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: 
Its a simple question of algebraic manipulation. Let $$a = (n^{6} + 6)^{1/6}, b = (n^{2} + 2)^{1/2}$$ and we need to calculate limit of $n(a - b)$ as $n \to \infty$. We can write $$f(n) = n(a - b) = \frac{n(a^{6} - b^{6})}{a^{5} + a^{4}b + a^{3}b^{2} + a^{2}b^{3} + ab^{4} + b^{5}}$$ Now $$a^{6} - b^{6} = n^{6} + 6 - (n^{2} + 2)^{3} = -2 - 6n^{4} - 12n^{2}$$ and hence the numerator $$n(a^{6} - b^{6}) = -6n^{5} - 12n^{3} - 2n$$ For denominator we need to express $a, b$ as $$a = n\left(1 + \frac{6}{n^{6}}\right)^{1/6}, b = n\left(1 + \frac{2}{n^{2}}\right)^{1/2}$$ It will now be seen that we can take $n^{5}$ common from each term in denominator and cancel this from the numerator. After this cancellation limit of each term in denominator is $1$ so that denominator tends to $6$. The numerator clearly tends to $-6$. The resulting limit of $f(n)$ is clearly $-6/6 = -1$.
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.