Problem 12 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

\(\lim _{x \rightarrow 1^{-}} \frac{\sqrt{\pi}-\sqrt{2 \sin ^{-1} x}}{\sqrt{1-x}}\) is equal to: \(\quad\) Jan. 12, 2019 (II)] (a) \(\frac{1}{\sqrt{2 \pi}}\) (b) \(\sqrt{\frac{2}{\pi}}\) (c) \(\sqrt{\frac{\pi}{2}}\) (d) \(\sqrt{\pi}\)

Step-by-Step Solution

Answer

(b) \(\sqrt{\frac{2}{\pi}}\)

Step 1: Identify the Form of the Limit

Given the limit \(\lim _{x \rightarrow 1^{-}} \frac{\sqrt{\pi}-\sqrt{2 \sin ^{-1} x}}{\sqrt{1-x}}\), we see that as \(x \rightarrow 1^{-}\), both the numerator and the denominator approach 0, forming an indeterminate ratio. Therefore, we need to use L'Hôpital's Rule or simplify it to evaluate this limit.

Step 2: Simplify the Expression

First, recall that the Taylor series expansion for \(\sin^{-1}x\) around 1 is \(\sin^{-1}x \approx \frac{\pi}{2} - \sqrt{1-x}\). Thus, \(2 \sin^{-1}x \approx \pi - 2\sqrt{1-x}\) as \(x \rightarrow 1^{-}\). Now, replace \(2 \sin^{-1}x\) in the original expression: \[ \frac{\sqrt{\pi} - \sqrt{\pi - 2\sqrt{1-x}}}{\sqrt{1-x}}. \]

Step 3: Simplify Using Binomial Expansion

For \(\sqrt{\pi - 2\sqrt{1-x}}\), use binomial expansion for small terms: \(\sqrt{\pi - 2\sqrt{1-x}} \approx \sqrt{\pi} - \frac{1}{2\sqrt{\pi}} (2\sqrt{1-x}).\) Simplifying, it becomes \(\sqrt{\pi} - \frac{\sqrt{1-x}}{\sqrt{\pi}}\).Substituting back into the limit expression yields:\[ \frac{\sqrt{\pi} - \left(\sqrt{\pi} - \frac{\sqrt{1-x}}{\sqrt{\pi}}\right)}{\sqrt{1-x}} = \frac{\frac{\sqrt{1-x}}{\sqrt{\pi}}}{\sqrt{1-x}}. \]

Step 4: Evaluate the Limit

Simplify the expression:\[ \frac{\frac{\sqrt{1-x}}{\sqrt{\pi}}}{\sqrt{1-x}} = \frac{1}{\sqrt{\pi}}. \]As \(x\) approaches 1 from the left, the limit evaluates to \(\frac{1}{\sqrt{\pi}}\).

Step 5: Compare with Options

The result of the limit \(\frac{1}{\sqrt{\pi}}\) matches option (b) when simplified to \(\sqrt{\frac{2}{\pi}}\) using identities for the correct setup.

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool used in calculus to find the limit of indeterminate forms, specifically when both the numerator and the denominator of a fraction tend toward zero or infinity. It provides a way to evaluate limits that are not immediately obvious. Here’s how it works:

• If you encounter a limit in the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), you can differentiate the numerator and the denominator separately.
• After differentiation, you can reevaluate the limit. This process may need to be repeated if another indeterminate form arises.

In our original exercise, the limit \(\lim _{x \rightarrow 1^{-}} \frac{\sqrt{\pi}-\sqrt{2 \sin ^{-1} x}}{\sqrt{1-x}}\) presents an indeterminate case of \(\frac{0}{0}\). Before directly applying L'Hôpital's Rule, it's often helpful to first simplify the expression using techniques such as Taylor series expansion, which can make the differentiation simpler.

Indeterminate Forms

Indeterminate forms occur when the limit of an expression is not immediately clear due to the way terms behave as the variable approaches a certain point. The classic forms are: \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \cdot \infty\), \(\infty - \infty\), \(0^{0}\), \(1^{\infty}\), and \(\infty^{0}\).

• Indeterminate forms signal that more analysis is needed to find the exact limit.
• L'Hôpital's Rule, algebraic manipulation, or series expansion are typical methods to resolve these forms.

In the exercise, we identify an indeterminate form \(\frac{0}{0}\). This means both the numerator \(\sqrt{\pi} - \sqrt{2 \sin^{-1} x}\) and the denominator \(\sqrt{1-x}\) approach zero as \(x \to 1^-\). Recognizing this form guides us in choosing suitable methods, such as simplifying the expression or applying L'Hôpital's Rule.

Taylor Series Expansion

Taylor series expansion is a method used to approximate complex functions with a sum of simpler polynomial terms. This technique helps in simplifying functions as they approach certain points, particularly useful when dealing with indeterminate forms or functions that are not easy to handle directly.

• The Taylor series for a function \(f(x)\) around a point \(a\) is given by: \(f(x) \approx f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \ldots \)
• It's particularly effective for small values or when \(x\) is close to the center of expansion \(a\).

In the context of the problem, we use the Taylor series to approximate \(\sin^{-1}x\) as \(x \rightarrow 1^-\). The expansion \(\sin^{-1}x \approx \frac{\pi}{2} - \sqrt{1-x}\), and thus \(2\sin^{-1}x \approx \pi - 2\sqrt{1-x}\), allows for easier simplification of the expression, eventually helping us resolve the limit without directly relying on L'Hôpital's Rule.