mathop {lim }limits_{x to 0} frac{{{x^3}cot x | vedclass

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(C) We evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{{x^3}\cot x}}{{1 - \cos x}}$.Using the identity $1 - \cos x = 2\sin^2(x/2)$ or mul

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) We evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{{x^3}\cot x}}{{1 - \cos x}}$.Using the identity $1 - \cos x = 2\sin^2(x/2)$ or multiplying by the conjugate:$\mathop {\lim }\limits_{x 	o 0} \frac{{{x^3}\cos x}}{{\sin x (1 - \cos x)}} = \mathop {\lim }\limits_{x 	o 0} \left( \frac{x}{\sin x} ight) \cdot \frac{x^2}{1 - \cos x} \cdot \cos x$.Since $\mathop {\lim }\limits_{x 	o 0} \frac{x}{\sin x} = 1$ and $\mathop {\lim }\limits_{x 	o 0} \frac{x^2}{1 - \cos x} = \mathop {\lim }\limits_{x 	o 0} \frac{x^2}{2\sin^2(x/2)} = \mathop {\lim }\limits_{x 	o 0} \frac{2}{( \sin(x/2) / (x/2) )^2} = 2$.Thus,the limit is $1 	imes 2 	imes \cos(0) = 1 	imes 2 	imes 1 = 2$.

$\mathop {\lim }\limits_{x \to 0} \frac{{{x^3}\cot x}}{{1 - \cos x}} = $

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