# Lim x→0 sin^3x/(sinx - x)

1. Oct 28, 2009

### Bohrok

$$\lim_{x\rightarrow 0} \frac{\sin^3x}{\sin x - x}$$

I've been trying to find this limit the hard way without l'Hôpital's rule, but no luck. I've tried different things like factoring the difference of cubes to get rid of the denominator, but I can't get out of an indeterminate form. I did get sin3x/(sinx + x) but that didn't help. that x by itself seems to be causing the problem. This one's got me a little frustrated but I must be missing something relatively simple...

2. Oct 28, 2009

### vwishndaetr

Even though you said you tried it without l'Hôpital's rule, did you actually try it?

You can keep using l'Hôpital's rule if you keep getting zero over zero or infinity over infinity.

Just a clue. Use l'Hôpital's rule twice, and see if you can simplify the fraction using the identity sin(x)^2 + cos(x)^2 = 1

3. Oct 28, 2009

### Bohrok

I did use l'Hôpital's rule just so I could find the actual limit, which was -6. I used the identity sin2x + cos2x = 1 but that still didn't help.

4. Oct 28, 2009

### vwishndaetr

Kinda confused.

Then what are you trying to do?

5. Oct 28, 2009

### Bohrok

Say I want to find limx→0tanx/x. I could find it this way
$$\lim_{x\rightarrow 0}\frac{\tan x}{x} = \lim_{x\rightarrow 0}\frac{\frac{\sin x}{\cos x}}{x} = \lim_{x\rightarrow 0} \frac{\sin x}{x}\times \frac{1}{\cos x} = 1\times \frac{1}{\cos 0} = 1$$

without using l'Hôpital's rule and using important limits like I used limx→0sinx/x = 1 above. I want to find the limit in the OP the "long way."

6. Oct 28, 2009

### Bohrok

I'm doing this limit just for fun; not a homework problem. In fact, I don't remember where I found this limit problem.

$$\frac{x^3}{\sin x - x}\times\frac{\sin^2x + x\sin x + x^2}{\sin^2x + x\sin x + x^2} = \frac{x^3 - \sin^3x + \sin^3x}{\sin^3x - x^3}(\sin^2x + x\sin x + x^2)$$

$$= \frac{\sin^3x -(\sin^3x - x^3)}{\sin^3x - x^3}(\sin^2x + x\sin x + x^2) = \left(\frac{\sin^3x}{\sin^3x - x^3} - 1\right)(\sin^2x + x\sin x + x^2)$$
Then I realized the right factor will make the whole expression 0... But I went a slightly different way that looked promising:

$$\frac{x^3}{\sin x - x} = \frac{x^3 - \sin^3x + \sin^3x}{\sin x - x} = \frac{-(\sin x - x)(x^2 + x\sin x + sin^2x)}{\sin x - x} ~+~ \frac{\sin^3x}{\sin x - x}$$

But I couldn't do anything with that last term.

Everything else I've tried was a dead end, or actually brought me back to what I started with and they're not really worth posting.