Prove by definition an example of limit prodcut/sum laws Here is the question:
Prove by definition of limit that if $x_n\to4$, $y_n\to-2$, then $3x_n+2y_n^2\to20$
What I can do is prove the product and sum laws for general case and use those arguments to prove my example but that seems too much work for this particular problem. Is there any simple way to prove such kind of examples?
 A: THEOREM., If $\{s_n\}_n^{\infty}$ and $\{t_n\}_n^{\infty}$ are sequences of real numbers, if $lim_{n \to \infty}\{sn\}=L$, 
and if $lim_{n \to \infty}\{t_n\}=M$, then $$lim_{n \to \infty}\{s_n+t_n\}=L+M$$. 
And 
THEOREM.If $\{s_n\}_n^{\infty}$ is a sequence of real numbers, if $c \in \mathbb{R}$ and if $lim_{n \to \infty}\{sn\}=L$, 
then $lim_{n \to \infty}\{cs_n\}=cL$.
(THEOREM from R.R.Goldberg Methods of Real Analysis)
$x_n\to4$, $y_n\to-2$, then $3x_n+2y_n^2\to20$
applying both theorem we get 
$\implies 3(4)+2(-2)^2=3(4)+2(4)=12+8=20$
Or 
$x_n\to4 \implies |x_n-4|<  \epsilon $ 
choose $ \epsilon = \frac{\epsilon'}{6}$
So that 
$ \implies |x_n-4|< \frac{\epsilon'}{6} $
$ \implies |3x_n-12|< \frac{3\epsilon'}{6}$
$\implies |3x_n-12|< \frac{\epsilon'}{2} $
$y_n\to-2$
$y_n\to-2 \implies |y_n-(-2)|<  \epsilon $
$\implies |y_n^2-(-2)^2|<  \epsilon^2$
$ \implies |y_n^2-4|<  \epsilon^2$
choose $ \epsilon = \frac{\sqrt\epsilon'}{2}$
$\implies |y_n^2-4|<  (\frac{\sqrt\epsilon'}{2})^2$
$\implies |y_n^2-4|<  (\frac{\epsilon'}{4})$
$\implies |2y_n^2-8|<  (2\frac{\epsilon'}{4})$
$\implies |2y_n^2-8|<  (\frac{\epsilon'}{2})$
$\implies |3x_n-12+2y_n^2-8|<  \frac{\epsilon'}{2}+\frac{\epsilon'}{2}$
$\implies |3x_n+2y_n^2-12-8|<  \epsilon'$
$\implies |3x_n+2y_n^2-20|<  \epsilon'$
$\implies 3x_n+2y_n^2 \to 20$
A: Since you already know what $x_n$ and $y_n$ converge to, try to split up $20$ so that $|3x_n+2y_n^2-20|$ breaks up nicely:
$$|3x_n+2y_n^2-20| = |3(x_n-4)+2(y_n^2-4)|\leq 3|x_n-4|+2|y_n-2||y_n+2|.$$
Now the quantities $|x_n-4|\to 0$ and $|y_n+2|\to 0$ as $n\to \infty$ by assumption. Moreover, $|y_n-2|$ remains bounded for large $n$ since $|y_n-2|=|y_n+2-4|\leq |y_n+2|+4$ (in particular, if we choose $n$ so large that $|y_n+2|<1$, then this quantity is bounded by $5$).
The remainder of the argument is just using the above estimates to cleverly choose our $N$; fix $\epsilon>0$. There are positive integers $N_1$ and $N_2$ such that
$$n\geq N_1\implies |x_n-4|<\epsilon/6$$
$$n\geq N_2\implies |y_n+2|<\min(1,\epsilon/20).$$
Taking $N=\max(N_1,N_2)$ and plugging things back into our original estimate will finish the proof.
The trickiest part of the proof was dealing with $y_n^2$, but if you recall the proof of the limit product law, the method is essentially the same there.
