# limit x->infinity

#### Petrus

##### Well-known member
$$\displaystyle f(x,y)=\frac{\ln(1+x^2y^2)}{x^4+y^4}$$
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
$$\displaystyle f(x,y)=\frac{\ln(1+x^2y^2)}{x^4+y^4}$$
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
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 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
Thanks! Now I see without finish the L'Hopital's rule as you need to do it 4 times 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
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,
$$\displaystyle |\pi\rangle$$