# Multivariable Calculus Limits

## Main Question or Discussion Point

lim of

cos((x^2 + y^2) - 1)/(x^2 + y^2)

as (x,y) approaches (0,0)

I have no clue how to tackle this problem. I tried to find the level set so at least I can have a clue of what the graph looks like, but then, I didn't know how to find the level sets either. If I set c = the equation, I have 2 unknowns so I cannot solve, and its not an obvious graph like a circle or something. On the other hand, I tried l'hopitale but that needs the derivative and what in the world am i taking a derivative in terms of since there are 2 variables?

you have an intermediate form of lahopital's theorey of the form 1/0
do you know how to do these?

Hurkyl
Staff Emeritus
Gold Member
I think he's missing a parenthesis -- the numerator is supposed to be cos(x²+y²) - 1. (P.S. 1/0 is not indeterminate, and AFAIK it's not L'Hôpital theory)

eutopia: what techniques have you seen used for similar problems? There is one in particular that makes this problem very simple.

its not an obvious graph like a circle or something.

MalleusScientiarum
For the limit to exist it has to exist regardless of the direction from which you are approaching the point. Parameterize lines passing through the origin and see if you can get that.

Hurkyl
Staff Emeritus
Gold Member
As you said, the limit has to exist (and be the same value) for any way you approach the origin -- just looking at the lines isn't good enough.

HallsofIvy
Homework Helper
"Lines through the origin", suggested by MalleusScientiarum, will help show that a limit does not exist by getting, hopefully, different limits on different lines. But they can't prove that a limit DOES exit (or find it) since even if the limit is the same along all lines, there might be other curves, not lines, passing through the origin that give a different limit.

The best way to handle ANY limit problem in more than one variable (going to (0,0) or (0,0,0), etc.) is to change to polar (spherical, etc.) coordinates since that way one variable, r (ρ, etc.) measures the distance to (0,0) directly! In this case, that's easy since x and y only appear in x2+ y2= r2.

The original function,
$$\frac{cos((x^2 + y^2) - 1)}{x^2 + y^2}$$
becomes
$$\frac{cos(r^2-1)}{r^2}$$
which clearly goes to infinity as r goes to 0.

Hurkyls suggested correction,
$$\frac{cos(x^2+y^2)-1}{x^2+y^2}$$
becomes
$$\frac{cos(r^2)-1}{r^2}$$
which now has only one variable and can be done by L'Hopital's rule. (The limit is 0.)