lim (x→1⁻) (√π - √2sin⁻¹x) / √(1-x) is equal to : (1) 1/√2 (2) √(π/2) (3) √(2/π) (4) √π

Q: Why do we multiply by the conjugate? Can't we just use L'Hopital's Rule?

L'Hopital's Rule is a powerful weapon in your arsenal, but differentiating $\sqrt{\sin^{-1} x}$ involves the chain rule and can get messy very quickly. Rationalization is often cleaner, keeps the algebraic structure intact, and helps you avoid potential calculation errors with complex derivatives.

Q: Why did we substitute $x = \cos \theta$?

Because $\cos^{-1} x$ is the 'problematic' term in our simplified expression. By setting $x = \cos \theta$, we transform the inverse trigonometric function into a simple variable $\theta$. This makes the limit much easier to handle, especially when paired with the standard limit $\lim_{u \to 0} \frac{\sin u}{u} = 1$.

Q: What happened to the $\sqrt{2}$ in the denominator?

When we simplified $\sqrt{1-\cos \theta}$ to $\sqrt{2\sin^2(\theta/2)}$, it became $\sqrt{2}\sin(\theta/2)$. That $\sqrt{2}$ is just a constant factor. We pulled it out of the limit to keep our workspace tidy, allowing us to focus on the core limit $\frac{\theta}{\sin(\theta/2)}$.

Question

$\lim_{x 	o 1^-} \frac{\sqrt{\pi} - \sqrt{2\sin^{-1} x}}{\sqrt{1-x}}$ is equal to :

Options:

(A) $\frac{1}{\sqrt{2}}$

(B) $\sqrt{\frac{\pi}{2}}$

(C) $\sqrt{\frac{2}{\pi}}$

(D) $\sqrt{\pi}$

Sigma Percentile · Accepted Answer

### The Limit of Your Potential: Mastering the Indeterminate Form

Welcome, JEE warrior. Today, we are not just solving a math problem; we are dissecting a classic trap.

Limits are the heartbeat of calculus, and this specific problem, $\lim_{x 	o 1^-} \frac{\sqrt{\pi} - \sqrt{2\sin^{-1} x}}{\sqrt{1-x}}$, is a perfect example of how a seemingly intimidating expression can be tamed with the right strategy.

### Phase 1: The Indeterminate Trap

Whenever you face a limit, your first step is always the same: direct substitution. It is the litmus test.

As $x 	o 1^-$, the numerator becomes $\sqrt{\pi} - \sqrt{2\sin^{-1}(1)}$. Since $\sin^{-1}(1) = \frac{\pi}{2}$, the numerator becomes $\sqrt{\pi} - \sqrt{2 \cdot \frac{\pi}{2}} = \sqrt{\pi} - \sqrt{\pi} = 0$.

The denominator is $\sqrt{1-1} = 0$. We have arrived at the $\frac{0}{0}$ indeterminate form. This is not a dead end; it is a green light indicating a hidden factor that must be canceled.

### Phase 2: The Surgical Strike (Rationalization)

When you see square roots in a limit, your brain should immediately scream: "Rationalize!" We need to get rid of those radicals by multiplying the numerator and the denominator by the conjugate: $(\sqrt{\pi} + \sqrt{2\sin^{-1} x})$.

Our expression transforms into:

$$\frac{(\sqrt{\pi} - \sqrt{2\sin^{-1} x})(\sqrt{\pi} + \sqrt{2\sin^{-1} x})}{\sqrt{1-x}(\sqrt{\pi} + \sqrt{2\sin^{-1} x})}$$

Using the identity $(a-b)(a+b) = a^2 - b^2$, the numerator simplifies to $\pi - 2\sin^{-1} x$. We factor out the $2$, giving us $2(\frac{\pi}{2} - \sin^{-1} x)$.

### Phase 3: The Elegant Identity

This is where the JEE examiner tests your intuition. We utilize the fundamental identity of inverse trigonometry: $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$.

Therefore, $\frac{\pi}{2} - \sin^{-1} x = \cos^{-1} x$. Our expression now looks much cleaner:

$$\lim_{x 	o 1^-} \frac{2\cos^{-1} x}{\sqrt{1-x}(\sqrt{\pi} + \sqrt{2\sin^{-1} x})}$$

Notice the term $(\sqrt{\pi} + \sqrt{2\sin^{-1} x})$ in the denominator. As $x 	o 1$, this term approaches $2\sqrt{\pi}$. We can pull this constant out of the limit, simplifying the expression to $\frac{1}{\sqrt{\pi}} \lim_{x 	o 1^-} \frac{\cos^{-1} x}{\sqrt{1-x}}$.

### Phase 4: The Final Transformation

To solve this, we use a substitution. Let $\cos^{-1} x = 	heta$, which implies $x = \cos 	heta$. As $x 	o 1^-$, $	heta 	o 0^+$.

Recall the half-angle identity: $1 - \cos 	heta = 2\sin^2(\frac{	heta}{2})$. The denominator becomes $\sqrt{2\sin^2(\frac{	heta}{2})} = \sqrt{2}\sin(\frac{	heta}{2})$.

Our limit is now:

$$\frac{1}{\sqrt{\pi}} \lim_{	heta 	o 0^+} \frac{	heta}{\sqrt{2}\sin(\frac{	heta}{2})}$$

To use the standard limit $\lim_{u 	o 0} \frac{\sin u}{u} = 1$, we manipulate the expression:

$$\frac{1}{\sqrt{\pi} \cdot \sqrt{2}} \lim_{	heta 	o 0^+} \frac{2 \cdot (\frac{	heta}{2})}{\sin(\frac{	heta}{2})} = \frac{2}{\sqrt{2\pi}} = \sqrt{\frac{2}{\pi}}$$

The complexity dissolves, leaving behind the final result: **$\sqrt{\frac{2}{\pi}}$**.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: $lim_{x \to 1^{-}} \frac{π - 2 s i n ^{- 1} x}{1 - x}$ is equal to :

Select Answer:

Visualized Solution

Identifying the Form of the Limit

Rationalizing the Numerator

Simplifying the Numerator

Applying Inverse Trig Identity

Evaluating the Constant Part

Substitution for the Final Limit

Using Half-Angle Identity

Final Calculation

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

The Limit of Your Potential

Phase 1

Phase 2

Phase 3

Phase 4

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