Evaluation of $\lim\limits_{n\to\infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2}) $ Could you, please, check if I solved it right.
\begin{align*}
\lim_{n \rightarrow \infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2})
&= \lim_{n \rightarrow \infty} \sqrt{n^2(1 + \frac1n)}
- \sqrt[3]{n^3(1 + \frac1n)})\\
&= \lim_{n \rightarrow \infty} (\sqrt{n^2} - \sqrt[3]{n^3})\\
&= \lim_{n \rightarrow \infty} (n - n) = 0.
\end{align*}
 A: $$\lim_{n \rightarrow \infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2})  $$
$$=\lim_{n \rightarrow \infty} n\left(1+\frac1n\right)^{\frac12}-n\left(1+\frac1n\right)^{\frac13}$$
$$=\lim_{n \rightarrow \infty} n\left(1+\frac1{2n}+O\left(\frac1{n^2}\right)\right)-n\left(1+\frac1{3n}+O\left(\frac1{n^2}\right)\right)$$ (Using  Taylor series or Generalized Binomial Theorem)
$$=\lim_{n \rightarrow \infty} \frac12+O\left(\frac1n\right)-\left(\frac13+O\left(\frac1n\right)\right)$$
$$=\frac12-\frac13$$

Alternatively, putting $h=\frac1n,$
$$ (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2})=\frac{(1+h)^{\frac12}-(1+h)^{\frac13}}h$$
$$\lim_{h\to0}\frac{(1+h)^{\frac12}-(1+h)^{\frac13}}h $$ can be handled  at least in three ways as follows:
$1: $ Using Taylor series,
$$\lim_{h\to0}\frac{(1+h)^{\frac12}-(1+h)^{\frac13}}h=\lim_{h\to0}\frac{1+\frac h2+O(h^2)-(1+\frac h3+O(h^2))}h$$
$$=\frac12-\frac13\text{ as }h\ne0\text{ as }h\to0$$
$2: $ $$\text{As }\lim_{h\to0}\frac{(1+h)^{\frac12}-(1+h)^{\frac13}}h \text{ is of the from }\frac\infty\infty,$$
Applying L'Hospital's Rule, $$\lim_{h\to0}\frac{(1+h)^{\frac12}-(1+h)^{\frac13}}h=\lim_{h\to0}\left(\frac12\cdot(1+h)^{\frac12-1}-\frac1{3}\cdot(1+h)^{\frac13-1}\right)=\frac12-\frac13$$
$3: $ Like Jyrki Lahtonen has approached,
Putting $1+h=y^{\text{lcm}(2,3)}=y^6$
$$\lim_{h\to0}\frac{(1+h)^{\frac12}-(1+h)^{\frac13}}h$$
$$=\lim_{y\to1}\frac{y^3-y^2}{y^6-1}$$
$$=\lim_{y\to1}\frac{y^2(y-1)}{(y-1)(y^5+y^4+y^3+y^2+y+1)}$$
$$=\lim_{y\to1}\frac{y^2}{y^5+y^4+y^3+y^2+y+1}\text{ as }y\ne1\text{ as } y\to1$$
$$= \frac{1^2}{1+1+1+1+1+1}$$
A: Others have spotted your error, and described ways of seeing that the limit is $1/6$. If you want to see a non-calculus way, then I suggest splitting it to two parts as follows:
$$
\sqrt{n^2+n}-\root3\of{n^3+n^2}=\left(\sqrt{n^2+n}-n\right)-\left(\root3\of{n^3+n^2}-n\right).
$$
This operation is valid, if you can show that both sequences on the r.h.s converge.
