The question: For each natural number, $n$, and $x \geq 2$, define $f_{n}(x) = \frac{1}{1+x^n}$. Find the function $f :[2,\infty) \rightarrow \mathbb{R}$ to which the sequence $\Big(f_{n} \Big)_{n=1}^{\infty}$ converges pint wise. Prove that the convergence is uniform.
My attempt:
Let $x \geq 2$. Let $\epsilon > 0$. Notice as $n$ gets large, $(1+x^{n}) \approx x^{n}$. Then, $\lim_{n \rightarrow \infty} \frac{1}{x^n} = 0$ Thus, $\lim_{n \rightarrow \infty} \frac{1}{1+x^n} = 0$.
Now to prove uniform convergence, we need to show $\forall \epsilon > 0$, there exists a $N \in \mathbb{N}$, such that $\forall n \geq N$, $|f_{n}(x) -f(x)| < \epsilon$. In other words, we show $| \frac{1}{1+x^{n}}| <\epsilon$.
Here is my scratch work:
We know that $x \geq 2$. This implies that $x^{n} \geq 2^{n}$. So, $x^{n} + 1 \geq 2^{n} +1$, which implies that $\frac{1}{1 + x^{n}} \leq \frac{1}{1 + 2^{n}} < \epsilon$. By algebra, if I choose $N = log_{2}(\frac{\epsilon -1}{\epsilon})$, then I will be able to get uniform convergence. Am I on the right track on this?
Thank you for your help.