<s>[INST] Prove $f(x)=x^2+3x$ is Continuous at $x=a$ I am attempting to give a $\delta,\epsilon$ proof that the following function, $f(x)=x^2+3x$ is continuous at $x=a$, where $a$ is any real number. I have the following:

Given any $\epsilon > 0$, find a $\delta > 0$ such that if $|x-a|<\delta$ then $|(x^2+3x)-((a^2)+3a)|<\epsilon\\ \\$
$\text{We have that:}\\ \\
\quad\quad \begin{align}|(x^2+3x)-((a^2)+3a)| &= |x-a||x+a|+3|x-a| <\epsilon\\
&=|x-a||x+a+3| <\epsilon \end{align}\\$
$\text{Attempt to Bound $|x+a+3|$: Let $\delta \leq 1$}\\ \\
\quad\quad \begin{align} &\Rightarrow |x-a|<1\\
&\Rightarrow -1 <x-a<1\\
&\Rightarrow a-1<x<a+1 ;\text{add $a+3$ in order to obtain $x+a+3$}\\ 
&\Rightarrow 2a+2<x+a+3<2a+4\\\end{align}$

The issue seems to be my attempt to bound $|x+a+3|$ because I am supposed to reach the following:

$\begin{align}|(x^2+3x)-((a^2)+3a)| &= |x-a||x+a|+3|x-a| <\epsilon\\
&=|x-a||x+a+3|< \epsilon\\
&=|x-a|*(2|a|+3)< \epsilon\\
&=|x-a|< \frac{\epsilon}{2|a|+3}\\ \\
&\text{Choose $\delta=min(1,\frac{\epsilon}{2|a|+3})$} \end{align}\\$

I am not exactly sure how we arrive at $2|a|+3$. An idea that I have is if $2a+2<x+a+3<2a+4$, and $a$ is a real number, then $2a+3$ is in between $2a+2$ and $2a+4$.
Any help would be greatly appreciated!
 [/INST] Use the sequential criterion for continuity at $a$. Let $(a_n)_{n=1}^\infty$ be a real sequence with limit $a$. Now, it suffices to show $\lim_{n\to\infty}f(a_n)=f(a)$.
$$\lim_{n\to\infty}f(a_n)=\lim_{n\to\infty}a_n(a_n+3)=\lim_{n\to\infty}a_n\cdot\lim_{n\to\infty}(a_n+3)=a(a+3)=f(a)$$
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