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.
 A: 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 $n$.
A: $$ \left( n + \frac{1}{6} \right)^6 < n^6 + n^5 + n^4 < \left( n + \frac{1}{6}+ \frac{7}{72n}    \right)^6 $$
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$$  $$
$$ \left( n + \frac{1}{6} \right)^6 =  n^6 + n^5 + \frac{5 n^4}{12} + \frac{5n^3}{54} + \frac{5n^2}{432} + \frac{n}{1296} + \frac{1}{46656} $$
Note that, for $n \geq 1,$ this is smaller than $n^6 + n^5 + n^4$
$$ $$
$$ \left( n + \frac{1}{6} \right)^6 =  n^6 + n^5 + \frac{5 n^4}{12}  \;+ \;\mbox{lower degree} $$
$$ \left( n + \frac{1}{6} \right)^5 =   n^5 + \frac{5 n^4}{6}  \;+ \;\mbox{lower degree}  $$
$$   \left( n + \frac{1}{6}+ \frac{7}{72n}    \right)^6 = \left( n + \frac{1}{6} \right)^6 + 6  \left( n + \frac{1}{6} \right)^5 \; \frac{7}{72n} + \mbox{lower degree}  $$
$$   \left( n + \frac{1}{6}+ \frac{7}{72n}    \right)^6 = \left( n + \frac{1}{6} \right)^6 +  \frac{7}{12n} \;  \left( n + \frac{1}{6} \right)^5 \;+ \mbox{lower degree}  $$
$$   \left( n + \frac{1}{6}+ \frac{7}{72n}    \right)^6 =  n^6 + n^5 + \frac{5 n^4}{12}+ \mbox{lower degree} + \frac{7 n^4}{12}+ \mbox{lower degree} $$
$$   \left( n + \frac{1}{6}+ \frac{7}{72n}    \right)^6 =  n^6 + n^5 + n^4 + \mbox{lower degree}  $$
