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The value of lim(xrarrpi)(tan(picos^(2)x...

The value of `lim_(xrarrpi)(tan(picos^(2)x))/(sin^(2)(2x))` is equal to

A

1

B

`pi`

C

`-(pi)/(4)`

D

`(pi)/(2)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the limit \( \lim_{x \to \pi} \frac{\tan(\pi \cos^2 x)}{\sin^2(2x)} \), we will follow these steps: ### Step 1: Substitute \( x = \pi \) First, we substitute \( x = \pi \) into the limit expression: \[ \tan(\pi \cos^2(\pi)) = \tan(\pi \cdot (-1)^2) = \tan(\pi) = 0 \] \[ \sin^2(2\pi) = \sin^2(0) = 0 \] This gives us the indeterminate form \( \frac{0}{0} \). ### Step 2: Apply L'Hôpital's Rule Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the denominator: \[ \lim_{x \to \pi} \frac{\tan(\pi \cos^2 x)}{\sin^2(2x)} = \lim_{x \to \pi} \frac{\frac{d}{dx}[\tan(\pi \cos^2 x)]}{\frac{d}{dx}[\sin^2(2x)]} \] ### Step 3: Differentiate the Numerator Using the chain rule, the derivative of the numerator is: \[ \frac{d}{dx}[\tan(\pi \cos^2 x)] = \sec^2(\pi \cos^2 x) \cdot \frac{d}{dx}[\pi \cos^2 x] \] \[ = \sec^2(\pi \cos^2 x) \cdot (-\pi \cdot 2 \cos x \sin x) = -2\pi \cos x \sin x \sec^2(\pi \cos^2 x) \] ### Step 4: Differentiate the Denominator The derivative of the denominator is: \[ \frac{d}{dx}[\sin^2(2x)] = 2 \sin(2x) \cdot \frac{d}{dx}[2x] = 4 \sin(2x) \cos(2x) \] ### Step 5: Substitute the Derivatives Back into the Limit Now substituting back into the limit gives: \[ \lim_{x \to \pi} \frac{-2\pi \cos x \sin x \sec^2(\pi \cos^2 x)}{4 \sin(2x) \cos(2x)} \] ### Step 6: Simplify the Limit We know that \( \sin(2x) = 2 \sin x \cos x \), so we can simplify: \[ = \lim_{x \to \pi} \frac{-2\pi \cos x \sin x \sec^2(\pi \cos^2 x)}{4 \cdot 2 \sin x \cos x \cos(2x)} \] \[ = \lim_{x \to \pi} \frac{-\pi \sec^2(\pi \cos^2 x)}{4 \cos(2x)} \] ### Step 7: Evaluate the Limit Now we substitute \( x = \pi \): \[ \sec^2(\pi \cos^2(\pi)) = \sec^2(\pi \cdot 1) = \sec^2(\pi) = 1 \] \[ \cos(2\pi) = 1 \] Thus, we have: \[ = \frac{-\pi \cdot 1}{4 \cdot 1} = -\frac{\pi}{4} \] ### Final Answer Therefore, the value of the limit is: \[ \boxed{-\frac{\pi}{4}} \]
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