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$f_n(x)=\frac{n}{n+x} $

is a sequence of functions with $f_n: \mathbb{R^+}\longrightarrow\mathbb{R}$

I have calculated that the pointwise limit is f(x)=1 for $x \in \mathbb{R^+}$

However i am unsure whether $f_n$ converge to f uniformly $\mathbb{R^+}$

whether $f_n$ converge to f uniformly $[0, 1]$

and what difference does the parameters make when solving the question?

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  • $\begingroup$ by Weierstrass M-test I would simply try to bound by some $M_n$. In this case $f_n(x)<=1$ for all x from [0,1] $\endgroup$ – Michael Mark Nov 1 '17 at 22:22
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The difference between $\mathbb R^+$ and $[0,1]$ comes from the fact that for any $n$, $\lim_{x\to \infty}f_n(x) = 0$ so $\sup_{x\in \mathbb R^+}|f_n(x)-1| = 1$ for every $n$. However, for $x\in[0,1]$ you can't take $x\to \infty$ and it can be shown that $\sup_{x\in[0,1]}|f_n(x)-1| \to 0$ as $n\to\infty$ (see szw1710's answer).

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  • $\begingroup$ I had a picture for this phenomenon, but I deleted it. I think, I will show it back. $\endgroup$ – szw1710 Nov 1 '17 at 22:57
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Observe that $|f_n(x)-1|=\frac{x}{n+x}\le \frac{x}{n}.$ Hence $f_n\to 1$ uniformly on any bounded subset of $\Bbb R_+$, because then $|f_n(x)-1|\le \frac{M}{n}$, where $M$ is an upper bound of our domain.

There is no uniform convergence on the whole half-line because if it was, almost all graphs of $f_n$ were in the horizontal strip between $1-\varepsilon$ and $1+\varepsilon$. See the picture and compare with spaceisdarkgreen's answer. enter image description here

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The best way I have found of looking at such problems is to instead look at sequences of functions converging pointwise to $0$. As you have found, the functions $f_{n}$ converge pointwise to $1$ on $\mathbb{R^+}$, so let's look at the functions $g_{n}$ given by $$g_{n}(x)=1-f_{n}(x)=1-\frac{n}{n+x}=\frac{x}{n+x} \, .$$

Showing that $f_{n}$ converges uniformly to $1$ on a set $U$ is equivalent to showing that $g_{n}$ converges uniformly to $0$ on $U$. The reason we make this change is because there is quite a straightforward method of checking whether or not a sequence of functions converges uniformly to $0$, and that is by looking at the maximum values of the functions.

Specifically, $g_{n}$ converges uniformly to $0$ on the set $U$ if and only if $$\sup_{x\in U} |g_{n}(x)|\rightarrow 0 \hspace{1em} \mbox{as}\hspace{0.5em} n\rightarrow\infty.$$

First let's take $U=[0,1]$. Given $y\in [0,1]$ and $n\in\mathbb{N}$, we know that $y\leqslant 1$ and $n\leqslant n+y$. Hence, $$|g_{n}(y)|=\frac{y}{n+y}\leqslant \frac{1}{n+y}\leqslant \frac{1}{n}. $$

Observe that the above bound does not depend on $y$, so $\sup_{x\in [0,1]}|g_{n}(x)|\leqslant 1/n$.

Since $1/n\rightarrow0$ as $n\rightarrow\infty$, we deduce that the $g_n$ converge uniformly to $0$ on $[0,1]$. In fact the above argument can be modified to show that the $g_n$ converge uniformly to $0$ on any bounded subset of $\mathbb{R}^{+}$. Therefore, the $f_{n}$ converge uniformly to $1$ on any bounded subset of $\mathbb{R}^{+}$.

If we now take $U=\mathbb{R}^{+}$, observe that $$\sup_{x\in \mathbb{R}^{+}} |g_{n}(x)|\geqslant |g_{n}(n)|=\frac{n}{n+n}=\frac{1}{2}.$$

This shows that the $g_{n}$ do not converge uniformly to $0$ on $\mathbb{R}^{+}$, whence the $f_{n}$ do not converge uniformly to $1$ on $\mathbb{R}^{+}$.

For more difficult functions, we can use calculus to compute the maximum values of the $g_{n}$ explicitly to see whether we have uniform convergence. For instance, you can check that in fact we have $\sup_{x\in \mathbb{R}^{+}} |g_{n}(x)|=1$ (hint: fix some $n\in\mathbb{N}$ and then take $x\rightarrow\infty$ in the definition of $g_{n}(x)$), which again shows that the $f_{n}$ do not converge uniformly to $1$ on $\mathbb{R}^{+}$.

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