Uniform convergence $f_n(x)=\frac{n+x}{1+nx}$

Question

Uniform convergence $f_n(x)=\frac{n+x}{1+nx}$ on $E$ and $G$, $E = [0,1], G = [1, +\infty)$

$f(x) = \lim_{ n \to \infty} \frac{n+x}{1+nx} = \frac{1}{x}$

$\lim_{ n \to \infty} \sup|f_n(x) - f(x)| = \lim_{ n \to \infty} \sup|\frac{x^2-1}{nx^2+x}|$

$g(x) = \frac{x^2-1}{nx^2+x}$ and then I've tried to find max of function g

I got $x = -\sqrt{n^2-1} - n$ and stuck. Am I right? What I've done wrong? How to research this?

Answer 1

Hint:

1. $f_n(0) = n$, which does not convergence pointwisely. Therefore what can you infer about uniform continuity on $E = [0,1]$ ? Another way to understand the problem: we have $f_n \in C^0(E)$ for every $n$ but $f \not\in C^0(E)$.
2. Now consider $x \in G = [1, +\infty)$ ; we have that $\frac{x^2 - 1}{nx^2 + x} < \epsilon \implies (\epsilon n -1)x^2 + \epsilon x + 1 > 0$. Now the discriminant is $\Delta = \epsilon^2 - 4\epsilon n + 4$. Take $N > \frac{\epsilon^2 + 4}{4\epsilon}$, then $\Delta < 0$ for every $n > N$, therefore yielding the uniform convergence.

Answer 2

Hint:

$g_n'(x)=0\iff$

$ x=\sqrt{n^2-1}-n=x_n$

as

$$\lim_{t\to 0^+}g_n(t)=+\infty$$

and $\lim_{n\to+\infty}x_n=0$,

$g_n$ is not bounded in$(0,1]$ and the convergence of $(f_n)$ is not uniform at $(0,1]$.

for $a\in (0,1]$ and large enough $n$,

$g_n$ is decreasing at $[a,1]$ and

$$sup_{[a,1]}g_n=g_n(a)$$ with goes to $0$ .

the convergence is uniform at $[a,1]$ but not at$(0,1]$.

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