Show uniform convergence of the sequence $f_n(x) = ( x-x^n/n)$ for $x \in [0,1]$

Question

Show uniform convergence of the sequence $(x-x^n/n)$ on $[0,1]$

To start with, i am not even able to see what the point was limit of this sequence should be. Little confused. like for $x=0$ I think the limit should be $0$ and for $x=1$ it should be $1$ and for values in between i think it should be $x$. Is that correct ? How do i further show its uniform convergence ?

Answer 1

The sequence of functions $f_n$: $\left(x-\frac{x^n}n\right)_n$ is point-wise convergent to the function $f\colon x\mapsto x$ and we have $$\forall x\in[0,1],\quad|f_n(x)-f(x)|=\frac{x^n}n\le \frac1n\xrightarrow{n\to\infty}0$$ so the convergence is uniform.

Answer 2

Your guess for the limit is $f : [0, 1] \to \mathbb{R}$, $f(x) = x$. Let's see if that works.

Let $\varepsilon > 0$, then

$$|f(x) - f_n(x)| = \left|x - (x - \frac{x^n}{n})\right| = \frac{x^n}{n} \leq \frac{1}{n} < \varepsilon$$

for $n \geq N = \lceil\varepsilon^{-1}\rceil$. So $f_n \to f$ uniformly.