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!
 A: $$\left|\frac{2n^3+n^2-2(n+2)^3}{(n+2)^3}\right|=\frac{11n^2+24n+16}{(n+2)^3}\le\frac{11n^2+n^2+n^2}{(n+3)^3}\le13\frac{n^2}{n^3}=\frac{13}n$$
Now, show the rightmost expression is less than $\;\epsilon\;$ for $\;n\;$ big enough and also show the first inequality is true for all $\;n\;$ greater than...?
A: Let$$L=\lim_{n\rightarrow\infty}\frac{2n^3+n^2}{(n+2)^3}$$ $$L=\lim_{n\rightarrow\infty}\frac{n^3(2+\frac1n)}{n^3(1+\frac{2}{n})^3}$$
$$L=\lim_{n\rightarrow\infty}\frac{(2+\frac1n)}{(1+\frac{2}{n})^3}$$
Now, as $n\rightarrow\infty$, $\frac1n\rightarrow0$
$$L=\frac{2+0}{1+0}$$
$$L=2$$
A: 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 $$ \lim_{n \to \infty}\frac{2n^3 + n^2}{(n+2)^3} = \lim_{u \to \infty} \frac{2(u-2)^3 + (u-2)^2}{u^3} = 2. $$
A: $$
\left|\frac{2n^3 +n^2}{(n+2)^3} - 2\right| \le \left|\left( \frac{n}{n+2} \right)^3 \left(2+\frac1{n}-2 \right)\right| + 2\left|\left( \frac{n}{n+2} \right)^3 -1 \right| \\
= \frac1{n+2}\left( \left| \left(\frac{n}{n+2} \right)^2 \right|+4\left|\left(\frac{n}{n+2} \right)^2 +\left(\frac{n}{n+2} \right) + 1 \right|\right)
\lt \frac{13}{n+2} \rightarrow 0
$$
A: $$a_{10}=\frac{2100}{12^3}=\frac{2.1}{1.2^3}$$
$$a_{100}=\frac{2010000}{102^3}=\frac{2.01}{1.02^3}$$
$$a_{1000}=\frac{2001000000}{1002^3}=\frac{2.01}{1.002^3}$$
$$a_{10000}=\frac{2000100000000}{10002^3}=\frac{2.0001}{1.0002^3}$$
$$\cdots$$
$$a_\infty=\frac21$$
It is enough to say that the fractional parts vanish.
A: $$lim_{n\to\infty} \frac{(2(n^3))+(n^2)}{(n+2)^3}$$
When we look at limits of quotients of polynomials as $n\to\infty$, we want to look at the largest degree of the polynomial. (it is the only one that is relevant to the limit as $n\to\infty$)
Thus $lim_{n\to\infty} \frac{(2(n^3))+(n^2)}{(n+2)^3}= lim_{n\to\infty} \frac{2(n^3)}{n^3}$
Which equals 2.
