# Calculus II

1. Feb 26, 2012

### Tiago1818

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 [Broken]

Last edited by a moderator: May 5, 2017
2. Feb 26, 2012

### SammyS

Staff Emeritus
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.

Last edited: Feb 26, 2012
3. Feb 26, 2012

### Tiago1818

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.

4. Feb 26, 2012

### micromass

Staff Emeritus
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.

5. Feb 26, 2012

### Tiago1818

I did it.

http://img825.imageshack.us/img825/237/digitalizar0029.png [Broken]

I do not know what to do now.

Last edited by a moderator: May 5, 2017
6. Feb 26, 2012

### Tiago1818

know yet, the answer is "2"

7. Feb 26, 2012

### I like Serena

8. Feb 26, 2012

### SammyS

Staff Emeritus
Hopefully, you know the limit: $\displaystyle \lim_{x\to0}\ \frac{\sin(x)}{x}\,.$

Multiply the numerator & denominator by 1 + cos(r) .

9. Feb 27, 2012

### Tiago1818

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 [Broken]

Last edited by a moderator: May 5, 2017
10. Feb 27, 2012

### HallsofIvy

Staff Emeritus
Two of the very first trig limits that most people learn in Calculus are
$$\lim_{x\to 0}\frac{sin(x)}{x}= 1$$
and
$$\lim_{x\to 0}\frac{1- cos(x)}{x}= 0$$