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$$\lim_{x\to 0}\frac{x^3}{\tan^3(2x)} $$

My textbook has an answering of $\frac{1}{8}$ and I'm quite confused on how they got that. Only thing that I could see to get an $8$ would be $2^3$ from $2x^3$.

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Just figured the question out! Far easier than I was making it out to be. – adeshina lawel Apr 29 '13 at 0:30

Try solving: $$\lim_{x\rightarrow 0}\frac{x}{\tan(2x)}$$by itself. That should give you an idea of how to tackle your problem.

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Okay, I'll do this. – adeshina lawel Apr 29 '13 at 0:20

You have $\tan(x) \sim x$ as $x\to 0$. So $\tan^3(2x)\sim 8x^3$ as $x\to 0$. Now draw your conclusion.

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What does ~ mean? – adeshina lawel Apr 29 '13 at 0:22
$f(x)\sim g(x)$ as $x\to a$ if $f(x)/g(x)\to 1$ as $x\to a$. – ncmathsadist Apr 29 '13 at 1:00

I'll assume that you know that $$\lim_{x\to 0}\frac{\text{sin}(x)}{x}=1$$ by knowing this you can show by doing a simple substitution $u=ax$ that$$\lim_{x\to 0}\frac{\text{sin}(ax)}{bx}=\frac{a}{b}$$this will also give you $$\lim_{x\to 0}\frac{bx}{\text{sin}(ax)}=\frac{1}{\lim_{x\to 0}\frac{\text{sin}(ax)}{bx}}=\frac{b}{a}$$knowing this and since $\text{tan}(x)=\frac{\text{sin}(x)}{\text{cos}(x)}$ you can show that $$\lim_{x\to 0}\frac{\text{tan}(x)}{x}=1$$ and $$\lim_{x\to 0}\frac{\text{tan}(ax)}{bx}=\frac{a}{b}$$ which implies that$$\lim_{x\to 0}\frac{bx}{\text{tan}(ax)}=\frac{b}{a}$$ that's all you need to solve your question and similar questions since$$\begin{align}\lim_{x\to 0}\frac{x^3}{\text{tan}^3(2x)}&=\lim_{x\to 0}\frac{x}{\text{tan}(2x)}\times \lim_{x\to 0}\frac{x}{\text{tan}(2x)}\times \lim_{x\to 0}\frac{x}{\text{tan}(2x)}\\&=\frac{1}{2}\times \frac{1}{2} \times \frac{1}{2}\\&=\frac{1}{8}.\end{align}$$

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$$\frac{x^3}{\tan^32x}=\frac{\cos^32x}8\left(\frac{2x}{\sin2x}\right)^3\xrightarrow [x\to0]{}\frac{1}{8}\cdot 1^3\ldots$$

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