To see that I would interpret $n=\sqrt{n^2}$ in the first summand and $n=\root3\of{n^3}$ in the second. Then use the usual tricks
$$
\sqrt a-\sqrt b=\frac{a-b}{\sqrt a+\sqrt b},
$$
and
$$
\root 3\of a-\root3\of b=\frac{a-b}{a^{2/3}+(ab)^{1/3}+b^{2/3}}.
$$
These come from the respective polynomial factorizations
$$
x^2-y^2=(x-y)(x+y),\qquad x^3-y^3=(x-y)(x^2+xy+y^2).
$$
A: Following your idea:
$$ (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2})
= \sqrt{n^2\left(1 + \frac1n\right)}
- \sqrt[3]{n^3\left(1 + \frac1n\right)}=$$
$$=n\left(\sqrt{1+\frac1n}-\sqrt[3]{1+\frac1n}\right)=\frac{\sqrt{1+\frac1n}-\sqrt[3]{1+\frac1n}}{\frac1n}$$
Looking at the above as function of the continuous variable $\,n\,$, we can use l'Hospital to try to calculate its limit:
$$\lim_{n\to\infty}\frac{\sqrt{1+\frac1n}-\sqrt[3]{1+\frac1n}}{\frac1n}\stackrel{\text{l'H}}=\lim_{n\to\infty}\left(\frac1{2\sqrt{1+\frac1n}}-\frac1{3\sqrt[3]{\left(1+\frac1n\right)^2}}\right)=\frac12-\frac13=\frac16$$
A: Here is a very calculus way to do the problem. (Your "taking" $n$ out is useful.) Write $h$ for $1/n$, and rewrite our limit  as
$$\lim_{h\to 0}\frac{(\sqrt{1+h} -1)-(\sqrt[3]{1+h}-1)}{h}, \tag{1}$$
because
$$\frac{\sqrt{1+h} -1}{h} \quad\text{and}\qquad \frac{\sqrt[3]{1+h}-1}{h}$$ 
are familiar expressions whose limits we can compute.
We recognize 
$$\lim_{h\to 0}\frac{\sqrt{1+h} -1}{h}$$ 
as the derivative of $\sqrt{1+x}$ at $x=0$, and 
$$\lim_{h\to 0}\frac{\sqrt[3]{1+h} -1}{h}$$ 
as the derivative of $\sqrt[3]{1+x}$ at $x=0$.
These two derivatives, evaluated at $0$, are $\frac{1}{2}$ and $\frac{1}{3}$ respectively. It follows that the limit (1) is equal to 
$\dfrac{1}{2}-\dfrac{1}{3}$.
A: You made a mistake when you applied the limit $n \to \infty$ to the $\left(1 + \frac 1n\right)$ terms but not other terms outside. This is not a valid limit application because the whole thing is still in an indeterminate $\infty - \infty$ form.
I can offer you a tedious method that does not require you to estimate the growth of the two terms in the difference. (In fact, if you expand out everything in $O(\ldots)$, you only need basic algebra and some fundamental limit theorems.)
Define
\begin{align*}
A_n & = \sqrt{n^2 + n} = \sqrt[6]{(n^2 + n)^3} = \sqrt[6]{n^6 + 3n^5 + 3n^4 + n^3}
= n\sqrt[6]{1 + O(1/n)}\\
B_n & = \sqrt[3]{n^3 + n^2} = \sqrt[6]{(n^3 + n^2)^2} = \sqrt[6]{n^6 + 2n^5 + n^4}
= n\sqrt[6]{1 + O(1/n)}.
\end{align*}
We want to find $\lim_{n\to\infty} A_n - B_n$.
Let us factor $A_n^6 - B_n^6$ as follows:
\begin{align*}
A_n^6 - B_n^6 & =
(A_n - B_n)(A_n^5 + A_n^4B_n + A_n^3B_n^2 + A_n^2B_n^3 + A_nB_n^4 + B_n^5) \\
\therefore A_n - B_n & =
\frac{A_n^6 - B_n^6}{A_n^5 + A_n^4B_n + A_n^3B_n^2 + A_n^2B_n^3 + A_nB_n^4 + B_n^5}
\end{align*}
We know that $A_n^6 - B_n^6 = n^5 + 2n^4 + n^3 = n^5(1 + O(1/n))$. All terms in the denominator are of the form $A^i_nB^{5-i}_n$, so
$$
A^i_nB^{5-i}_n = \sqrt[6]{(n^6 + 3n^5 + 3n^4 + n^3)^i(n^6 + 2n^5 + n^4)^{5-i}}
= n^5\sqrt[6]{1 + O(1/n)}.
