We have the sequence of functions $$f_n:\mathbb{R}\rightarrow \mathbb{R} , \ \ \ \ \ f_n(x):=\sqrt[n]{x^2+1}$$ for $n\in \mathbb{N}$. Check if we have a pointwise or a uniform convergence.
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I have done the following :
First we have to calculate the limit of $f_n$ : $$f(x)=\lim_{n\rightarrow \infty}f_n(x)=\lim_{n\rightarrow \infty}\sqrt[n]{x^2+1}=1$$
So $f_n(x)$ converges to $f(x)$ pointwise.
Then we have to find an upper bound of $|f_n(x)-f(x)|= |\sqrt[n]{x^2+1}-1|$ that tends to zero, right?
We have \begin{align*}|\sqrt[n]{x^2+1}-1|&=\sqrt[n]{x^2+1}-1 \ \ \ \text{ since } x^2+1>1 \\ &\leq \sqrt[n]{x^2+1}\end{align*} The last sequence doesn't tend to zero. So we have to find an other upper bound, right?
Or do we use here the $\epsilon$-definition ?
The convergence is not uniform because $$\forall n\in\mathbb{N}: \lim_{x\rightarrow \infty }\left|f_n(x)-f(x)\right|=\lim_{x\rightarrow \infty }\left|\sqrt[n]{x^2+1}-1\right| =\infty $$i.e. it is impossible to find an upper bound to the expression that depends only on $n$ but is independent of $x$.