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 ?
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.