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

#### bubbers

1. The problem statement, all variables and given/known data

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

2. Relevant equations

r^2=x^2+y^2

3. 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?

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#### jbunniii

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.

#### SammyS

Staff Emeritus
Homework Helper
Gold Member
1. The problem statement, all variables and given/known data

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

2. Relevant equations

r^2=x^2+y^2

3. 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:

#### LCKurtz

Homework Helper
Gold Member
1. The problem statement, all variables and given/known data

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

2. Relevant equations

r^2=x^2+y^2

3. 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?

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