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here is the limit I'm trying to find out:

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

Since it is an indeterminate form, I simply applied l'Hopital's Rule and I ended up with:

$\lim_{x\rightarrow 0} \frac{x^3}{\tan^3(2x)} = \lim_{x\rightarrow 0}\frac{6\cos^3(2x)}{48\cos^3(2x)} = \frac{6}{48} = 0.125$

Unfortuntely, as far as I've tried, I haven't been able to solve this limit without using l'Hopital's Rule. Is it possibile to algebrically manipulate the equation so to have a determinate form?

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    @AlexChamberlain take for example: $\lim_{x\to 0} \frac{\sin x}{x}$. When you're applying l'Hopital's rule, you're using the fact that $(\sin x)' = \cos x$ but it's the consequence of $\frac{\sin x}{x} \to 1$ when $x\to 0$.2012-06-12

7 Answers 7

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Use $\lim_{x \to 0} \frac{\sin x}{x} = 1$:

$\lim_{x \to 0} \frac{x^3}{\tan^3 2x} = \lim_{x\to 0} \left( \frac{(2x)^3}{\sin^3 2x} \cdot \frac{\cos^3 2x}{8} \right) \stackrel{[1 \cdot \frac{1}{8}]}{=} \frac{1}{8}$

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    Good,good. +1 now!2012-06-12
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$\lim_{x \to 0} \frac{x^3}{\tan (2x)^3}=\lim_{x \to 0} \frac{x^3}{(2x)^3}=\frac18$

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    I've deleted the last few comments. Let's keep everything civil here.2012-06-12
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$ \lim_{x \rightarrow 0 }\frac{x^3}{\tan^3 (2x) } = \lim_{x \rightarrow 0 } \left ( \frac{2x}{\tan (2x) } \right )^3 \frac{1}{2^3} = \frac{1}{2^3}$ $ $

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$\tan x =x +o(x)$ then $\tan (2x) \sim 2x$ , then : $\frac{x^3}{\tan^3 x} \sim \frac{x^3}{8x^3} \sim \frac 18$

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Another idea using $\,\,\displaystyle{\frac{\sin x}{x}\underset{x\to 0}{\longrightarrow}1\,,\,\cos kx\underset{x\to 0}\longrightarrow 1\,\,(k=\text{a constant})\,\,,\,\sin 2x=2\sin x\cos x}$:

$\frac{x^3}{\tan^3 2x}=\frac{x^3}{\frac{\sin^32x}{\cos^32x}}=\cos^32x\frac{x^3}{\left(2\sin x\cos x\right)^3}=\frac{1}{8}\frac{\cos^32x}{\cos^3x}\left(\frac{x}{\sin x}\right)^3\underset{x\to 0}\longrightarrow \frac{1}{8}\cdot\frac{1}{1}\cdot 1^3=\frac{1}{8}$

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We may resort to $\sin(x) and solve it elementarily. By Squeeze's theorem we get that:

$\lim_{x\rightarrow0}\frac{x^3 \cos^3(2x)}{{(2x)}^3}\leq \lim_{x\rightarrow0}\frac{x^3}{\tan^3(2x)}\leq \lim_{x\rightarrow0}\frac{x^3}{(2x)^3}$

Therefore, taking also into account the symmetry the limit is $\frac{1}{8}$.

The proof is complete.

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$\lim_{x\rightarrow 0} \frac{x^3}{\tan^3(2x)}$=$\lim_{x\rightarrow 0} \frac{x^3}{\frac{\sin^3(2x)}{\cos^3(2x)}}$=$\lim_{x\rightarrow 0} \frac{x^3\cos^3(2x)}{\sin^3(2x)}$=$\lim_{x\rightarrow 0}\frac{x\cdot x\cdot x \cdot\cos^3(2x)}{\sin(2x)\cdot \sin(2x)\cdot\sin(2x)}$=$\lim_{x\rightarrow 0}\frac{2x\cdot 2x\cdot 2x \cdot \frac{1}{8} \cos^3(2x)}{\sin(2x)\cdot\sin(2x)\cdot\sin(2x)}$=$|\lim_{x\rightarrow 0}\frac{\sin x}{x}=1$$\Rightarrow$ $\lim_{x\rightarrow 0}\frac{x}{\sin x}=1$ $\Rightarrow$ $\lim_{x\rightarrow 0}\frac{2x}{\sin 2x}=1$|=$\lim_{x\rightarrow 0}\frac{2x}{\sin(2x)}\cdot\lim_{x\rightarrow 0}\frac{2x}{\sin(2x)}\cdot\lim_{x\rightarrow 0}\frac{2x}{\sin(2x)}\cdot\frac{1}{8}\lim_{x\rightarrow 0}{\cos^3(2x)}$=$1\cdot 1\cdot 1\cdot \frac{1}{8}\lim_{x\rightarrow 0}{\cos^3(2x)}$=$\frac{1}{8}\lim_{x\rightarrow 0}{\cos^3(2x)}$=$\frac{1}{8}{\cos[\lim_{x\rightarrow 0}(2x)]^3}$=$\frac{1}{8}\cos0$=$\frac{1}{8}\cdot 1$=$\frac{1}{8}$