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

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Homework Statement



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

Homework Equations



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?
 

Answers and Replies

  • #2
jbunniii
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What went wrong when you tried to use L'Hospital's rule? Where did you get stuck?
 
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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.
 
  • #4
SammyS
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Homework Statement



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

Homework Equations



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:
[itex]\displaystyle \lim_{r\,\to\,0^{+}}\frac{\sin(r^2)}{r^2}[/itex]

[itex]\displaystyle =\lim_{r\,\to\,0^{+}}\frac{d(\sin(r^2))/dr}{d(r^2)/dr}[/itex]​
This is not what you have.
 
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  • #5
LCKurtz
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Homework Statement



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

Homework Equations



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?
 

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