Otdfbt

For this type of problem,
my preferred weapon
is the generalized binomial theorem
in the form
$(1+x)^a
=1+ax+a(a-1)x^2/2+O(x^3)
$.

Often the simpler form
$(1+x)^a
=1+ax+O(x^2)
$
is enough.

So,

$beginarray\
sqrt[3]n^3+3n^2-sqrtn^2+2n
&=nsqrt[3]1+3/n-nsqrt1+2/n\
&=n(1+(3/n)(1/3)+O(1/n^2))-n(1+(2/n)(1/2)+O(1/n^2))\
&=n(1+1/n+O(1/n^2))-n(1+1/n+O(1/n^2))\
&=n+1+O(1/n)-n-1-O(1/n)\
&=O(1/n)\
&to 0
qquadtextas n to infty\
endarray
$

Note that
if what was wanted
was $n$ times the difference,
the additional term in the expansion
would have been needed.

You might try this
for practice.

Another way to solve the problem is the following, where no Differential Calculus is needed.

By elementary algebraic manipulation of radicals, you get:

$$ beginsplit sqrt[3]n^3 + 3n^2 - sqrtn^2 + 2n &= underbracesqrt[6](n^3 + 3n^2)^2_=:sqrt[6]a - underbracesqrt[6](n^2 + 2n)^3_=:sqrt[6]b \ & = fraca - bsqrt[6]a^5 + sqrt[6]a^4b + sqrt[6]a^3b^2 + sqrt[6]a^2b^3 + sqrt[6]ab^4 + sqrt[6]b^5 \ &approx frac-3n^46n^5 \ &= - frac12n endsplit $$

with the latter member going to $0$ as $n to infty$.

First let's get rid of the square root:

$$lim_n to infty (sqrt[3]n^3-3n^2-sqrtn^2+2n)=lim_n to infty fracsqrt[3](n^3+3n^2)^2-(n^2+2n)sqrt[3](n^3+3n^2)+sqrtn^2+2n\= -lim_n to infty fracn^2+2n-sqrt[3](n^3+3n^2)^2sqrt[3](n^3+3n^2)+sqrtn^2+2n.$$

Now let's get rid of the cube root:

$$-lim_n to infty fracn^2+2n-sqrt[3](n^3+3n^2)^2sqrt[3](n^3+3n^2)+sqrtn^2+2n$$

$$=-lim_n to infty frac(n^2+2n)^3-(n^3+3n^2)^2(sqrt[3](n^3+3n^2)+sqrtn^2+2n)((n^2+2n)^2+(n^2+2n)sqrt[3](n^3+3n^2)^2+sqrt[3](n^3+3n^2)^4).$$

The numerator yields $-3n^4$ and a bunch of lower-order terms, and the denominator yields $6n^5$ and a bunch of lower order terms, so dividing numerator and denominator by $n^5$ shows that the answer is $0$.

Use the following fact: when $xto 0$,
$$(1+x)^alpha-1simalpha x.$$
So
$$sqrt[3]n^3+3n^2-sqrtn^2+2n=nleft(sqrt[3]1+frac3n-1-left(sqrt1+frac2n-1right)right)$$
$$=fracsqrt[3]1+frac3n-1-left(sqrt1+frac2n-1right)frac1n,$$
use the above fact:
$$sqrt[3]1+frac3n-1 sim frac1n,sqrt1+frac2n-1sim frac1n,$$
and this implies
$$lim_n to infty (sqrt[3]n^3+3n^2-sqrtn^2+2n )=0.$$

In fact:
$$sqrt[3]n^3+3n^2-sqrtn^2+2nsim -frac12n,ntoinfty.$$

For positive $n$, let $a=sqrt[3]n^3+3n^2$, and let $b=sqrtn^2+2n$.

Note that $a < n+1$ and
beginalign*
3n +1 &= (n+1)^3-a^3\[4pt]
&= bigl((n+1)-abigr)bigl((n+1)^2+a(n+1)+a^2)\[4pt]
&> bigl((n+1)-abigr)(n+1)^2\[4pt]
endalign*
hence $0 < n+1-a < largefrac3n+1(n+1)^2$, which implies $displaystylelim_nto infty;(n+1)-a = 0$.

Also, $b < n+1$ and
beginalign*
1&=(n+1)^2-b^2\[4pt]
&= bigl((n+1)-bbigr)bigl((n+1)+bbigr)\[4pt]
&> bigl((n+1)-bbigr)(n+1)\[4pt]
endalign*
hence $0 < (n+1)-b < largefrac1n+1$, which implies $displaystylelim_nto infty;(n+1)-b = 0$.

Then we get
$$lim_nto infty;a-b = lim_nto infty;bigl(a - (n+1)bigr) + bigl((n+1)-bbigr) = 0 + 0 = 0$$

There is a way to use directly the two binomial formulas you mentioned. It only needs a little trick:

$$sqrt[3]n^3+3n^2-sqrtn^2+2n = sqrt[3]n^3+3n^2colorblue-n + n-sqrtn^2+2n$$
Now, consider
begineqnarray* sqrt[3]n^3+3n^2-n
& stackreln=sqrt[3]n^3= & fracn^3+3n^2 - n^3(sqrt[3]n^3+3n^2)^2 + nsqrt[3]n^3+3n^2+ n^2\& = & frac3left(sqrt[3]1+frac3nright)^2 + sqrt[3]1+frac3n + 1\
& stackreln to inftylongrightarrow & 1
endeqnarray*

Similarly, you get $n - sqrtn^2+2n stackreln to inftylongrightarrow -1$.

Hence, you get

$$lim_n to infty (sqrt[3]n^3+3n^2-sqrtn^2+2n ) = lim_n to infty (sqrt[3]n^3+3n^2-n ) + lim_n to infty (n-sqrtn^2+2n ) = 1-1 = 0$$