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Question: Evaluate the limit \(\underset{x\to 0}{\mathop{\lim }}\,\left[ \dfrac{{{\sin }^{-1}}x}{x} \right]\) ...

Evaluate the limit lim⁡x→0 [sin⁡−1xx]\underset{x\to 0}{\mathop{\lim }}\,\left[ \dfrac{{{\sin }^{-1}}x}{x} \right] (where [∗]\left[ * \right]denotes the greatest integer function).

Explanation

Solution

Hint: First we have to check if the limit is in an indeterminate form. If it is indeterminate, then we have to use L'Hopital's rule to solve this limit.

Complete step-by-step answer:
The limit given in the question is lim⁡x→0 [sin⁡−1xx]\underset{x\to 0}{\mathop{\lim }}\,\left[ \dfrac{{{\sin }^{-1}}x}{x} \right].
By looking at the limit, it is clear that we can’t apply limits here directly. We will get an indeterminate form as below,
lim⁡x→0 [sin⁡−1xx] ⇒sin⁡−100 ⇒00 \begin{aligned} & \underset{x\to 0}{\mathop{\lim }}\,\left[ \dfrac{{{\sin }^{-1}}x}{x} \right] \\\ & \Rightarrow \dfrac{{{\sin }^{-1}}0}{0} \\\ & \Rightarrow \dfrac{0}{0} \\\ \end{aligned}
The L’ Hopital’s rule is used for evaluating the limits of indeterminate forms. The indeterminate form should be of the form 00\dfrac{0}{0} or ∞∞\dfrac{\infty }{\infty }. All other forms must be converted to these forms in order to apply the L’ Hopital’s rule.
So, we will have to use the L’ Hopital’s rule to solve this limit. It can be explained as below.
Consider a limit of the form lim⁡x→a [f(x)g(x)]\underset{x\to a}{\mathop{\lim }}\,\left[ \dfrac{f\left( x \right)}{g\left( x \right)} \right]. According to the L’ Hopital’s rule, this limit can be solved by differentiating the numerator and the denominator separately and then applying the limits to it,
lim⁡x→a [f′(x)g′(x)]\underset{x\to a}{\mathop{\lim }}\,\left[ \dfrac{f'\left( x \right)}{g'\left( x \right)} \right].
It doesn’t change the value of the limit.
Therefore, we need to compute the derivatives of the numerator sin⁡−1x{{\sin }^{-1}}x and the denominator xx separately.
We know that the derivative of sin⁡−1x{{\sin }^{-1}}x is given by ddx(sin⁡−1x)=11−x2\dfrac{d}{dx}({{\sin }^{-1}}x)=\dfrac{1}{\sqrt{1-{{x}^{2}}}}.
Also, we know that the derivative of xx is given by ddx(x)=1\dfrac{d}{dx}(x)=1.
So, we can substitute these in the limits as below,

& \Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\left[ \dfrac{\dfrac{1}{\sqrt{1-{{x}^{2}}}}}{1} \right] \\\ & \Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\left[ \dfrac{1}{\sqrt{1-{{x}^{2}}}} \right] \\\ \end{aligned}$$ Now applying the limit we get, $$\begin{aligned} & \Rightarrow \left[\dfrac{1}{\sqrt{1-0}} \right]\\\ & \Rightarrow \left[\dfrac{1}{\sqrt{1}} \right]\\\ & \Rightarrow \left[\dfrac{1}{1}\right] \\\ & \Rightarrow 1 ({\because {\text {greatest integer function}}}) \\\ \end{aligned}$$ Thus we have obtained the value of $\underset{x\to 0}{\mathop{\lim }}\,\left[ \dfrac{{{\sin }^{-1}}x}{x} \right]$ and the correct answer as $$1$$. Note: Whenever this type of question is asked, the first step would be to check if it is an indeterminate form. If it is in the indeterminate form $$\dfrac{0}{0}$$ or $$\dfrac{\infty }{\infty }$$ and then the L’Hopital’s rule can be applied. If it is any other indeterminate form, then we must be able to convert it into the form $$\dfrac{0}{0}$$ or $$\dfrac{\infty }{\infty }$$ in order to apply the L’Hopital’s rule.