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limit x->infinity

Petrus

Well-known member
Feb 21, 2013
739
\(\displaystyle f(x,y)=\frac{\ln(1+x^2y^2)}{x^4+y^4}\)
decide if it got a limit if \(\displaystyle x^2+y^2=\infty\). if so calculate it.

well I go to polar form and we got
\(\displaystyle \lim_{r^2->\infty}\frac{\ln(1+r^2\cos^2(\theta)r^2\sin^2( \theta))}{r^4\cos^2(\theta)^4+r^4\sin^4(\theta)}\)
we see both approach to limit but the bottom will go alot faster so it will be equal to zero, is this wrong to say like this?

Regards,
\(\displaystyle |\pi\rangle\)
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
\(\displaystyle f(x,y)=\frac{\ln(1+x^2y^2)}{x^4+y^4}\)
decide if it got a limit if \(\displaystyle x^2+y^2=\infty\). if so calculate it.

well I go to polar form and we got
\(\displaystyle \lim_{r^2->\infty}\frac{\ln(1+r^2\cos^2(\theta)r^2\sin^2( \theta))}{r^4\cos^2(\theta)^4+r^4\sin^4(\theta)}\)
we see both approach to limit but the bottom will go alot faster so it will be equal to zero, is this wrong to say like this?

Regards,
\(\displaystyle |\pi\rangle\)
Hi Petrus, :)

I would use the L'Hopital's rule over the variable \(r\).
 

Petrus

Well-known member
Feb 21, 2013
739
Hi Petrus, :)

I would use the L'Hopital's rule over the variable \(r\).
Thanks! Now I see without finish the L'Hopital's rule as you need to do it 4 times :p we will end with a \(\displaystyle 24\cos^2(\theta)sin^2(\theta)\) at top and we will keep have some r at bottom which will make bottom to \(\displaystyle \infty\) so it will equal to 0!:) I got one question that I would like to know.

is it allowed to say \(\displaystyle r^2->\infty\) is same as \(\displaystyle r->\infty\) when you calculate limits?

Regards,
\(\displaystyle |\pi\rangle\)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,779
Thanks! Now I see without finish the L'Hopital's rule as you need to do it 4 times :p we will end with a \(\displaystyle 24\cos^2(\theta)sin^2(\theta)\) at top and we will keep have some r at bottom which will make bottom to \(\displaystyle \infty\) so it will equal to 0!:) I got one question that I would like to know.
Alternatively, you can use the upper bound $\ln(1+u) \le \sqrt u$.

Btw, do you need 4 applications of l'Hospital's rule? It seems I only need 1.

is it allowed to say \(\displaystyle r^2->\infty\) is same as \(\displaystyle r->\infty\) when you calculate limits?
It's unusual, but I don't think it's wrong.
It fits into the definition of a limit.
 

Petrus

Well-known member
Feb 21, 2013
739
Alternatively, you can use the upper estimate $\ln(1+u) \le \sqrt u$.

Btw, do you need 4 applications of l'Hospital's rule? It seems I only need 1.



It's unusual, but I don't think it's wrong.
It fits into the definition of a limit.
you are correct! we will be able to divide \(\displaystyle 4r^3\) on top and bottom! Thanks!

Regards,
\(\displaystyle |\pi\rangle\)