Uniform convergence proof

Question

How can I show that the function $ f_n(x)=\frac{nx}{1+nx} $ converges uniformly on the interval $ x \in [1, \infty ) $ I have already proven the pointwise limit to be $ f(x)=1 $

I am working with the definition $ \forall \epsilon \gt0 \exists N \in \mathbb{N} $ such that $ \forall x \in [1, \infty) $ and $ \forall n \ge N \mid \frac{nx}{1+nx} - 1 \mid \lt \epsilon $

Do I need to produce a value of N or a value of $ \epsilon $?

Answer 1

I assume you are seeking after a proof that the function $f: x \mapsto 1$ on $[1, \infty[$ is the uniform limit of the sequence $(f_{n})$ because of your title and you saying that you have worked out that $f$ is the pointwise limit.

Let $\varepsilon > 0$. If $N \geq 1,$ then $f_{n} \to f$ uniformly on $[1, \infty[$ iff $$\bigg| \frac{nx}{1+nx} - 1 \bigg| = \frac{1}{1+nx} \leq \frac{1}{1+N} < \varepsilon,$$ which means that $$N > \frac{1}{\varepsilon} - 1.$$ So taking $N := \lceil \frac{1}{\varepsilon} - 1 \rceil + 1$ suffices.

Answer 2

We have $f_n'(x) = {n \over (nx+1)^2}$, hence $|f_n'(x)| \le 1$ for all $x \ge 1$. Then the mean value theorem shows that $|f_n(x)-f_n(y)| \le |x-y|$ from which uniform continuity follows.

Answer 3

If you want to show uniform convergence of the sequence of function $f_n$, Use M-test. That is $M_n=\sup\{|f_n(x)-f(x)|:x\in[1,\infty)\} $,where $f$ is the pointwise limit. Now $f_n\rightarrow f$ uniformly on the given domain iff $M_n\rightarrow 0$ as $n\rightarrow\infty$. Here $M_n=\frac{1}{n+1}(\rightarrow0)$.