}{dx}\left( \ln \left( x \right) \right)}{\dfrac{d}{dx}\left( \csc x \right)} \right)\]
We know that the formulas of derivatives that are
\[\begin{align}
  & \dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x} \\
 & \dfrac{d}{dx}\left( \csc x \right)=-\csc x.\cot x \\
\end{align}\]
By using the above formulas in RHL we get
\[\begin{align}
  & \Rightarrow \ln \left( RHL \right)=\displaystyle \lim_{x \to {{0}^{+}}}\left( \dfrac{\dfrac{1}{x}}{-\csc x.\cot x} \right) \\
 & \Rightarrow \ln \left( RHL \right)=-\displaystyle \lim_{x \to {{0}^{+}}}\left( \dfrac{\sin x.\tan x}{x} \right) \\
\end{align}\]
Here we can see that the RHS is getting an indeterminate form.
So, by applying the L – hopital’s rule again we get
\[\Rightarrow \ln \left( RHL \right)=-\displaystyle \lim_{x \to {{0}^{+}}}\left( \dfrac{\dfrac{d}{dx}\left( \sin x.\tan x \right)}{\dfrac{d}{dx}\left( x \right)} \right)\]
We know that the chain rule of derivatives that is
\[\dfrac{d}{dx}\left( f\left( x \right)\times g\left( x \right) \right)={f}'\left( x \right)\times g\left( x \right)+f\left( x \right)\times {g}'\left( x \right)\]
By using the chain rule in above equation we get
\[\Rightarrow \ln \left( RHL \right)=-\displaystyle \lim_{x \to {{0}^{+}}}\left( \dfrac{\cos x.\tan x+\sin x.{{\sec }^{2}}x}{1} \right)\]
Now, by expanding the limit we get
\[\begin{align}
  & \Rightarrow \ln \left( RHL \right)=-\left( \cos 0.\tan 0+\sin 0.{{\sec }^{2}}0 \right) \\
 & \Rightarrow \ln \left( RHL \right)=-0 \\
 & \Rightarrow RHL={{e}^{-0}}=1 \\
\end{align}\]
Here we can see that both LHL and RHS both exist and are equal to 1
Therefore we can conclude that
 \[\therefore \displaystyle \lim_{x \to 0}{{\left| x \right|}^{\sin x}}=1\]

So, the correct answer is “Option B”.

Note: Students may stop the problem in the middle and give the wrong answer.
We got the LHL initially as
\[\begin{align}
  & \Rightarrow LHL={{\left( -0 \right)}^{\sin 0}} \\
 & \Rightarrow LHL={{0}^{0}} \\
\end{align}\]
Here, it is the undetermined form. So, we need to apply the logarithm function on both sides of LHL and use the L – hospital rule to find the value of LHL.
But students may stop the solution there and gives the answer as the limit \[\displaystyle \lim_{x \to 0}{{\left| x \right|}^{\sin x}}\] does not exist
But when we get an undetermined form then we need to go for alternatives to find the required limit.