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

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

& x,x>0 \\\ & -x,x<0 \\\ \end{aligned} \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{aligned} & x,x>0 \\\ & -x,x<0 \\\ \end{aligned} \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{aligned} & \Rightarrow LHL={{\left( -0 \right)}^{\sin 0}} \\\ & \Rightarrow LHL={{0}^{0}} \\\ \end{aligned}$$ We know that $${{0}^{0}}$$ is an indeterminate form So, let us apply the logarithm function for LHL then we get $$\begin{aligned} & \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{aligned}$$ We know that the formula of logarithms that is $$\Rightarrow \ln \left( {{z}^{b}} \right)=b\ln z$$ By using this formula to above equation we get $$\begin{aligned} & \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{aligned}$$ We know that the L – hopital’s rule is given as $$\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{aligned} & \dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x} \\\ & \dfrac{d}{dx}\left( \csc x \right)=-\csc x.\cot x \\\ \end{aligned}$$ By using the above formulas in LHL we get $$\begin{aligned} & \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{aligned}$$ Here we can see that the RHS is getting an indeterminate form. So, by applying the L – hopital’s rule again we get $$\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{aligned} & \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{aligned}$$ Therefore, we can see that the value of LHL is 1 Now let us find the value of RHL then we get $$\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{aligned} & \Rightarrow RHL={{\left( 0 \right)}^{\sin 0}} \\\ & \Rightarrow RHL={{0}^{0}} \\\ \end{aligned}$$ We know that $${{0}^{0}}$$ is an indeterminate form So, let us apply the logarithm function for RHL then we get $$\begin{aligned} & \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{aligned}$$ We know that the formula of logarithms that is $$\Rightarrow \ln \left( {{z}^{b}} \right)=b\ln z$$ By using this formula to above equation we get $$\begin{aligned} & \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{aligned}$$ We know that the L – hopital’s rule is given as $$\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{aligned} & \dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x} \\\ & \dfrac{d}{dx}\left( \csc x \right)=-\csc x.\cot x \\\ \end{aligned}$$ By using the above formulas in RHL we get $$\begin{aligned} & \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{aligned}$$ Here we can see that the RHS is getting an indeterminate form. So, by applying the L – hopital’s rule again we get $$\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{aligned} & \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{aligned}$$ Here we can see that both LHL and RHS both exist and are equal to 1 Therefore we can conclude that $$\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{aligned} & \Rightarrow LHL={{\left( -0 \right)}^{\sin 0}} \\\ & \Rightarrow LHL={{0}^{0}} \\\ \end{aligned}$$ Here, it is the undetermined form. So, we need to apply the logarithm function on both sides of LHL and use the L – hospital rule to find the value of LHL. But students may stop the solution there and gives the answer as the limit $$\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.