Define $f_n: \mathbb{R} \to \mathbb{R}$ by $f_n(x)= \dfrac{n+\cos(x)}{10n+\cos^2(x)}$.

(a) Find the limit $f(x) = \lim\limits_{n \to \infty} f_n(x)$

(b) Does $f_n\to f$ converge uniformly on $\mathbb R$?

Not quite sure how to start this? I can figure out the limit but not sure how to show whether it converges uniformly...

  • $\begingroup$ What do you have for $f$? Then what is $\sup_{x \in \mathbb{R}} \left| f_n(x) - f(x) \right| = \|f_n - f\|_\infty$? $\endgroup$ – Jon Warneke Dec 2 '15 at 2:04
  • $\begingroup$ For all $x$ we have $(n-1)/(10n+1)\leq f_n(x)\leq (n+1)/10n.$ $\endgroup$ – DanielWainfleet Dec 2 '15 at 16:21

When , $x=(2n-1)\frac{\pi}{2}$ then $\cos x=0$ and then $f_n(x)\to \frac{1}{10}$.

When , $x\not=(2n-1)\frac{\pi}{2}$ then $0<|\cos x|\le 1$. Then , $f_n(x)\to \frac{1}{10}$.

Hence , $\displaystyle f(x)=\lim_n f_n(x)=\frac{1}{10}$.

Now , $\displaystyle \sup_n|f_n(x)-f(x)|\to 0$ as $n\to \infty$. So , the sequence of function $\{f_n(x)\}$ converges uniformly.


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