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For $n=1,2,3,\dots,$ and $|x| < 1$ I need to prove that $\frac{x}{1+nx^2}$ converges uniformly to zero function. How ?. For $|x| > 1$ it is easy.

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Let $f_n(x)=\frac{x}{1+nx^2}$ then we have $$f'_n(x)=\frac{1-nx^2}{(1+nx^2)^2}=0\iff x=n^{-1/2}:=x_n$$ hence $$||f_n||_\infty=f_n(x_n)=\frac{1}{2}x_n\to0$$ so we have the uniform convergence to $0$.

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  • $\begingroup$ This is really awesome.. As in Weirstrass M test I was aware that I should make some bound for each $f_n$ like $|f_n|\leq M_n$ and then see if $\sum M_n < \infty$.. But i was not sure how to choose that bound But now your idea of considering derivative test to choose supremum is very helpful now... Thank you again! :) $\endgroup$ – user87543 Nov 23 '13 at 17:42
  • $\begingroup$ You're welcome. $\endgroup$ – user63181 Nov 23 '13 at 18:41
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You have $${x\over 1 + nx^2} = {1\over \sqrt{n}} {{\sqrt{n}x\over 1 + nx^2}} $$ The function $x\mapsto{x\over 1 + x^2}$ is bounded; let $M$ be the supremum of its absolute value. Then you have $$\left|{x\over 1 + nx^2}\right| \le {M\over \sqrt{n}}.$$

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$f_n(0)=0\to0$ and for $x\ne0,$ $|f_n(x)|=\left|\dfrac{x}{1+nx^2}\right|\le\left|\dfrac{1}{nx}\right|\to0$ as $n\to\infty$

So, $f(x)=\displaystyle\lim_{n\to\infty}f(x)\equiv0$ on $(-1,1).$ Let $\forall~n\ge1,$$$M_n=\displaystyle\sup_{|x|<1}\left|\dfrac{x}{1+nx^2}-0\right|=\displaystyle\sup_{|x|<1}\left|\dfrac{x}{1+nx^2}\right|=\displaystyle\sup_{|x|<1}\dfrac{|x|}{1+nx^2}$$

By the Jensen's inequality, for $x\ne0,$ $$\dfrac{\dfrac{1}{|x|}+\dfrac{nx^2}{|x|}}{2}\ge\sqrt n\implies0\le\dfrac{|x|}{1+nx^2}\le\dfrac{1}{2\sqrt n}\to0$$

Consequently by Squeeze theorem $M_n\to0.$

Thus $f_n\rightrightarrows f$ on $(-1,1).$

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