What is the following limit I have been working hard on this limit:
$$(n^6 + n^5 + n^4)^{1/6} - n$$
as n goes to infinity. I have no clue of what i need to do. I always get indeterminante forms.

Observe that 
$$(n^6 + n^5 + n^4)^{1/6} - n = n(1 + \frac{1}{n} + \frac{1}{n^2})^{1/6} - n$$
Use the fact that the first term of Taylor series of $$(1+ x)^{\alpha} \sim 1 + \alpha x$$
in the neighborhood of $0$. This means we can choose $x = \frac{1}{n} + \frac{1}{n^2}$ and $\alpha = \frac{1}{6}$ to say that the first equation behaves as $$(n^6 + n^5 + n^4)^{1/6} - n \sim n(1 + \frac{1}{6}(\frac{1}{n} + \frac{1}{n^2})) - n$$
Now, we get
$$(n^6 + n^5 + n^4)^{1/6} - n \sim \frac{n}{6}(\frac{1}{n} + \frac{1}{n^2})$$
which finally gives
$$(n^6 + n^5 + n^4)^{1/6} - n \sim \frac{1}{6}$$ for large $n$.
