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$$f_n(x)=\frac{nx^2}{1+nx^2}$$

In the case of non-uniform convergence, find subintervals on which the convergence is uniform and prove your conjecture.

I'm thinking that because both the numerator and denominator are increasing at the same rate as n approaches infinity, then $lim_{n \to \infty}f_n(x)=1$ for all $x \in (0, \infty]$.

For the uniform convergence piece, I'm thinking that $f_n(x)$ converges uniformly on the interval $[\eta,\infty)$, but i'm kind of confused on that part. The definition of uniform convergence, geometrically, says that if we make a vertical "collar" of radius $\epsilon$ around the graph of the limit function $f$, the graphs of all functions in the sequence after the $N$th one must lie within the "collar". But i feel like no matter how large you make $\epsilon$, you can fit in any $f_n(x)$.

What I'm thinking for the uniform convergence proof:

Fix $\eta$ such that $0<\eta<\infty$. Claim that $(f_n)$ converges uniformly on $[\eta, \infty)$ to $f(x)=1$. Fix $\epsilon>0$ and choose $N \in \mathbb{N}$ such that, $N> ?$. Not sure what to pick for $N$.

I'm trying to choose $N$ based on, $d(f_n(x),f(x))<\epsilon$, but i'm struggling to find something to use.

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  • $\begingroup$ Check Dini's theorem (en.wikipedia.org/wiki/Dini%27s_theorem). Using it, I think you can argue that on any compact interval that does not include zero the convergence is uniform (and implied by pointwise). $\endgroup$ – Jan Dec 20 '18 at 2:03
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$\displaystyle f_n(x)=\frac{nx^2}{1+nx^2}=\dfrac{x^2}{\frac{1}{n}+x^2}\to\begin{cases}0 &x=0\\1 &x\neq0\end{cases}$

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  • $\begingroup$ I'm confused by what you are telling me here... how does $\frac{nx^2}{1+nx^2} = \frac{x^2}{\frac{1}{n}+x^2}$? $\endgroup$ – big_math_boy Dec 20 '18 at 2:02
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    $\begingroup$ dividing the numerator and denominator by $n$, $n\neq0$. $\endgroup$ – Yadati Kiran Dec 20 '18 at 2:03
  • $\begingroup$ oh duh, and this is for point-wise convergence, not uniform convergence? $\endgroup$ – big_math_boy Dec 20 '18 at 2:07
  • $\begingroup$ No, $f_n\to f$ uniformly for $x\neq0$ i.e. $x\in(0,a),a\in\mathbb{R}$. $\endgroup$ – Yadati Kiran Dec 20 '18 at 2:08
  • $\begingroup$ What $N$ do you recommend I use to prove this? $\endgroup$ – big_math_boy Dec 20 '18 at 15:53

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