Evaluate: cos left[ lim_{x rightarrow infty} | vedclass

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Question

(B) Let $L = \cos \left[ \lim_{x \rightarrow \infty} \frac{2 \pi |x| + \pi x}{|x| - 3x} + \lim_{x \rightarrow 0} \frac{\cos \left( \frac{\pi}{2} \cos^

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(B) Let $L = \cos \left[ \lim_{x ightarrow \infty} \frac{2 \pi |x| + \pi x}{|x| - 3x} + \lim_{x ightarrow 0} \frac{\cos \left( \frac{\pi}{2} \cos^2 x ight)}{x^2} ight]$.For the first limit,as $x ightarrow \infty$,$|x| = x$. Thus,$\lim_{x ightarrow \infty} \frac{2 \pi x + \pi x}{x - 3x} = \lim_{x ightarrow \infty} \frac{3 \pi x}{-2x} = -\frac{3 \pi}{2}$.For the second limit,$\lim_{x ightarrow 0} \frac{\cos \left( \frac{\pi}{2} \cos^2 x ight)}{x^2} = \lim_{x ightarrow 0} \frac{\cos \left( \frac{\pi}{2} (1 - \sin^2 x) ight)}{x^2} = \lim_{x ightarrow 0} \frac{\cos \left( \frac{\pi}{2} - \frac{\pi}{2} \sin^2 x ight)}{x^2}$.Using $\cos \left( \frac{\pi}{2} - 	heta ight) = \sin 	heta$,we get $\lim_{x ightarrow 0} \frac{\sin \left( \frac{\pi}{2} \sin^2 x ight)}{x^2}$.Since $\sin 	heta \approx 	heta$ for small $	heta$,this becomes $\lim_{x ightarrow 0} \frac{\frac{\pi}{2} \sin^2 x}{x^2} = \frac{\pi}{2} 	imes (1)^2 = \frac{\pi}{2}$.Substituting these values back into the expression: $L = \cos \left( -\frac{3 \pi}{2} + \frac{\pi}{2} ight) = \cos(-\pi) = -1$.

Evaluate: $\cos \left[ \lim_{x \rightarrow \infty} \frac{2 \pi |x| + \pi x}{|x| - 3x} + \lim_{x \rightarrow 0} \frac{\cos \left( \frac{\pi}{2} \cos^2 x \right)}{x^2} \right]$

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