The value of mathop {lim }limits_{x to 0} fra | vedclass

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(C) We know that $\mathop {\lim }\limits_{u 	o 0} \frac{1 - \cos u}{u^2} = \frac{1}{2}$ and $\mathop {\lim }\limits_{x 	o 0} \frac{	an x - x}{x^3}

Vedclass · Accepted Answer

$\frac{3}{8}$

Vedclass · Answer

(C) We know that $\mathop {\lim }\limits_{u 	o 0} \frac{1 - \cos u}{u^2} = \frac{1}{2}$ and $\mathop {\lim }\limits_{x 	o 0} \frac{	an x - x}{x^3} = \frac{1}{3}$.Given expression: $L = \mathop {\lim }\limits_{x 	o 0} \frac{1 - \cos (1 - \cos x)}{x(	an x - x)}$.Multiply and divide by $(1 - \cos x)^2$ and $x^4$:$L = \mathop {\lim }\limits_{x 	o 0} \left[ \frac{1 - \cos (1 - \cos x)}{(1 - \cos x)^2} ight] 	imes \left[ \frac{(1 - \cos x)^2}{x^4} ight] 	imes \left[ \frac{x^4}{x(	an x - x)} ight]$.$L = \frac{1}{2} 	imes \left( \mathop {\lim }\limits_{x 	o 0} \frac{1 - \cos x}{x^2} ight)^2 	imes \left( \mathop {\lim }\limits_{x 	o 0} \frac{x^3}{	an x - x} ight)$.$L = \frac{1}{2} 	imes \left( \frac{1}{2} ight)^2 	imes \left( \frac{1}{1/3} ight) = \frac{1}{2} 	imes \frac{1}{4} 	imes 3 = \frac{3}{8}$.

The value of $\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos (1 - \cos x)}}{{x\tan x - {x^2}}}$ is:

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