$$
Therefore,
\begin{align*}
\lim_{n\to\infty} A_n - B_n
& = \lim_{n\to\infty}\frac{n^5(1 + O(1/n))}{6n^5\sqrt[6]{1 + O(1/n)}} \\
& = \frac 16
\end{align*}
A: Although it seems like a pain, you can multiply by "the conjugate" here, too. But you want $a^6-b^6$ on too, so multiply by $a^5+a^4b+...+b^5$ on top and bottom. The too cancels out nicely. The bottom seems bad, but remember that every term on the bottom grows like $n^2$, and there are six terms, thus giving 1/6.
A: Forget your last term and Just set $x = 1/n$ in the second
The first term of the series is $x/2 - x/3=x/6.$
Since $x=1/n,$ the result is just $1/6.$
A: You want
$\lim_{n \rightarrow \infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2})
$.
More generally,
consider
$f_{a, c}(n)
=\sqrt[a]{n^a+n^{a-c}}
$
where
$a> 0$
and
$a > c > 0$.
I will use the
generalized binomial theorem
and generalized binomial coefficients.
We have
$\begin{array}\\
f_{a, c}(n)
&=\sqrt[a]{n^a+n^{a-c}}\\
&=n\sqrt[a]{1+n^{-c}}\\
&=n(1+n^{-c})^{1/a}\\
&=n\sum_{k=0}^{\infty} \binom{1/a}{k}n^{-ck}
\qquad\text{(generalized binomial theorem)}\\
&=n\sum_{k=0}^{\infty} \dfrac{\prod_{j=0}^{k-1}(1-aj)}{a^kk!}n^{-ck}
\qquad\text{(see (**) below)}\\
&=n\left(1+\sum_{k=1}^{\infty} \dfrac{\prod_{j=0}^{k-1}(1-aj)}{a^kk!n^{ck}}\right)\\
&=n+\sum_{k=1}^{\infty} \dfrac{\prod_{j=0}^{k-1}(1-aj)}{a^{k}k!n^{ck-1}}\\
&=n+\dfrac{1}{an^{c-1}}+\dfrac{1-a}{2a^2n^{2c-1}}+\dfrac{(1-a)(1-2a)}{6a^3n^{3c-1}}+...\\
\end{array}
$
If $c=1$,
$f_{a, 1}(n)
=\sqrt[a]{n^a+n^{a-1}}
=n+\dfrac{1}{a}+\dfrac{1-a}{2a^2n}+\dfrac{(1-a)(1-2a)}{6a^3n^{2}}+...
$.
In particular
$f_{2, 1}(n)
=\sqrt{n^2+n}
=n+\dfrac{1}{2}-\dfrac{1}{8n}+\dfrac{3}{48n^{2}}+...
=n+\dfrac{1}{2}-\dfrac{1}{8n}+\dfrac{1}{16n^{2}}+...
$
and
$f_{3, 1}(n)
=\sqrt[3]{n^3+n^{2}}
=n+\dfrac{1}{3}-\dfrac{2}{18n}+\dfrac{10}{162n^{2}}+...
=n+\dfrac{1}{3}-\dfrac{1}{9n}+\dfrac{5}{81n^{2}}+...
$.
Therefore the limit is
$\lim_{n \to \infty} ((n+\dfrac{1}{2}+O(1/n))-(n+\dfrac{1}{3}+O(1/n)))
=\dfrac16
$.
Note:
In this kind of computation,
I compute first and
worry about convergence later
(or never).
**
We have,
since
$\binom{r}{k}
=\dfrac{\prod_{j=0}^{k-1}(r-j)}{k!}
$,
$\begin{array}\\
\binom{1/a}{k}
&=\dfrac{\prod_{j=0}^{k-1}(\frac1{a}-j)}{k!}\\
&=\dfrac{\prod_{j=0}^{k-1}(1-aj)}{a^kk!}\\
\end{array}
$
