Consider the sequence of functions $f_{1}, f_{2}, \ldots$ on $[0,1]$ defined by $f_{k}(x) = x^{k}$ for all $x \in [0,1]$ and $k \in \mathbb{N}$. Show that this sequence converges pointwise to the following function: $$ f(x) = \begin{cases} 0 & 0 \leq x < 1 \\ 1 & x = 1 \end{cases} $$
Scrap:
Case 1: $0 \leq x < 1$
$$
|f_{k}(x) - f(x)| \\
|f_{k}(x) - 0| \\
|f_{k}(x)| \\
|x^{k}| \\
|x|^{k} \\
x^{k}
$$
So
$$
x^{k} < \epsilon \\
log(x^{k}) < log(\epsilon) \\
k*log(x) < log(\epsilon) \\
k > \frac{log(\epsilon)}{log(x)}
$$
Case 2: x = 1 $$ |f_{k}(x) - f(x)| \\ |1^{k} - 1| \\ |1 - 1| \\ |0| \\ 0 \\ $$ So $$ 0 < \epsilon $$ This is already true by definition.
Proof:
Let $x \in [0,1]$
Let $\epsilon > 0$
Choose $N = \frac{log(\epsilon)}{log(x)}$
Then for every $k > N$ we have:
Case 1: $0 \leq x < 1$ $$ |f_{k}(x) - f(x)| \\ |f_{k}(x) - 0| \\ |f_{k}(x)| \\ |x^{k}| \\ |x|^{k} \\ x^{k} $$
But I am now stuck, since I do not how to properly rewrite this to use the chosen k which seems to be correct in the scrap paper (my equality seems to not be correct).
There is no need to invoke logarithms here. If $0\leqslant x<1$ then $x^n-x^{n-1}=x^n(1-x)\geqslant 0$, so the sequence $\{x_n\}$ is decreasing. As the sequence is bounded below, it converges to some $y\in\mathbb R$. Let $\lim_{n\to\infty}x^n=y$. Then $$ xy = x\lim_{n\to\infty} x^n = \lim_{n\to\infty} x^{n+1}=\lim_{n\to\infty}x^n=y. $$ As $xy=y$, we have $(x-1)y=0$, so $y=0$.