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*}

$$\lim_{n \rightarrow \infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2})  $$
$$=\lim_{n \rightarrow \infty} n\left(1+\frac1n\right)^{\frac12}-n\left(1+\frac1n\right)^{\frac13}$$
$$=\lim_{n \rightarrow \infty} n\left(1+\frac1{2n}+O\left(\frac1{n^2}\right)\right)-n\left(1+\frac1{3n}+O\left(\frac1{n^2}\right)\right)$$ (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 $1+h=y^{\text{lcm}(2,3)}=y^6$
$$\lim_{h\to0}\frac{(1+h)^{\frac12}-(1+h)^{\frac13}}h$$
$$=\lim_{y\to1}\frac{y^3-y^2}{y^6-1}$$
$$=\lim_{y\to1}\frac{y^2(y-1)}{(y-1)(y^5+y^4+y^3+y^2+y+1)}$$
$$=\lim_{y\to1}\frac{y^2}{y^5+y^4+y^3+y^2+y+1}\text{ as }y\ne1\text{ as } y\to1$$
$$= \frac{1^2}{1+1+1+1+1+1}$$
