How to show that $f(x)=x^n$ converges pointwise in domain $D=[0,1)$.

Question

Suppose $f_n:\mathbb{R}\to\mathbb{R}$ is defined by $f_n(x)=x^n$ where $n\in\mathbb{N}$, and $D=[0,1)$. I wish to show that $f(x)=x^n$ converges to $f_n(x)=0$ pointwise as $n\to\infty$, i.e.

$\forall\epsilon>0,\forall x\in D, \exists N\in\mathbb{N} \text{ s.t. } \forall n\ge N, |f_n(x)-f(x)|<\epsilon$

My attempt:

Suppose $\epsilon>0$ and $x\in D$. Choose $N=\left\lceil \log_{x} (\epsilon) \right\rceil$. Letting $n\ge N$, we have

$$|f_n(x)-f(x)|=|x^n-0|=x^n\le x^N\le x^{\log_{x} (\epsilon)}=\epsilon.$$

Is that right? I am not sure about my choice of $N$ here.

Answer 1

That choice of $N$ could work, but for a simpler path see the hint below.

Hint: Since the result is trivial for the real number $0$, suppose that $b\in(0,1)$ is arbitrary. Since $b\in(0,1)$, we have that $(f_n(b))=(b^n)$ is a decreasing sequence that we know is bounded by $1$. So by the Monotone Convergence Theorem, $(b^n)$ converges to $inf\{b^k: k\in\mathbb{N}\}$. Show that the infimum is $0$.

You won't need to worry about finding an $N$ for this arbitrary $b$ since convergence will imply its existence.

Answer 2

I believe that the best choice would be to take the natural logarithm, since you can have problems when taking a logarithm using x as a base.

The case $x = 0$ is trivial. Suppose 0 < x < 1, then $\dfrac{1}{x} > 1$. Let $n_0 > \dfrac{ln\bigg[\dfrac{1}{\varepsilon}\bigg]}{ln\bigg[\dfrac{1}{x}\bigg]}$. Then

$$n > n_0 \Rightarrow n \cdot ln\bigg[\dfrac{1}{x}\bigg] > ln\bigg[\dfrac{1}{\varepsilon}\bigg]$$

$$\Rightarrow \dfrac{1}{x^n} > \dfrac{1}{\varepsilon}$$

$$\Rightarrow x^n \ = \ \mid x^n - 0 \mid \ < \ \dfrac{1}{\varepsilon}. $$

Therefore, the sequence of functions $x^n \longrightarrow 0$ pointwise.