Explore whether or not the sequence converges uniformly

Question

Explore whether or not the sequence converges uniformly on $E = [0, 2] $. $$f_n(x) = \sqrt[n]{1+x^n} $$

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I tried to find $\displaystyle f(x) = \lim_{n \to \infty} \sqrt[n]{1+x^n} = \lim_{n \to \infty} e^{\frac1n \cdot \ln(1+x^n)}$ and stuck here. At first glance I thought the limit may depend on whether $x < 1$ or $x > 1$ but I'm not sure about that. I would appreciate any help

Answer 1

Consider both $x \in [0,1]$ where $f_n(x) \to 1$ and $x \in [1,2]$ where $f_n(x) \to x$.

For a further hint note that for $x > 1$

$$\sqrt[n]{1+x^n } - x = \frac{(1+x^n) - x^n}{(\sqrt[n]{1+x^n })^{n-1} + (\sqrt[n]{1+x^n })^{n-2}x + \ldots + x^{n-1} } \leqslant \frac{1}{n}$$

Answer 2

I'll get you started by finding the pointwise limits: You are right to examine $$ f_n(x)=\exp(1/n\log(1+x^n)) $$

First find the pointwise limits: For $0\leq x<1$, clearly $$ f_n(x)\to1 $$ by continuity of $\exp$ and $\log$.

If $x=1$, $$ \lim_{n\to\infty}f_n(1)=\lim_{n\to\infty}2^{1/n}=1 $$

If $x>1$, note $\log(1+x^n)\sim n\log(x)$ as $n\to \infty$, so that the limit should be $x$. Indeed, note that by the mean value theorem, $$ |\log(1+x^n)-\log(x^n)|=1/y $$ for $x^n<y<1+x^n$, so that $$ |\log(1+x^n)-\log(x^n)|\leq \frac{1}{x^n}\to 0 $$ See if you can modify some of these estimates to make them uniform.