Proof for uniform convergence of sequence of functions

Question

I was given this problem:

These are my calculations and I'm asking for verification:

Pointwise limit:

$\lim_{n \to \infty} f_{n}(x) = \lim_{n \to \infty} \frac{x^{2n}}{1+x^{2n}} = \lim_{n \to \infty} \frac{x^{n}}{\frac{1}{x^n}+x^{n}} = 1$

Uniform convergence:

$\mid f_{n}(x)-f(x)\mid = \mid f_{n}(x) - 1\mid = \mid\frac{x^{2n}}{1+x^{2n}} -1 \mid= \mid \frac{x^{n}}{\frac{1}{x^n}+x^{n}} - \frac{\frac{1}{x^n}+x^{n}}{\frac{1}{x^n}+x^{n}}\mid = \mid -\frac{\frac{1}{x^n}}{\frac{1}{x^n}+x^{n}}\mid = \frac{\frac{1}{x^n}}{\frac{1}{x^n}+x^{n}} \leq \frac{1}{x^n}$

Thus:

$\lim_{n \to \infty} sup\{\mid f_{n}(x)-f(x)\mid : x \in [R, \infty)\} = \lim_{n \to \infty} \frac{1}{x^n} = 0.$

From this follows that $f_n(x)$ is uniform convergent

Answer 1

You have written "$|f_n(x) - f(x)| = f_n(x) - 1$". That is not correct. The RHS is negative.

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To do it, an easy way is to note that for every $n \in \Bbb N$, the function $f_n$ is increasing on $[R, \infty)$. And also that $f_n(x) < 1$ for every $n \in \Bbb N$ and $x \le R$.

Now, you know that $$\lim_{n\to\infty} f_n(R) = 1.$$ So, given any $\epsilon > 0$, choose $N \in \Bbb N$ such that $|f_n(R) - 1| < \epsilon$ for all $n \ge N$.
Since $f_n(R) \le f_n(x) < 1$ for all $n \in \Bbb N$ and $x > R$, it follows that $$|f_n(x) - 1| < \epsilon,$$ for all $x > R$ and $n \in \Bbb N$, as desired.

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EDIT: This is after the post was edited.
You write

$$\lim_{n \to \infty} \sup\{\mid f_{n}(x)-f(x)\mid : x \in [R, \infty)\} = \lim_{n \to \infty} \frac{1}{\color{#FF0000}x^n} = 0.$$

It should actually be

$$\lim_{n \to \infty} \sup\{\mid f_{n}(x)-f(x)\mid : x \in [R, \infty)\} = \lim_{n \to \infty} \frac{1}{\color{#FF0000}R^n} = 0.$$

Answer 2

For all $x \in [R, +\infty)$, we find that, since $R > 1$, therefore we have $$ \begin{align} \lim_{n \to \infty} f_n(x) &= \lim_{n \to \infty} \frac{ x^{2n} }{ 1 + x^{2n} } \\ &= \lim_{n \to \infty} \frac{1}{ \frac{1}{x^{2n}} + 1} \\ &= \frac{1}{0+1} \qquad [\mbox{ as $x \geq R > 1$, so $\lim_{n \to \infty} \frac{1}{x^{2n}} = 0$ } ]\\ &= 1. \end{align} $$ Now let $f \colon [R, +\infty) \rightarrow \mathbb{R}$ be defined by the formula $$ f(x) \colon= 1 \qquad \mbox{for all } x \in [R, +\infty). \tag{0} $$ Then the sequence $\left( f_n \right)_{n \in \mathbb{N}}$ converges pointwise to the function $f$ on $[R, +\infty)$.

Let us now check if this convergence is uniform.

We note that, for all $n \in \mathbb{N}$ and for all $x \in [R, +\infty)$, we have $$ \begin{align} \left\lvert f_n(x) - f(x) \right\rvert &= \left\lvert \frac{x^{2n}}{1+x^{2n}} - 1 \right\rvert \\ &= \left\lvert \frac{-1}{1+x^{2n}} \right\rvert \\ &= \frac{1}{\left\lvert 1+x^{2n} \right\rvert } \\ &= \frac{1}{ 1+x^{2n} } \\ &< \frac{1}{x^{2n} } \\ &\leq \frac{1}{R^{2n}}. \tag{1} \end{align} $$

Now as $R > 1$, so $$ \lim_{n \to \infty} \frac{1}{R^{2n}} = 0. $$

Thus from (2) we can conclude that, given a real number $\varepsilon > 0$, we can find a natural number $N = N(\varepsilon)$ such that $$ \left\lvert \frac{1}{R^{2n}} - 0 \right\rvert = \frac{1}{R^{2n}} < \varepsilon $$ for any natural number $n > N$. In fact, we can take $N$ to be any natural number greater than the quantity $$ \begin{cases} \frac{ - \ln \varepsilon }{ \ln R} \ \mbox{ if } \varepsilon \neq 1, \\ 1 \ \mbox{ if } \varepsilon = 1. \end{cases} $$

So using (1) we can conclude that, given a real number $\varepsilon > 0$, we can find a natural number $N$ such that $$ \left\lvert f_n(x) - f(x) \right\rvert < \varepsilon $$ for all $ x \in [R, +\infty)$ and for any natural number $n > N$.

Hence the sequence $\left( f_n \right)_{n \in \mathbb{N}}$ indeed converges uniformly to the function $f$ defined by the formula (0) above.