How to prove a sequence of a function converges uniformly? For $n \in \mathbb{N}$, define the formula, $$f_n(x)= \frac{x}{2n^2x^2+8},\quad x \in [0,1].$$ Prove that the sequence $f_n$ converges uniformly on $[0,1]$, as $n \to \infty$.
I know that the definition says $f_n$ converges uniformly to $f$ if given $\forall \epsilon \gt 0$, $\forall n \geq N$, such that $|f_n(x) - f(x)| \lt \epsilon, \forall n \geq N$ and $\forall x \in [0,1].$ 
I looked first at the pointwise convergence and found that $$\lim_{n \rightarrow \infty} \frac{x}{2n^2x^2+8} = 0, \forall x \in [0,1].$$
So how do I use this to choose an $n \geq N$ such that $|f_n(x) - f(x)| \lt \epsilon$ ?
Right now, I have 
"proof: Let $\epsilon > 0, \exists N \in \mathbb{N}$ such that, n $\geq N \Rightarrow \frac{1}{2n^2+8} \lt \epsilon$,
by $|f_n(x) - 0| = |\frac{x}{2n^2x^2+8}| \leq |\frac{x^2}{2n^2x^2+8}| \leq \frac{1}{2n^2x^2+8} \;\;\;\; \forall x \in [0,1].$
Since $\lim_{n \rightarrow \infty} \frac{x}{2n^2x^2+8} = 0, \forall x \in [0,1]$, $f_n(x)$ will converge uniformly to $0$ on $[0,1]$."
Is this correct? Am I missing something? Is something not correct? I'm unsure about my choice of $N$. Please & thanks!
 A: Another way to maximize : 
$\dfrac{x}{2n^2x^2+8}\leq \dfrac{x}{8nx}=\dfrac{1}{8n}$ where we used the AM-GM inequality 
:http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means to show  
$\dfrac{2n^2x^2+8}{2} \geq \sqrt {2n^2x^2 \cdot 8} \ \ \ \ $  Applying this easy inequality is often a good first way to find a supremum for the function if you have a summation involved. 
So since the function converges pointwise to $f(x)=0$ We have the following result:
$\lim_{n \to \infty}\sup_{0\leq x\leq 1} |f_n(x)-f(x)|=\lim_{n \to \infty}|\dfrac{1}{8n}|=0$
A: You can write
$$f_n(x)={1\over n} g(n\>x)\qquad(0\leq x\leq 1)$$
with
$$g(t)={t\over 2t^2+8}\qquad(0\leq t<\infty)\ .$$
Since $g(t)$ converges to $0$ for $t\to\infty$ we already can say that $g$ is bounded. A quantitative estimate can be obtained as follows: For $t>0$ one has
$$0<g(t)={1\over 8}\ {2\over{t\over 2}+{2\over t}}\leq{1\over 8}\ .$$
Therefore we now have
$$|f_n(x)|\leq{1\over 8n}\qquad(n\geq1,\ 0\leq x\leq1)\ ,$$
which shows that the $f_n$ converge uniformly to $0$ on $[0,1]$.
A: related problem: (I), (II), (III). Here is a systematic technique. In order to find $\sup_{0\leq x\leq 1} |f_n(x)-f(x)| $, you need to maximize the function $\Big|\frac{x}{2n^2x^2+8}\Big|$ over the interval $[0,1]$. Now, let
$$ g(x)=\frac{x}{2n^2x^2+8} \implies g'(x) = \frac{4-n^2 x^2}{(2n^2x^2+8)^2}=0 \implies x=\frac{2}{n}$$
gives the max of the function $g(x)$ which is $g(2/n)=1/8n$. You can check this by checking the sign of $g''(x)$ which should be $< 0$. Hence we have 
$$ \sup_{0\leq x\leq 1} |f_n(x)-f(x)|= \sup_{0\leq x\leq 1} \Big|\frac{x}{2n^2x^2+8}\Big|=\frac{1}{8n}  < \epsilon. $$
A: Pointwise convergence is not enough to say that the function converges uniformly. Here, $$f_n(x)= \frac{x}{2n^2x^2+8},x \in [0,1]$$ has pointwise convergence to $f(x)=0$, so by definition $$|f_n(x)-f(x)|=\bigg|\frac{x}{2n^2x^2+8}-0\bigg|= \bigg|\frac{x}{2n^2x^2+8}\bigg|< \frac{1}{8n}$$
This shows that that the function is uniformly convergent.
