# Using polar coordinates, show that lim (x,y)->(0,0) [sin(x^2+y^2)]/[x^2+y^2] = 1

bubbers

## Homework Statement

Using polar coordinates, show that lim (x,y)->(0,0) [sin(x^2+y^2)]/[x^2+y^2] = 1

r^2=x^2+y^2

## The Attempt at a Solution

I was able to get the limit into polar coordinates:

lim r->0^+ [sin(r^2)]/r^2

but I'm not sure how to take this limit. I tried L'Hospital's, but it was just messy and useless seeming. So, any hints?

Homework Helper
Gold Member
What went wrong when you tried to use L'Hospital's rule? Where did you get stuck?

bubbers
I got:

lim r->0^+ [sin(r^2)(2r)-r^2cos(r^2)(2r)]/r^4

which simplifies to:

lim r->0^+ [2sin(r^2)-2r^2cos(r^2)]/r^3

but that's still 0/0 form, so it got me nowhere useful as far as I can see.

Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

Using polar coordinates, show that lim (x,y)->(0,0) [sin(x^2+y^2)]/[x^2+y^2] = 1

r^2=x^2+y^2

## The Attempt at a Solution

I was able to get the limit into polar coordinates:

lim r->0^+ [sin(r^2)]/r^2

but I'm not sure how to take this limit. I tried L'Hospital's, but it was just messy and useless seeming. So, any hints?
Applying L'Hôpital's rule gives you:
$\displaystyle \lim_{r\,\to\,0^{+}}\frac{\sin(r^2)}{r^2}$

$\displaystyle =\lim_{r\,\to\,0^{+}}\frac{d(\sin(r^2))/dr}{d(r^2)/dr}$​
This is not what you have.

Last edited:
Homework Helper
Gold Member

## Homework Statement

Using polar coordinates, show that lim (x,y)->(0,0) [sin(x^2+y^2)]/[x^2+y^2] = 1

r^2=x^2+y^2

## The Attempt at a Solution

I was able to get the limit into polar coordinates:

lim r->0^+ [sin(r^2)]/r^2

If you let ##\theta=r^2## does the limit become anything you have already studied?