# Limit finding of an indeterminate form

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?

• L'Hopital's rule is a really useful tool, why not use it? – Alex Chamberlain Jun 12 '12 at 9:00
• @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 $x\to 0$. – qoqosz Jun 12 '12 at 9:04

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

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

• What? I thought it was $1$. – Gigili Jun 12 '12 at 9:06
• @Gigili It is, but you can't correct 1 character... – Alex Chamberlain Jun 12 '12 at 9:07
• @Gigili just a typo :) – qoqosz Jun 12 '12 at 9:09
• Good,good. +1 now! – Gigili Jun 12 '12 at 9:11

$$\lim_{x \to 0} \frac{x^3}{\tan (2x)^3}=\lim_{x \to 0} \frac{x^3}{(2x)^3}=\frac18$$

• There is no justification in that line what so ever! – Alex Chamberlain Jun 12 '12 at 9:08
• @AlexChamberlain: only equivalence of $\tan x$ and $x$ is used around $0$ - should it be justified? Certainly, better choice than l'Hopital's rule. – Ilya Jun 12 '12 at 9:11
• 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 $\lim\limits_{x\to 0}$. – Brian M. Scott Jun 12 '12 at 9:16
• -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 $\lim_{n \to \infty} (1+1/n)^n = \lim_{n \to \infty} (1+0)^n = 1$. – Hans Lundmark Jun 12 '12 at 10:16
• 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. – Hans Lundmark Jun 12 '12 at 12:24

$$\lim_{x \rightarrow 0 }\frac{x^3}{\tan^3 (2x) } = \lim_{x \rightarrow 0 } \left ( \frac{2x}{\tan (2x) } \right )^3 \frac{1}{2^3} = \frac{1}{2^3}$$ 

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

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

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

We may resort to $\sin(x)<x<\tan(x),\space 0< x <\frac{\pi}{2}$ 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.

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