# Find the limit of the sequence of functions$f_n(x) = \frac{x^n-1}{x^n+1}$ & determine if it is uniform convergence on (1,$\infty$) & on (2,$\infty$).

Find the limit of the sequence of functions$f_n(x) = \frac{x^n-1}{x^n+1}$ and determine if it is uniform convergence on the intervals (1,$\infty$) and on (2,$\infty$).

my thoughts:

if x > 1 $\large \lim_{n \rightarrow \infty}f_n(x) = \frac{x^n-1}{x^n+1}$ $\large \lim_{n \rightarrow \infty}\frac{1-\frac{1}{x^n}}{1+\frac{1}{x^n}} = 1$

$\large f_n(x)$ converges uniformly if $\forall \epsilon > 0 \exists N : \forall n > N, \forall x < x < \infty \implies |\frac{x^n-1}{x^n+1}-1|$

$\large = |\frac{-2}{x^n+1}| = \frac{2}{x^n+1} < \frac{2}{x^n} < \epsilon$

$\frac{2}{x^n} < \epsilon$ therefore $\large x^n > \frac{2}{e}$ and $\large nlnx > ln(\frac{2}{e})$ and $n > \frac{\large ln2 - ln\epsilon}{lnx}$ so N = $\frac{\large ln2 - ln\epsilon}{lnx}$

if x > 1 then $x^n+1 > 2$ so $n>N=\frac{\large ln2 - ln\epsilon}{lnx}$ whenever x > 1 so it is uniformly convergent on (2,$\infty$)

$f_n(x)$ is not uniformly convergent on (1,$\infty$) if $\exists \epsilon > 0 : \forall N \exists n > N \exists x \epsilon 1 < x < \infty : |f_n(x) - 1| \geq \epsilon$

$\large \frac{2}{x^n+1} \geq \epsilon$ for x = 1 and n = 1, $\large \frac{2}{x^n+1} = 1 \geq \epsilon$

so $f_n(x)$ is not uniformly convergent on (1,$\infty$)

The sequence $f_n$ does not converge to $1$ for $x=1$; it is $0$ at $x=1$.

Recall that a uniformly convergent sequence of continuous functions converges to a continuous function. Since the terms $f_n(x)$ are continuous on $[1,\infty)$, but $f$ is discontinuous at $1$, then the convergence cannot be uniform on the interval $[1,\infty)$.

In fact, the convergence is not uniform on the open interval $(1,\infty)$. Note that we can take $\epsilon =\frac{1}{e+1}>0$. Then, for $x\in(1,\infty)$, we can take $x=1+1/n$. Since as $n\to \infty$, $f_n\to (e-1)/(e+1)$, then for all $N$, there exist an $n>N$, such than $|f_n(x)-1|>\epsilon=\frac{1}{e+1}$. And we conclude that the convergence is not uniform on the open interval.

If we look at convergence on any half-closed interval $[a,\infty)$, then we have

$$\left|\frac{x^n-1}{x^n+1}-1\right|=\frac{2}{x^n+1}<\frac{2}{a^n}<\epsilon$$

whenever $n>\frac{\log(2/\epsilon)}{\log a}$. And we have uniform convergence on any half-open interval $[a,\infty)$, $a>1$.

• Please let me know how I can improve my answer. I really want to give you the best answer I can. - Mark Commented Nov 27, 2015 at 18:35