# Uniform convergence of $\frac{n\cdot \sin \left(x\right)}{1+n\cdot \cos \left(x\right)}$

Check uniform convergence of $$f_n(x) =\frac{n\cdot \sin \left(x\right)}{1+n\cdot \cos \left(x\right)}$$ on $$[-a;a$$] where $$0
ok so firstly I check pointwise convergence $$\lim_{n\rightarrow \infty} f_n(x) = \tan(x)$$ Ok. So now I can check uniform convergence: $$\sup_{x \in [-a;a]} |f_n(x) - f(x) | = ... \\ = \sup_{x \in [-a;a]} |\tan(x)| \cdot \left| \frac{1}{1+n\cdot \cos(x)}\right|$$

1. If $$\forall \epsilon >0.$$ $$a-\epsilon < \pi/2$$ then $$\sup_{x \in [-a;a]} |\tan(x)| \cdot \left| \frac{1}{1+n\cdot \cos(x)}\right| = +\infty$$ so then $$f_n$$ doesn't uniformly converge. But how to deal with other cases? anywhere close to $$\pi/2$$ I have the same result...

Note that for $$0:
$$\sup_{x \in [-a;a]} |\tan(x)| \cdot \left| \frac{1}{1+n\cdot \cos(x)}\right|= |\tan(a)| \cdot \left| \frac{1}{1+n\cdot \cos(a)}\right| \to 0$$ for $$n\to\infty$$.
So it is uniformly convergent. You point about $$a$$ being close to $$\pi/2$$ is not relevant because $$a$$ is fixed.
• because $|\tan(a)| \cdot \left| \frac{1}{1+n\cdot \cos(a)}\right| = c < \infty$, right? – trolley Apr 3 at 19:30