What is the following limit

Question

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.

Answer 1

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 $n$.

Answer 2

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

$$ \left( n + \frac{1}{6} \right)^6 = n^6 + n^5 + \frac{5 n^4}{12} \;+ \;\mbox{lower degree} $$

$$ \left( n + \frac{1}{6} \right)^5 = n^5 + \frac{5 n^4}{6} \;+ \;\mbox{lower degree} $$

$$ \left( n + \frac{1}{6}+ \frac{7}{72n} \right)^6 = \left( n + \frac{1}{6} \right)^6 + 6 \left( n + \frac{1}{6} \right)^5 \; \frac{7}{72n} + \mbox{lower degree} $$ $$ \left( n + \frac{1}{6}+ \frac{7}{72n} \right)^6 = \left( n + \frac{1}{6} \right)^6 + \frac{7}{12n} \; \left( n + \frac{1}{6} \right)^5 \;+ \mbox{lower degree} $$

$$ \left( n + \frac{1}{6}+ \frac{7}{72n} \right)^6 = n^6 + n^5 + \frac{5 n^4}{12}+ \mbox{lower degree} + \frac{7 n^4}{12}+ \mbox{lower degree} $$ $$ \left( n + \frac{1}{6}+ \frac{7}{72n} \right)^6 = n^6 + n^5 + n^4 + \mbox{lower degree} $$