[College Calc 01-03. Limits of Functions] #03

Claim: $\lim_{x\rightarrow 0} \dfrac{\tan^nx-\sin^nx}{x^{n+2}}=\dfrac{n}{2}\;, \;\;\;\;\; \forall n\in\mathbb{R}$ Proof: $\begin{aligned} \lim_{x\rightarrow 0} \dfrac{\tan^nx-\sin^nx}{x^{n+2}} &=\lim_{x\rightarrow 0} \dfrac{\tan^nx(1-\cos^nx)}{x^{n+2}} \\ \\ &=\lim_{x\rightarrow 0}\dfrac{\tan^nx}{x^n}\cdot \dfrac{1-\cos^nx}{x^2}\\ \\ &=\lim_{x\rightarrow 0}\dfrac{\tan^nx}{x^n}\cdot \lim_{x\rightarrow 0}\dfrac{1-\cos^nx}{x^2} \\ \\ &=1\cdot \lim_{x\rightarrow 0}\dfrac{\frac{\mathrm{d}}{\mathrm{d}x}(1-\cos^nx)}{\frac{\mathrm{d}}{\mathrm{d}x}(x^2)} \\ \\ &=\lim_{x\rightarrow 0}\dfrac{n\cos^{n-1}x\sin x}{2x}\\ \\ &=\lim_{x\rightarrow 0}\dfrac{n\cos^{n-1}x}{2}\cdot \lim_{x\rightarrow 0}\frac{\sin x}{x} \\ \\ &=\frac{n}{2} \end{aligned}$

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In the above problem, $n=\dfrac{2}{3}$ . So the limit is, $\begin{aligned} q=\lim_{x\rightarrow 0} \dfrac{\tan^{\frac{2}{3}}x-\sin^{\frac{2}{3}}x}{x^p\sin (5x)} &=\lim_{x\rightarrow 0} \dfrac{\tan^{\frac{2}{3}}x-\sin^{\frac{2}{3}}x}{x^{\frac{8}{3}}} \cdot \frac{x^{\frac{8}{3}}}{x^p\cdot \dfrac{\sin (5x)}{5x}\cdot 5x} \\ &=\dfrac{\frac{2}{3}}{2}\cdot \dfrac{\displaystyle \lim_{x\to 0}\;x^{\frac{5}{3}-p}}{5\cdot \displaystyle \lim_{x\to 0}\;\dfrac{\sin (5x)}{5x}} \\ &=\frac{1}{3}\cdot \frac{1}{5} \lim_{x\to 0}\;x^{\frac{5}{3}-p} \end{aligned}$ Since $q$ is non-zero, $\dfrac{5}{3}-p=0\implies p=\dfrac{5}{3}\implies q=\dfrac{1}{15}$

Therefore, $120(p+q)=120\left(\dfrac{5}{3}+\dfrac{1}{15}\right)=\boxed{208}$

$\lim_{x\to 0} \frac{\sqrt[3]{\tan^2 x} - \sqrt[3]{\sin^2 x}}{x^p \sin (5x)} \\ \, \\ = \lim_{x\to 0} \frac{(\sqrt[3]{\tan x}-\sqrt[3]{\sin x}){\color{#3D99F6}(\sqrt[3]{\tan x}+\sqrt[3]{\sin x})}}{x^p \cdot 5x} \\ \, \\ = \lim_{x\to 0}\frac{{\color{#3D99F6}2\sqrt[3]{x}}(\sqrt[3]{\tan x}-\sqrt[3]{\sin x})(\sqrt[3]{\tan^2 x}+\sqrt[3]{\tan x}\sqrt[3]{\sin x}+\sqrt[3]{\sin^2 x})}{x^p \cdot 5x \cdot {\color{#E81990}(\sqrt[3]{\tan^2 x}+\sqrt[3]{\tan x}\sqrt[3]{\sin x}+\sqrt[3]{\sin^2 x})}} \\ \, \\ =\lim_{x\to 0}\frac{2\sqrt[3]{x}(\tan x-\sin x)}{x^p \cdot 5x \cdot {\color{#E81990}3\sqrt[3]{x^2}}} \\ \, \\ = \lim_{x\to 0} \frac{2\sqrt[3]{x} \cdot {\color{goldenrod}\tan x}{\color{limegreen}(1-\cos x)}}{x^p \cdot 5x \cdot 3\sqrt[3]{x^2}} \\ \, \\ = \lim_{x\to 0} \frac{2\cdot x^{1/3} \cdot {\color{goldenrod}x} \cdot {\color{limegreen}\frac{1}{2}x^2}}{x^p \cdot 5x \cdot 3x^{2/3}} \\ \, \\ = \lim_{x\to 0} \frac{x^{5/3}}{15x^p} = q$

Since $q\neq 0$ , we have $p=5/3$ and thus $q=1/15.$ Therefore, $120(p+q) = 200 + 8 = 208.$

P.S. I would love someone to try and solve the entire thing with l'hopital.

I WA the problem ~~

By taylor expanding the function $\displaystyle \frac{\sqrt[3]{\tan^2 x}-\sqrt[3]{\sin^2 x}}{\sin (5x)}$ , we get $\frac{x^{\frac{5}{3}}}{15}+\frac{79x^{\frac{11}{3}}}{270}+\frac{42647x^{\frac{17}{3}}}{48600}+\frac{4300659x^{\frac{23}{3}}}{18370800}+O(x^{\frac{25}{3}})$ which leads us to choosing $p=\frac{5}{3}$ and $q=\frac{1}{15}$ .