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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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    L'Hopital's rule is a really useful tool, why not use it?2012-06-12
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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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    What? I thought it was $1$.2012-06-12
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    @Gigili It is, but you can't correct 1 character...2012-06-12
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    @Gigili just a typo :)2012-06-12
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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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    There is no justification in that line what so ever!2012-06-12
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    @AlexChamberlain: only equivalence of $\tan x$ and $x$ is used around $0$ - should it be justified? Certainly, better choice than l'Hopital's rule.2012-06-12
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    Both equalities are false as written: the limit on the left is not the function in the middle, and the function in the middle is not the number $1/8$. The middle item requires $\lim\limits_{x\to 0}$.2012-06-12
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    @Ilya YES! L'Hopital's rule is a proven theorem of Mathematical analysis. The statement above is a hand wavy guess at what a limit could be.2012-06-12
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    @AlexChamberlain: "hand wavy guess"? TIL.2012-06-12
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    -1. It's quite misleading to write the solution like this (making approximations without showing the error terms), even though strictly speaking the equalitites are true. If you show this to students, they will soon start making mistakes like $\lim_{n \to \infty} (1+1/n)^n = \lim_{n \to \infty} (1+0)^n = 1$.2012-06-12
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    @HansLundmark: I didn't know I was that popular. I won't show this to students, don't worry. I refuse to accept responsibility for their mistakes.2012-06-12
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    Well, the person asking this question seems to study calculus, so you've already showed it to at least one student...2012-06-12
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    My point is that some justification is needed for why replacing $\tan x$ by $x$ gives the correct result in this case, but not for $\lim_{x \to 0}(x - \tan x)/x^3$, for example.2012-06-12
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    @HansLundmark: There are much more important things to be worried about, I'm pretty sure a student at this level knows we're using the fact that $lim_{x \to 0} \frac{x}{\tan x}=1$. I'm fond of short answers especially when they have the potential to achieve what's being asked immediately. I'm more worried about the situation when a student writes about a page for a question like this. My answer is short and complete, and I think most people know it's awesome like that.2012-06-12
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    I solved **this** problem, not the one you mentioned. I won't add anything to my answer.2012-06-12
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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$ $\lim_{x\rightarrow 0}\frac{x}{\sin x}=1$ $\Rightarrow$ $\lim_{x\rightarrow 0}\frac{2x}{\sin 2x}=1$|=$\lim_{x\rightarrow 0}\frac{2x}{\sin(2x)}\cdot\lim_{x\rightarrow 0}\frac{2x}{\sin(2x)}\cdot\lim_{x\rightarrow 0}\frac{2x}{\sin(2x)}\cdot\frac{1}{8}\lim_{x\rightarrow 0}{\cos^3(2x)}$=$1\cdot 1\cdot 1\cdot \frac{1}{8}\lim_{x\rightarrow 0}{\cos^3(2x)}$=$\frac{1}{8}\lim_{x\rightarrow 0}{\cos^3(2x)}$=$\frac{1}{8}{\cos[\lim_{x\rightarrow 0}(2x)]^3}$=$\frac{1}{8}\cos0$=$\frac{1}{8}\cdot 1$=$\frac{1}{8}$