Question: Find $\lim\_{n \to \infty}n\left[(n^6+6)^{1/6}-(n^2+2)^{1/2}\right]$.

I thought that expression in brackets becomes $0$ and consequently the answer will be $0$, but no, answer is $-2$, whereas wolfram shows $-1$. Can somebody explain, please?

Answer: 
$$(n^6+6)^{1/6} \sim n+\frac{1}{n^5}+O\left(\frac{1}{n^{11}}\right)$$
$$(n^2+2)^{1/2} \sim n+\frac{1}{n}+O\left(\frac{1}{n^{3}}\right)$$
Therefore the difference is $\sim - n\times \frac{1}{n} = -1$
The problem is that the term in brackets *does* go to $0$, but the thing it's multiplied by ($n$) goes to infinity, so you have the "indeterminate form" $0 \cdot \infty$. One trick for addressing this is to turn it into the form $\frac{\infty}{\infty}$ by inverting the first term, i.e.,

$$
\lim\_{n \to \infty}n\left((n^6+6)^{1/6}-(n^2+2)^{1/2}\right) = \lim\_{n \to \infty}\frac{(n^6+6)^{1/6}-(n^2+2)^{1/2}}{1/n}
$$
which you can then work on via L'hopital's Rule.