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Question: limx→ 0 sin (π cos2x)/x2 is equal to...

limx→ 0 sin (π cos2x)/x2 is equal to

A

-π

B

π

C

π/2

D

1

Answer

π

Explanation

Solution

The problem asks to evaluate the limit: lim⁡x→0sin⁡(πcos⁡2x)x2\lim_{x \to 0} \frac{\sin(\pi \cos^2 x)}{x^2}

Step 1: Identify the indeterminate form. As x→0x \to 0, cos⁡x→cos⁡0=1\cos x \to \cos 0 = 1. So, the argument of sine, πcos⁡2x→π(1)2=π\pi \cos^2 x \to \pi (1)^2 = \pi. The numerator sin⁡(πcos⁡2x)→sin⁡(π)=0\sin(\pi \cos^2 x) \to \sin(\pi) = 0. The denominator x2→0x^2 \to 0. Thus, the limit is of the indeterminate form 00\frac{0}{0}.

Step 2: Use trigonometric identities. We know the identity cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 x. Substitute this into the argument of the sine function: lim⁡x→0sin⁡(π(1−sin⁡2x))x2\lim_{x \to 0} \frac{\sin(\pi (1 - \sin^2 x))}{x^2} lim⁡x→0sin⁡(π−πsin⁡2x)x2\lim_{x \to 0} \frac{\sin(\pi - \pi \sin^2 x)}{x^2} Now, use the trigonometric identity sin⁡(π−θ)=sin⁡(θ)\sin(\pi - \theta) = \sin(\theta). Let θ=πsin⁡2x\theta = \pi \sin^2 x. lim⁡x→0sin⁡(πsin⁡2x)x2\lim_{x \to 0} \frac{\sin(\pi \sin^2 x)}{x^2}

Step 3: Apply standard limit formula. We use the standard limit lim⁡y→0sin⁡yy=1\lim_{y \to 0} \frac{\sin y}{y} = 1. To apply this, we need to manipulate the expression by multiplying and dividing by the argument of the sine function, which is πsin⁡2x\pi \sin^2 x. As x→0x \to 0, sin⁡x→0\sin x \to 0, so πsin⁡2x→0\pi \sin^2 x \to 0. Thus, we can treat πsin⁡2x\pi \sin^2 x as yy in the standard limit formula. lim⁡x→0sin⁡(πsin⁡2x)πsin⁡2x⋅πsin⁡2xx2\lim_{x \to 0} \frac{\sin(\pi \sin^2 x)}{\pi \sin^2 x} \cdot \frac{\pi \sin^2 x}{x^2} Now, evaluate the two parts of the product: The first part: lim⁡x→0sin⁡(πsin⁡2x)πsin⁡2x=1(since πsin⁡2x→0 as x→0)\lim_{x \to 0} \frac{\sin(\pi \sin^2 x)}{\pi \sin^2 x} = 1 \quad \left(\text{since } \pi \sin^2 x \to 0 \text{ as } x \to 0\right) The second part: lim⁡x→0πsin⁡2xx2=πlim⁡x→0(sin⁡xx)2\lim_{x \to 0} \frac{\pi \sin^2 x}{x^2} = \pi \lim_{x \to 0} \left( \frac{\sin x}{x} \right)^2 We know that lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1. So, π(1)2=π\pi (1)^2 = \pi

Step 4: Combine the results. Multiplying the results from the two parts: 1⋅π=π1 \cdot \pi = \pi

The value of the limit is π\pi.

Explanation of the solution: The limit is of the 00\frac{0}{0} indeterminate form. By using the identity cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 x, the argument of the sine function becomes π(1−sin⁡2x)=π−πsin⁡2x\pi(1 - \sin^2 x) = \pi - \pi \sin^2 x. Using sin⁡(π−θ)=sin⁡(θ)\sin(\pi - \theta) = \sin(\theta), the expression simplifies to sin⁡(πsin⁡2x)x2\frac{\sin(\pi \sin^2 x)}{x^2}. To apply the standard limit lim⁡y→0sin⁡yy=1\lim_{y \to 0} \frac{\sin y}{y} = 1, we multiply and divide by πsin⁡2x\pi \sin^2 x. This transforms the limit into lim⁡x→0(sin⁡(πsin⁡2x)πsin⁡2x)⋅(πsin⁡2xx2)\lim_{x \to 0} \left( \frac{\sin(\pi \sin^2 x)}{\pi \sin^2 x} \right) \cdot \left( \pi \frac{\sin^2 x}{x^2} \right). The first part evaluates to 1, and the second part simplifies to πlim⁡x→0(sin⁡xx)2=π(1)2=π\pi \lim_{x \to 0} \left(\frac{\sin x}{x}\right)^2 = \pi (1)^2 = \pi. Thus, the overall limit is π\pi.