mathop {lim }limits_{x to 0} frac{{5sin x + x | vedclass

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Question

(B) To evaluate the limit $\mathop {\lim }\limits_{x 	o 0} \frac{{5\sin x + x\cos x}}{{2	an x - {x^2}}}$,we divide both the numerator and the denomi

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) To evaluate the limit $\mathop {\lim }\limits_{x 	o 0} \frac{{5\sin x + x\cos x}}{{2	an x - {x^2}}}$,we divide both the numerator and the denominator by $x$:$\mathop {\lim }\limits_{x 	o 0} \frac{{\frac{5\sin x}{x} + \frac{x\cos x}{x}}}{{\frac{2	an x}{x} - \frac{x^2}{x}}}$$= \mathop {\lim }\limits_{x 	o 0} \frac{{5(\frac{\sin x}{x}) + \cos x}}{{2(\frac{	an x}{x}) - x}}$Using the standard limits $\mathop {\lim }\limits_{x 	o 0} \frac{\sin x}{x} = 1$ and $\mathop {\lim }\limits_{x 	o 0} \frac{	an x}{x} = 1$,we get:$= \frac{5(1) + \cos(0)}{2(1) - 0} = \frac{5 + 1}{2} = \frac{6}{2} = 3$

$\mathop {\lim }\limits_{x \to 0} \frac{{5\sin x + x\cos x}}{{2\tan x - {x^2}}}$ is

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