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MystifiedBoy @MystifiedBoy

October 2022 1 36 Report

Evaluate the following limit
[tex]\boxed{\sf{\lim_{x \to 0} \dfrac{\sqrt{cos(x)} - \sqrt[3]{cos(x)}}{sin^{2}}}}[/tex]
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$\Large \mathbb{PROBLEM:}$

$\textsf{Evaluate the following limit:}$

$\Large \qquad \displaystyle \sf \lim_{x \to 0} \dfrac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^{2} x} \\ \:$

$\Large \mathbb{SOLUTION:}$

$\quad \displaystyle \sf \lim_{x \to 0} \dfrac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^{2} x}$

$\textsf{We observe that in the present form, the limit}$ $\textsf{becomes }\frac{0}{0}\textsf{ which is indeterminate.}$

$\sf \quad Let\ \sqrt[6]{\cos x} = u \longrightarrow \cos x = u^6$

$\implies \sf \sqrt{\cos x} = \sqrt{u^6} = u^3$

$\implies \sf \sqrt[3]{\cos x} = \sqrt[3]{u^6} = u^2$

$\implies \sf \sin^2 x = 1 - \cos^2 x = 1 - u^{12}$

$\textsf{As x approaches the value 0, u approaches the}$ $\textsf{value 1. The given limit becomes}$

$\quad \displaystyle \sf \lim_{u \to 1} \dfrac{u^3 - u^2}{1 - u^{12}}$

$\displaystyle \sf = \lim_{u \to 1} \dfrac{u^2(u - 1)}{(1 - u^6)(1 + u^6)}$

$\displaystyle \sf = \lim_{u \to 1} \dfrac{u^2(u - 1)}{(1 - u^3)(1 + u^3)(1 + u^6)}$

$\displaystyle \sf = \lim_{u \to 1} \dfrac{-u^2\cancel{(1-u)}}{\cancel{(1 - u)}(1 + u + u^2)(1 + u^3)(1 + u^6)}$

$\displaystyle \sf = \lim_{u \to 1} \dfrac{-u^2}{(1 + u + u^2)(1 + u^3)(1 + u^6)}$

$\sf = \dfrac{-1}{(1 + 1 + 1)(1 + 1)(1 + 1)}$

$\sf = \dfrac{-1}{3(2)(2)}$

$= \boxed{\sf -\dfrac{1}{12}} \Longleftarrow\textsf{Answer}$

maureen22658 Pa help po sa english

maureen22658 Pa help po sa english.

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