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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

$\lim _{y \rightarrow 0} \frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4} = $

$\mathop {\lim }\limits_{x \to \infty } (\sqrt {{x^2} + 8x + 3} - \sqrt {{x^2} + 4x + 3} ) = $

$\mathop {\lim}\limits_{x \to 1} \left[ {\left[ {\frac{4}{{{x^2} - {x^{ - 1}}}} - \frac{{1 - 3x + {x^2}}}{{1 - {x^3}}}} \right]^{ - 1} + \frac{{3 \cdot ({x^4} - 1)}}{{{x^3} - {x^{ - 1}}}}} \right] = $

If $[ \cdot ]$ denotes the greatest integer function,then evaluate the limit: $\lim _{x \rightarrow \frac{\pi^{+}}{2}} \frac{[\sin x]-[\cos x]+1}{2}$

$\mathop {\lim }\limits_{n \to \infty } {({4^n} + {5^n})^{1/n}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module