lim (x-0) (sin(pi cos^2 x)) / x^2 is equal to (A) -pi (B) pi (C) pi/2 (D) 1

Q: Wait, why can't I just use L'Hôpital's Rule here? It seems much faster.

You absolutely can! L'Hôpital's Rule is a powerful weapon in your arsenal. If you differentiate the numerator and denominator twice, you will arrive at the same result. However, mastering the algebraic manipulation of trigonometric identities is crucial for JEE Advanced because sometimes the derivatives become so complex that they lead to algebraic errors. Always keep the standard limit method as your primary, elegant approach.

Q: Why did we change $\sin(\pi - \pi \sin^2 x)$ to $\sin(\pi \sin^2 x)$?

That is a great observation! We used the allied angle formula $\sin(\pi - \theta) = \sin \theta$. By setting $\theta = \pi \sin^2 x$, we effectively shifted the argument of the sine function to something that approaches zero as $x$ approaches zero. This is the key step that allows us to apply the standard limit $\lim_{u \to 0} \frac{\sin u}{u} = 1$.

Q: Does the $\pi$ inside the sine function affect the limit value?

Yes, it does! The $\pi$ acts as a scaling factor. When we multiply and divide by $\pi \sin^2 x$, that $\pi$ remains in the expression. It doesn't vanish; it propagates through the limit calculation and becomes the final coefficient of your answer. Never lose track of constants during your algebraic manipulations!

Q: Can I just replace $\sin x$ with $x$ when $x$ is small?

Yes, you are referring to the small-angle approximation $\sin x \approx x$. This is perfectly valid in the context of limits as $x \to 0$. In fact, it is a shortcut! If you replace $\sin x$ with $x$, the expression simplifies to $\frac{\sin(\pi x^2)}{x^2}$, which directly leads to $\pi$ using the standard limit. It's a great way to verify your work.

Q: Why is the final answer $\pi$ and not $1$?

It is easy to get distracted by the $\sin(\theta)/\theta$ part, which equals $1$. But remember, we had an extra $\pi$ factor from the argument of the sine function. When you split the limit, you are left with $\lim_{x \to 0} \frac{\pi \sin^2 x}{x^2}$. Since $\lim_{x \to 0} (\frac{\sin x}{x})^2 = 1$, you are left with $\pi \cdot 1 = \pi$. Don't let the constants hide in plain sight!

Question

$\lim_{x 	o 0} \frac{\sin(\pi \cos^2 x)}{x^2}$ is equal to

Options:

(A) $-\pi$

(B) $\pi$

(C) $\pi/2$

(D) 1

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE excellence. Today, we are going to dissect a problem that, at first glance, looks like a standard trigonometric limit, but beneath the surface, it is a beautiful exercise in algebraic manipulation and the art of 'forcing' a standard form.

We are looking at the limit:

$$ \lim_{x 	o 0} \frac{\sin(\pi \cos^2 x)}{x^2} $$

### The Indeterminate Gateway

Whenever you encounter a limit, your first duty is to test the waters. What happens when $x$ approaches $0$?

If we substitute $x = 0$ directly into our expression, the numerator becomes $\sin(\pi \cos^2(0)) = \sin(\pi \cdot 1) = \sin(\pi) = 0$. The denominator is simply $0^2 = 0$.

We have arrived at the classic $0/0$ indeterminate form. This is not a dead end; it is an invitation to find the hidden ratio the function approaches at the origin.

### The Trigonometric Transformation

Now, look at the argument of the sine function: $\pi \cos^2 x$. This is the 'trap.' We cannot easily apply the standard limit $\lim_{u 	o 0} \frac{\sin u}{u} = 1$ because the argument is approaching $\pi$, not $0$.

We need to transform this using the fundamental identity $\cos^2 x = 1 - \sin^2 x$. By substituting this into our expression, we get:

$$ \lim_{x 	o 0} \frac{\sin[\pi(1 - \sin^2 x)]}{x^2} = \lim_{x 	o 0} \frac{\sin(\pi - \pi \sin^2 x)}{x^2} $$

Suddenly, the structure changes. We have the form $\sin(\pi - 	heta)$, where $	heta = \pi \sin^2 x$. Using the allied angle formula $\sin(\pi - 	heta) = \sin 	heta$, our expression simplifies beautifully to:

$$ \lim_{x 	o 0} \frac{\sin(\pi \sin^2 x)}{x^2} $$

### The Art of Forcing the Limit

Observe the argument of the sine function: $\pi \sin^2 x$. As $x 	o 0$, this argument also approaches $0$. This is exactly what we need.

To use the standard limit $\lim_{u 	o 0} \frac{\sin u}{u} = 1$, the denominator must match the argument of the sine function perfectly. Currently, our denominator is just $x^2$.

We perform the 'JEE maneuver'—multiply and divide by the term we desire:

$$ \lim_{x 	o 0} \left[ \frac{\sin(\pi \sin^2 x)}{\pi \sin^2 x} \cdot \frac{\pi \sin^2 x}{x^2} ight] $$

### The Elegant Cancellation

We can now split this into two separate limits using the product rule:

$$ \left( \lim_{x 	o 0} \frac{\sin(\pi \sin^2 x)}{\pi \sin^2 x} ight) \cdot \left( \lim_{x 	o 0} \frac{\pi \sin^2 x}{x^2} ight) $$

The first part is now in the perfect standard form. As $x 	o 0$, the argument $\pi \sin^2 x$ also goes to $0$, so the first limit is simply $1$.

For the second part, we can pull the constant $\pi$ out and rewrite the limit as:

$$ \pi \cdot \left( \lim_{x 	o 0} \frac{\sin x}{x} ight)^2 $$

Since $\lim_{x 	o 0} \frac{\sin x}{x} = 1$, the entire second part becomes $\pi \cdot (1)^2 = \pi$. Multiplying our two results together, we get $1 \cdot \pi = \pi$.

### Final Result

The complexity of the trigonometric function collapses into the elegant constant. The final answer is:

**$\pi$**

Animated Solution for Mathematics - Limits, Continuity and Differentiability: $lim_{x \to 0} \frac{s i n ( π c o s ^{2} x )}{x ^{2}}$ is equal to

Select Answer:

Visualized Solution

The Limit Challenge: $lim_{x \to 0} \frac{s i n ( π c o s ^{2} x )}{x ^{2}}$

Checking the Indeterminate Form

The Trigonometric Tool: $cos^{2} x = 1 - sin^{2} x$

Substituting the Identity

Applying $sin (π - θ) = sin θ$

The Simplified Limit Expression

The Standard Limit: $lim_{θ \to 0} \frac{s i n θ}{θ} = 1$

Multiplying and Dividing by $π sin^{2} x$

Splitting and Solving the Limits

The Final Answer is $π$

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

Analyzing the Setup

The Indeterminate Gateway

The Trigonometric Transformation

The Art of Forcing the Limit

The Elegant Cancellation

Final Result

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Animated Solution for Mathematics - Limits, Continuity and Differentiability: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limx→0​x2sin(πcos2x)​ is equal to

Select Answer:

Visualized Solution

The Limit Challenge: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limx→0​x2sin(πcos2x)​

Checking the Indeterminate Form

The Trigonometric Tool: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cos2x=1−sin2x

Substituting the Identity

Applying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin(π−θ)=sinθ

The Simplified Limit Expression

The Standard Limit: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limθ→0​θsinθ​=1

Multiplying and Dividing by .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } πsin2x