Uniform convergence, and how to show it?

Question

We've just been introduced to uniform convergence, and the method presented is to take the supremum of the absolute value of the difference between the limit function and the function in the sequence, and show that this is less than $\epsilon$ for $n \ge N$ regardless of $x$.

I think I may have misunderstood something, for I am looking at a problem, and i don't know how to apply above? Essentially, we have a sequence of functions $\frac{1-(x/4)^n}{1+(2/x)^n}$ for $x \in (2,4)$: show uniform convergence towards $1$.... but that function does go towards 1, for all x, as $x/4 < 1$and $2/x< 1$? So is this pointwise convergence I've shown, or is it also uniform?

Answer 1

Formally:

$$\left|\frac{1-\left(\frac x4\right)^n}{1+\left(\frac2x\right)^n}-1\right|=\frac{\left(\frac2x\right)^n+\left(\frac x4\right)^n}{1+\left(\frac2x\right)^n}\xrightarrow[n\to\infty]{}0$$

and the above doesn't depend on a specific $\;x\in(2,4)\;$

More formally: you want to show that for any $\;\epsilon >0\;$ there exists $\;N\in\Bbb N\;$ s.t.

$$n>N\implies |f_n(x)-1|<\epsilon\;,\;\;\text{for any}\;\;x\in (2,4)$$

Well, use the first part and the fact that the num erator on the right side of the equality converges to zero whether the denominator approaches $\;1\;$

Answer 2

The convergence is not uniform on $(2,4).$ Let $$(1)\,\,\,\,f_n(x) = \frac{1-(x/4)^n}{1+(2/x)^n}.$$ We know $f_n(x) \to 1$ pointwise on $(2,4).$ Suppose the convergence is uniform. Crucial fact: Then for any sequence $x_n$ in $(2,4),$ we have $f(x_n) \to 1.$* Plugging $x_n = 4-1/n$ into (1), the numerators then equal $1-(1-(1/4n))^n \to 1-e^{-1/4}.$ The denominators $\to 1.$ Thus $f_n(4-1/n)\to 1- e^{-1/4}\ne 1$ contradiction.

*Suppose $f_n \to f$ uniformly on some set $E.$ Then $\lim_{n\to \infty} \sup_{E}|f_n-f| = 0.$ Let $x_n$ be any sequence in $E.$ Then $$|f_n(x_n) - f(x_n)| \le \sup_{E}|f_n-f|\to 0.$$