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:

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

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

 

[twoflower]

1. For your numerator:
Set:
[tex]a=\sqrt[4]{n+2}[/tex]
[tex]b=\sqrt[4]{n+1}[/tex]
Show that (for example by polynomial division):
[tex]a^{4}-b^{4}=(a-b)(a^{3}+a^{2}b+ab^{2}+b^{3})[/tex]
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:

[tex]
\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)
[/tex]

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

 

[Dark Knight]

please... i don't 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:

[tex]\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}})[/tex]

Can be bounded above since [itex](n+3)^\frac{1}{3}>(n)^\frac{1}{3}[/itex]
[tex] \lim_{n\rightarrow\infty} ((n+2)^{\frac{1}{4}}-(n+1)^{\frac{1}{4}})(n+3)^{\frac{2}{3}}[/tex]
But that's equal to
[tex]\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})}[/tex]
which is bounded above by
[tex]\lim_{n \rightarrow\infty} \frac{(n+3)^{\frac{2}{3}}}{4(n+1)^\frac{3}{4}}[/tex]
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:

[tex]
\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)
[/tex]

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

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:
[tex]\frac{(n+3)^{\frac{2}{3}}+(n+3)^{\frac{1}{3}}n^{\frac{1}{3}}+n^{\frac{2}{3}}}{3((n+2)^{\frac{3}{4}}+(n+2)^{\frac{2}{4}}(n+1)^{\frac{1}{4}}+(n+2)^{\frac{1}{4}}(n+1)^{\frac{2}{4}}+(n+1)^{\frac{3}{4}})}[/tex]
This is easy to evaluate, the leading order behavior is [tex]\frac{1}{4n^{\frac{1}{12}}}, n\to\infty[/tex]