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I am investigating the pointwise an uniform convergence of the sequence of the following function:

$$ f_n(x) = \frac{1}{1+nx^2}, x\in \mathbb{R} $$

Thanks to previous answers, I now know that:

$$ \lim_{n \to \infty}{f_n(x)} = \lim_{n \to \infty}{\frac{1}{1+nx^2}} \\ = 1 \text{ for } x = 0, \\ \text{ and } 0 \text{ for } 0 \lt x \le \infty $$

Which tells me that $f_n(x)$ converges pointwise for $x = 1$ and for $x = 0$. Now, how do I determine if $f_n(x)$ converges uniformly or not?

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You are wrong. For each $x\in\mathbb R$,$$\lim_{n\to\infty}\frac1{1+nx^2}=\begin{cases}1&\text{ if }x=0\\0&\text{ otherwise.}\end{cases}$$And the convergence is not uniform because a uniformly convergent sequence of continuous functions converges to a continuous function.

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  • $\begingroup$ My mistake, sorry. But how do I prove tbe convergence step by step? $\endgroup$ – Omari Celestine Jan 24 '18 at 19:09
  • $\begingroup$ @OmariCelestine If $x=0$, then $\frac1{1+nx^2}=1$ and therefore the limi is $1$. Otherwise, $\lim_{n\to\infty}1+nx^2=+\infty$ and therefore $\lim_{n\to\infty}\frac1{1+nx^2}=0$. $\endgroup$ – José Carlos Santos Jan 24 '18 at 20:58
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The limit of $f$ is not continuous at $0$ so the convergence is not uniform.

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You can see the non-uniform convergence in below illusion: enter image description here

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