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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: $$ \sup_{x \in [-a;a]} |f_n(x) - f(x) | = ... \\ = \sup_{x \in [-a;a]} |\tan(x)| \cdot \left| \frac{1}{1+n\cdot \cos(x)}\right| $$

  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...
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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.

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  • $\begingroup$ because $|\tan(a)| \cdot \left| \frac{1}{1+n\cdot \cos(a)}\right| = c < \infty$, right? $\endgroup$ – trolley Apr 3 at 19:30
  • $\begingroup$ It updated my reply. Being bounded is not enough it needs to tend to zero. $\endgroup$ – maxmilgram Apr 3 at 19:34

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