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.
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...]