# Finding limit of function in 2 variables

1. Oct 8, 2007

### TheSaxon

1. The problem statement, all variables and given/known data
lim as (x,y)-->(0,0) of sin(x^2+y^2)/(x^2+y^2)

Questions: Does limit exist and if so, what is it.

2. Relevant equations

3. The attempt at a solution

1. The professor instructed us to convert to polar coordinates to see if result depends on theta. If it does no limit.

Conversion:

lim as r-->0 of sin(r^2 cos^2(theta) + r^2 sin^2(theta)) / (r^2 cos^2(theta) + r^2 sin^2(theta))

This I think reduces to sin(r^2)/r^2. This implies that the limit exists, correct(since the answer doesn't depend on theta)? However, this doesn't help me with actually finding the limit, as plugging in 0 still yields indeterminate 0/0. What do i do?

Last edited by a moderator: Oct 8, 2007
2. Oct 8, 2007

### bob1182006

If you notice you can make a substitution!

say u=x^2+y^2

$$lim_{(x,y)->(0,0)}\frac{\sin(x^2+y^2)}{x^2+y^2}$$
as (x,y)->(0,0) u->(0)
$$lim_{u->0} \frac{\sin u}{u}$$

and well you should know sin u/u by heart xD

also you could have done the substitution from sin(r^2)/r^2 actually it's almost the same thing since (x,y)->(0,0) (r,theta)->(0,0)

and theta is nowhere in there so you have lim r->0 sin(r^2)/r^2

Last edited: Oct 8, 2007
3. Oct 8, 2007

### TheSaxon

i guess im just dumb, i didnt know that lim of sin u / u is 1. im taking math after a 2 year hiatus :(

4. Oct 8, 2007

### bob1182006

o well it's just a limit that sometimes comes up and when it does it's useful to know it. Also it's part of the proof of derivative of sin x.

also when you have a limit like sin(r^2)/r^2 it's a calculus 1 type limit so you can use L'Hospital's rule if you get an indeterminate form, but using it here you get 0.

5. Oct 8, 2007

### TheSaxon

dont u get 1? sin(r^2)/r^2 L'hopitaled becomes (cos(r^2)2r)/2r = cos(r^2) = 1?

Anyway, thanks for help

Last edited by a moderator: Oct 8, 2007
6. Oct 8, 2007

### bob1182006

o whops sorry i did the derivative 2x and got (2rsinr^2)/2r ><

but yes you could do it multiple ways but the substitution would be fastest since you can do it straight away.

7. Oct 9, 2007

### HallsofIvy

Staff Emeritus
You don't need to use L'Hopital. You know from Calc I that
$$\lim_{x\rightarrow 0}\frac{sin(x)}{x}= 1$$
Just think of r2 as x.