lim limits _{x to 0} frac{{{{(sin x - tan x)} | vedclass

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(C) To evaluate the limit $L = \lim \limits _{x 	o 0} \frac{{{{(\sin x - 	an x)}^2} - {{(1 - \cos 2x)}^4} + {x^5}}}{{7\cdot{{({{	an }^{ - 1}}x)}^7}

Vedclass · Accepted Answer

$1/3$

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(C) To evaluate the limit $L = \lim \limits _{x 	o 0} \frac{{{{(\sin x - 	an x)}^2} - {{(1 - \cos 2x)}^4} + {x^5}}}{{7\cdot{{({{	an }^{ - 1}}x)}^7}\, + {{({{\sin }^{ - 1}}x)}^6}+ 3{{\sin }^5}x}}$,we analyze the leading terms as $x 	o 0$.Numerator:$(\sin x - 	an x)^2 = (x - \frac{x^3}{6} - (x + \frac{x^3}{3}))^2 + O(x^8) = (-\frac{x^3}{2})^2 + O(x^8) = \frac{x^6}{4} + O(x^8)$.$(1 - \cos 2x)^4 = (1 - (1 - \frac{(2x)^2}{2}))^4 = (2x^2)^4 = 16x^8$.Thus,the numerator is $\frac{x^6}{4} - 16x^8 + x^5 \approx x^5$ as $x 	o 0$.Denominator:$7(	an^{-1} x)^7 + (\sin^{-1} x)^6 + 3(\sin x)^5 \approx 7x^7 + x^6 + 3x^5$.The leading term is $3x^5$.Limit $L = \lim \limits _{x 	o 0} \frac{x^5}{3x^5} = \frac{1}{3}$.

$\lim \limits _{x \to 0} \frac{{{{(\sin x - \tan x)}^2} - {{(1 - \cos 2x)}^4} + {x^5}}}{{7\cdot{{({{\tan }^{ - 1}}x)}^7}\, + {{({{\sin }^{ - 1}}x)}^6}+ 3{{\sin }^5}x}}$ is equal to

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