Limit with cube and fourth root

[twoflower]

Hi guys, I have another limit I can't move with. Well, I guess it goes to zero, but can't show a bulletproof evidence:


\lim_{n \rightarrow \infty} \frac{ \sqrt[4]{n + 2} - \sqrt[4]{n + 1}}{ \sqrt[3]{n + 3} - \sqrt[3]{n}}


Even after I got rid of denominator, I can't find some known lemma to show that this limit is really 0. Will somebody help me to find any?

Thank you.

[arildno]

1. For your numerator:
Set:
a=\sqrt[4]{n+2}
b=\sqrt[4]{n+1}
Show that (for example by polynomial division):
a^{4}-b^{4}=(a-b)(a^{3}+a^{2}b+ab^{2}+b^{3})
2. Use a similar technique for your denominator.
It should now be quite simple to evaluate your limit.

[twoflower]

1. For your numerator:
Set:
a=\sqrt[4]{n+2}
b=\sqrt[4]{n+1}
Show that (for example by polynomial division):
a^{4}-b^{4}=(a-b)(a^{3}+a^{2}b+ab^{2}+b^{3})
2. Use a similar technique for your denominator.
It should now be quite simple to evaluate your limit.

Yes I used this method to get rid of denominator, hence I got this:


\frac{1}{3} \lim_{n \rightarrow \infty} \left( \sqrt[4]{n+2}.\sqrt[3]{(n+3)^2} + \sqrt[4]{n+2}.\sqrt[3]{n^2+3n} + \sqrt[4]{n+2}.\sqrt[3]{n^2} - \sqrt[4]{n+1}.\sqrt[3]{(n+3)^2} - \sqrt[4]{n+1}.\sqrt[3]{n^2+3n} - \sqrt[4]{n+1}.\sqrt[3]{n^2} \right)


I tried to divide it somehow with the largest term but the only thing I got were undeterminate forms like 0. \infty

[Dark Knight]

please....... i dont know how to solve the limits that its denominator is from 8th or 5th or...etc degree!!!! what should i do to solve such a quesion?!!!

[NateTG]

Let's take a second look at that limit:

\lim_{n\rightarrow\infty} ((n+2)^{\frac{1}{4}}-(n+1)^{\frac{1}{4}})\frac{1}{3}((n+3)^{\frac{2}{3} }+(n)^{\frac{1}{3}}(n+3)^{\frac{1}{3}}+n^{\frac{2} {3}})

Can be bounded above since (n+3)^\frac{1}{3}>(n)^\frac{1}{3}
\lim_{n\rightarrow\infty} ((n+2)^{\frac{1}{4}}-(n+1)^{\frac{1}{4}})(n+3)^{\frac{2}{3}}
But that's equal to
\lim_{n\rightarrow\infty} \frac{(n+3)^{\frac{2}{3}}}{((n+2)^\frac{1}{4}+(n+1 )^\frac{1}{4})((n+2)^\frac{1}{2}+(n+1)^\frac{1}{2} )}
which is bounded above by
\lim_{n \rightarrow\infty} \frac{(n+3)^{\frac{2}{3}}}{4(n+1)^\frac{3}{4}}
Which has a larger exponet on bottom, so it goes to zero.

[arildno]

Yes I used this method to get rid of denominator, hence I got this:


\frac{1}{3} \lim_{n \rightarrow \infty} \left( \sqrt[4]{n+2}.\sqrt[3]{(n+3)^2} + \sqrt[4]{n+2}.\sqrt[3]{n^2+3n} + \sqrt[4]{n+2}.\sqrt[3]{n^2} - \sqrt[4]{n+1}.\sqrt[3]{(n+3)^2} - \sqrt[4]{n+1}.\sqrt[3]{n^2+3n} - \sqrt[4]{n+1}.\sqrt[3]{n^2} \right)


I tried to divide it somehow with the largest term but the only thing I got were undeterminate forms like 0. \infty
But this is not at all what you should do!!
You should get rid of the difference both in the denominator AND numerator.
Then you'll end up with:
\frac{(n+3)^{\frac{2}{3}}+(n+3)^{\frac{1}{3}}n^{\f rac{1}{3}}+n^{\frac{2}{3}}}{3((n+2)^{\frac{3}{4}}+ (n+2)^{\frac{2}{4}}(n+1)^{\frac{1}{4}}+(n+2)^{\fra c{1}{4}}(n+1)^{\frac{2}{4}}+(n+1)^{\frac{3}{4}})}
This is easy to evaluate, the leading order behavior is \frac{1}{4n^{\frac{1}{12}}}, n\to\infty