$f_{n}(x)=(x+\frac{1}{n})^2$, $x\in \mathbb{R}$, converges pointwise to $f(x)=x^2$

Question

I am trying to prove that $\{f_{n}(x)\}_{n\in \mathbb{N}}$, where $f_{n}(x)=(x+\frac{1}{n})^2$, and $x\in \mathbb{R}$. Converges pointwise to $f(x)=x^2$, $x\in \mathbb{R}$. Is the following correct?

Consider arbitrary $\epsilon>0$. Consider arbitrary $x\in \mathbb{R}.$ I now argue that there exists $N \in \mathbb{N}$ such that $n>N$ $\implies$ $|(x+\frac{1}{n})^2-x^2|<\epsilon$.

First, suppose that $x\geq 0$. Then $(x+\frac{1}{n})^2$ is monotone decreasing for all $n\in \mathbb{N}$, and bounded below by $x^2$. Thus, there must exist an $n^*$ big enough such that $n>n^*$ $\implies$ $(x+\frac{1}{n})^2<x^2+\frac{\epsilon}{2}$. Set $N=n^*$. Then, $n>N \implies$ $|(x+\frac{1}{n})^2-x^2|<x^2+\frac{\epsilon}{2}-x^2<\epsilon$

Second, suppose that $x<0$. Then $(x+\frac{1}{n})^2$ is monotone increasing for all $n\in \mathbb{N}$, and bounded above by $x^2$. Thus, there must exist an $n^*$ big enough such that $n>n^*$ $\implies$ $(x+\frac{1}{n})^2>x^2-\frac{\epsilon}{2}$. Set $N=n^*$. Then, $n>N \implies$ $|(x+\frac{1}{n})^2-x^2|=x^2-(x+\frac{1}{n})^2<x^2-x^2+\frac{\epsilon}{2}<\epsilon$

Answer 1

For each fixed $x\in\mathbb{R}$, the expression $f_{n}(x)$ behaves in the same way as a sequence depending on $n$.

Having said that, you can apply the arithmetic properties of sequences to obtain the desired result: \begin{align*} \lim_{n\to\infty}f_{n}(x) & = \lim_{n\to\infty}\left(x + \frac{1}{n}\right)^{2}\\\\ & = \lim_{n\to\infty}\left(x^{2} + \frac{2x}{n} + \frac{1}{n^{2}}\right)\\\\ & = x^{2} + 0 + 0 = x^{2} \end{align*}

Hopefully this helps!