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I tried solving the question by substituting $x=r\cos(\theta)$ and $y= r\sin(\theta)$ , but even that is not helping me in any way.is this the correct way to proceed? Or should i follow any other method?

Marja
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joey lang
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  • "i tried solving the question by substituting x=rcosx=rcos and y=rsin" Sorry but this sounds as if you DID NOT try, actually. If you have, please provide the details of your (unsuccessful) computations. – Did Aug 07 '17 at 09:48
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    hi. i really tried to proceed with the method i mentioned above. i got stuck in a step with too many sin and cos thetas and didn't know how to proceed from there. – joey lang Aug 07 '17 at 09:52
  • Which part of "please provide the details of your (unsuccessful) computations" do you fail to understand? – Did Aug 07 '17 at 09:54

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It helps ! Let $x=r \cos t$ and $y =r \sin t$. Then $r=\sqrt{x^2+y^2}$. Now show that

$|f(x,y)=|f(r \cos t, r \sin t)| \le 5r=5 \sqrt{x^2+y^2}$

Fred
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  • hi.thank you for answering. can you please provide the answer with all the steps , because i am still not able to get it? – joey lang Aug 07 '17 at 09:46
  • @Fred it should be $|f( r\cos t, r\sin t)|\leq 5r$. – Mundron Schmidt Aug 07 '17 at 10:01
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    @joey lang: plug in the polar coordinates and use $|\cos t|\leq 1$ and $\sin t|\leq 1$. You get $|f(r\cos t,r\sin t)|\leq 5r$. You have therefore $|f(x,y)-f(0,0)|=|f(x,y)|\leq 5\sqrt{x^2+y^2}<5\delta$. Now choose $\delta>0$ such that $5\delta<0.01$. Get it? – Mundron Schmidt Aug 07 '17 at 10:04
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    Hey mundron. Thanks a lot. I Got the answer now – joey lang Aug 07 '17 at 10:14