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Let $\{f_{n}\}_{n\geq 1}$ be a sequence of function given by $f_{n}(x)=\frac{1}{x}+\frac{1}{n}$. Does $f$ converge pointwise on $\mathbb{R}\setminus\{0\}$? Does $f$ converge uniformly $\mathbb{R}\setminus\{0\}$?

I think yes, to both since.

Pointwise convergence: $\{f_{n}(x)\}_{n\geq 1}$ converges for every $x\in\mathbb{R}\setminus\{0\}$ to $f(x)=\frac{1}{x}$.

Uniform convergence: Let $\epsilon > 0$ and take $N=\frac{1}{\epsilon}$. Then $|f_{n}(x)-f(x)|=\frac{1}{n}<\frac{1}{N}=\epsilon$ whenever $n>N$, which holds for every $x\in\mathbb{R}\setminus\{0\}$.

Alternatively; $\lim_{n\rightarrow\infty}\sup\{|f_{n}(x)-f(x)|\mid x\in\mathbb{R}\setminus\{0\}\}=0$.

Is this correctly done ?

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    $\begingroup$ Looks good! Some people like $N$ to be a natural number, so you might say $N > \frac 1\epsilon$. $\endgroup$
    – Umberto P.
    Commented Apr 23, 2014 at 16:41
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    $\begingroup$ Yeah, that's fine. A good line of attack is to always try and show pointwise convergence first (since this is usually easiest) and if it is not pointwise convergent, then it will not be uniform convergent (as uniform convergence is stronger than pointwise convergence and thus all uniform convergent sequences are pointwise convergent - so by contraposition, a non-pointwise convergent sequence is also not uniform convergent!). $\endgroup$ Commented Apr 23, 2014 at 16:44

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Yes, your solution is correct.

You may wish to observe the following more general fact: if $f$ is any real-valued function and $(a_n)$ is a sequence of real numbers converging to zero, then the functions $f_n=f+a_n$ converge to $f$ uniformly.

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