Solving Limit Using Polar Coordinates

Tiago1818
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Staff was trying to understand a matter of calculation. I hope someone can explain me in detail how to solve this limit using polar coordinates:

http://img36.imageshack.us/img36/2667/semttulokej.png
 
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Tiago1818 said:
Staff was trying to understand a matter of calculation. I hope someone can explain me in detail how to solve this limit using polar coordinates:

<img src="http://latex.codecogs.com/gif.latex?\lim_{(x,y)\to(0,0)} \frac{sen(x^2 + y^2)}{1-cos\sqrt[]{x^2 + y^2}}" title="\lim_{(x,y)\to(0,0)} \frac{sen(x^2 + y^2)}{1-cos\sqrt[]{x^2 + y^2}}" />
Hello Tiago1818. Welcome to PF !

Here's an attempt to reconstruct what was in your image.

\displaystyle \lim_{(x,y)\to(0,0)}\ \ \frac{\sin(x^2 + y^2)}{1-\cos\sqrt{x^2 + y^2}}

The first thing to do is to change your expression to polar coordinates.

According to the rules of this Forum, you need to show some effort before we can help you.
 
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I've done it, but do not know how to continue. I wanted to see the resolution step by step, with comments about what is being done, of course if you can.I am Brazilian and I'm using google translator, so what I am saying may seem a bit confusing.
 
Tiago1818 said:
I wanted to see the resolution step by step, with comments about what is being done, of course if you can.

We will do no such thing.

YOU need to show what you tried first, then we will GUIDE you in the right direction. We do not simply give out the answers.
 
I did it.

http://img825.imageshack.us/img825/237/digitalizar0029.png

I do not know what to do now.
 
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know yet, the answer is "2"
 
Hopefully, you know the limit: \displaystyle \lim_{x\to0}\ \frac{\sin(x)}{x}\,.

Multiply the numerator & denominator by 1 + cos(r) .
 
I did so, using L'Hospital. He had no knowledge of this theorem. Thanks for the tip.

http://img6.imageshack.us/img6/4122/digitalizar0030b.jpg
 
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  • #10
Two of the very first trig limits that most people learn in Calculus are
\lim_{x\to 0}\frac{sin(x)}{x}= 1
and
\lim_{x\to 0}\frac{1- cos(x)}{x}= 0
 
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