# Find Limit

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

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

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?

CompuChip
Homework Helper
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.

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

dynamicsolo
Homework Helper
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...]

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Gib Z
Homework Helper
exploit the heck out of

lim u->0 (sin u)/u = 1 .
Which, again, is just another form of applying small angle approximation/l'hopital/taylor series =D