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 $0<a< \frac{\pi}{2}$
ok so firstly I check pointwise convergence
$$ \lim_{n\rightarrow \infty} f_n(x) = \tan(x) $$
Ok. So now I can check uniform convergence:
$$ \sup_{x \in [-a;a]} |f_n(x) - f(x) | = ... \\ = \sup_{x \in [-a;a]} |\tan(x)| \cdot \left| \frac{1}{1+n\cdot \cos(x)}\right| $$
- 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 $0<a<\pi/2$:
$$\sup_{x \in [-a;a]} |\tan(x)| \cdot \left| \frac{1}{1+n\cdot \cos(x)}\right|= |\tan(a)| \cdot \left| \frac{1}{1+n\cdot \cos(a)}\right| \to 0$$ 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.