# Find Limit

1. Jan 4, 2008

### SonyDvDPro

Find $$\mathop {\lim }\limits_{x \to 0} \frac{{\tan (nx) - n\tan (x)}} {{n\sin (x) - \sin (nx)}}$$

2. Jan 4, 2008

### CompuChip

This looks like homework.
Taylor expansion? L'Hôpitals rule?

3. Jan 4, 2008

### nanoWatt

I'm taking a guess here, but couldn't tan(nx) be extracted to sin(nx) / cos(nx) ?

Or is that complicating things, or just wrong?

4. Jan 4, 2008

### CompuChip

That's right and possible though I doubt it will simplify things (but unless you know the Taylor expansion / small argument approximation / derivative and what more you may need to calculate the limit for the tangent, you can use tan = sin/cos to find them). But I think it is time for the TS to show some work.

Last edited: Jan 4, 2008
5. Jan 4, 2008

### Gib Z

Small angle approximation, Taylor expansion and L'Hopital's rule is all exactly the same :( Well, at least here it is.

6. Jan 4, 2008

### dynamicsolo

Another approach, after rewriting the tangent terms as sine/cosine terms, would be to divide top and bottom by x and, with multiplications above and below by n in appropriate places, exploit the heck out of

lim u->0 (sin u)/u = 1 .

[EDIT: Ah-haha! Not so simple as this one looks. This approach does nothing to fix the problem with the denominator. And l'Hopital doesn't fare much better, as you'll get 0 - 0 in the denominator endlessly. I'm checking to see if there's a "nice" way to eliminate the difference in the denominator...]

Last edited: Jan 5, 2008
7. Jan 5, 2008

### Gib Z

Which, again, is just another form of applying small angle approximation/l'hopital/taylor series =D