# Two variable limit question (Question/Answer included)

1. May 23, 2012

### s3a

1. The problem statement, all variables and given/known data
Problem:
Find the limit if it exists, or state that it does not exist for: lim (x,y)→(0,0) [2x^3 + 6y^3]/[x^2 + y^2].

lim (x,y)→(0,0) [2x^3 + 6y^3]/[x^2 + y^2] = 0

2. Relevant equations
r^2 = x^2 + y^2

3. The attempt at a solution
I know I can replace (x,y)→(0,0) with r→0 and x^2 + y^2 with r^2 but I am confused about what to do for the numerator.

Also, I don't think this is formally/mathematically acceptable but, if I replace the denominator with r^2 and keep the numerator with the x and y and then break the fractions to 2x^3 /r^2 + 6y^3 /r^2 and then I have, in each summand, the top cubic competing with the bottom square as both the numerator's and denominator's variables increase according to the limit so the top "wins" and the answer is 0.

While I wonder if my thinking in the above paragraph is correct, I would still like to know what the formal/mathematical way of approaching this problem is.

Any help in solving this problem would be greatly appreciated!

2. May 23, 2012

### HallsofIvy

Staff Emeritus
You do the same with the numerator. In polar coordinates, $x= r cos(\therta)$ and $y= r sin(\theta)$ so $2x^3+ 6y^3= 2r^3 cos^3(\theta)+ 6r^3 sin^3(\theta)= r^3(2cos^3(\theta)+ 6sin^3(\theta))$ and
$$\frac{2x^3+ 6y^3}{x^2+ y^2}= \frac{r^3(2cos^3(\theta)+ 6sin^3(\theta))}{r^2}$$
$$= r(2cos^3(\theta)+ 6sin^3(\theta)$$
which goes to 0 as r goes to 0 no matter what $\theta$ is.

3. May 23, 2012

### s3a

That makes a lot of sense! Thanks!