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*}

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 
$\dfrac{1}{2}-\dfrac{1}{3}$.
