<s>[INST] Prove that the limit of $a_n = \frac{2n^3+n^2}{(n+2)^3}$ is $2$. if $a_n = \dfrac{2n^3+n^2}{(n+2)^3}$, prove that the limit of $a_n$ (as $n$ tends to infinite) is $2$ using the definition of a limit. 
My attempt was $\left |\dfrac{(2(n^3))+(n^2)}{(n+2)^3} - 2\right |=\left |\dfrac{(2(n^3))+(n^2)-2((n+2)^3)}{(n+2)^3}\right |$ and tried to simplify it as much as I could (though I couldn't get very far) then set it to less than Epsilon. 
I'm not sure if that was the right approach though. Would anyone be able to help me out on this? Any help is greatly appreciated. Thank you!
 [/INST] we will make a change of variable $u = n + 2, n = u - 2.$ then 
$\begin{align}\frac{2n^3 + n^2}{(n+2)^3} &= \frac{2(u-2)^3 + (u-2)^2}{u^3}\\
& = \dfrac{2(u^3 - 6u^2+12u -8) +(u^2-4u+4)}{u^3}\\
&= 2 -\dfrac{11}{u}+\dfrac{20}{u^2}-\dfrac{12}{u^3}\end{align}$
pick an $\epsilon < 0.1.$ choose $N$ big enough so that for $u \ge N$ we have $$\dfrac{1}{u^3} \le \dfrac{1}{u^2}\le \dfrac{1}{u} \le\dfrac{\epsilon}{43} < 0.1$$
now $$|\frac{2(u-2)^3 + (u-2)^2}{u^3} -2|=  |-\dfrac{11}{u}+\dfrac{20}{u^2}-\dfrac{12}{u^3}| \le \frac{43}{u} \le \epsilon$$
therefore $$ \lim_{n \to \infty}\frac{2n^3 + n^2}{(n+2)^3} = \lim_{u \to \infty} \frac{2(u-2)^3 + (u-2)^2}{u^3} = 2. $$
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