<s>[INST] Evaluating Convergence (Uniform) Hi Guys was attempting this question and was wondering if I was doing the question correctly?
Determine whether or not the sequence of functions is uniformly convergent:-
$$g_n:(0,1)\to \mathbb{R}$$
$$g_n(x) = \frac{n^3+1}{n^3x^2+1}, x\in(0,1)$$
Checking point wise convergence first 
$$\lim_{n\to \infty}g_n(x) =  \lim_{n\to \infty}\frac{n^3+1}{n^3x^2+1}$$
Dividing by $n^3$gives the following :-
$$\lim_{n\to \infty}g_n(x) =  \lim_{n\to \infty}\frac{1+\frac{1}{n^3}}{x^2+\frac{1}{n^3}}$$
Taking the Limit as n $\to \infty$ gives the following
$$\lim_{n\to \infty}g_n(x) =  \frac{1+\frac{1}{\infty^3}}{x^2+\frac{1}{\infty^3}}$$
$$\lim_{n\to \infty}g_n(x) =  \frac{1+0}{x^2+1} = \frac{1}{x^2}$$
Therefore by point wise convergence the sequence of functions converges to the previous function.
In order to determine the uniform convergence we must analyze the follwing
$$M_n =  sup|f_n(x)-f(x)|,x\in \mathbb{R}$$
$$|f_n(x)-f(x)|$$
$$|\frac{n^3+1}{n^3x^2+1} - \frac{1}{x^2}|$$
$$\frac{(n^3x^2+x^2)-(n^3x^2+1)}{(n^3x^2+1)(x^2)}$$
$$|\frac{x^2-1}{(n^3x^2+1)(x^2)}|$$
The mod gives
$$\frac{x^2+1}{(n^3x^2+1)(x^2)}$$
is it accurate to say the following when checking to see uniform convergence
$$\frac{x^2+1}{(n^3x^2+1)(x^2)} < \frac{1}{n^3}$$
$$\lim{n \to \infty} $$
Therefore I can conlclude that $$sup|f_n(x)-f(x)|\to 0$$
Therefore the function is uniformly convergent? Oh am i wrong in my evaluation?
 [/INST] For $ x\in (0,1)$  and $ n$ large enough,
$$g_n(x)=|f_n(x)-\frac{1}{x^2}|=\frac{|x^2-1|}{(n^3x^2+1)x^2}$$
Now take the sequence $ (x_n) $ such that
$$n^3x_n^2=1$$ or
$$x_n=n^{-\frac 32}=\frac{1}{n^{\frac 32}}$$
Then,  $ x_n\in (0,1)$ and
$$|f_n(x_n)-f(x_n)|=\frac{|n^{-3}-1|}{2n^{-3}}$$
$$=\frac 12|1-n^{3}| \to +\infty$$
But
$$|f_n(x_n)-f(x_n)|\le  \sup_{(0,1)}|f_n-f|$$
thus
$$\lim_{n\to+\infty}\sup_{(0,1)}|f_n-f|=+\infty$$
The convergence is not uniform at $(0,1)$.
It is uniform at $(a,1) $ with $0<a<1$.
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