# Limits of Multivariable Functions

1. Nov 1, 2016

### CoolDude420

1. The problem statement, all variables and given/known data
Find the following limit:

2. Relevant equations

3. The attempt at a solution
My lecturer has said that rational functions which are a ratio of two polynomials are continuous on R^2. He also said that the limits of continuous functions can be computed by direct substitution.

The function here f(x,y) = x^2+y^2-x^3-xy^2/x^2+y^2 is a rational function(ratio of two polynomials) so it should be continuous on R^2, which means we can compute the limit by direct substitution. But if you try and sub in 0 for x and y, you get 0/0 which means the function is not continuous at (0,0)???

( I know I am meant to factor them and then do it but I don't understand why it isn't working like this straight up if the rules say it should)

2. Nov 1, 2016

### BvU

When you sub in $\epsilon_x$ and $\epsilon_y$ things might go better. But factoring shows that straight away.

3. Nov 1, 2016

### andrewkirk

That doesn't sound right. The function $f:\mathbb R^2\to \mathbb R$ such that $f(x,y)=1/x$ is a ratio of two polynomials, but is not continuous. Additional constraints are needed on the types of polynomial for the statement to be true. I suspect (but have not checked it) that a sufficient constraint might be the requirement that the lowest order of any term in the numerator is no lower than the lowest order of any term in the denominator.
That also sounds wrong. The function $g:[0,\infty)\to\mathbb R$ such that $g(0)=0$ and, for nonzero $x,\ g(x)=x/\log x$, is continuous but its value at $x=0$ cannot be calculated by direct substitution. I suppose it depends on exactly what is meant by 'direct substitution' though.

Last edited: Nov 1, 2016
4. Nov 2, 2016

### SammyS

Staff Emeritus
I suppose he could mean that any rational function is continuous over its domain. In the case of this rational function, the domain is ℝ2\(0,0).

However, you are correct about the function not being continuous at the origin, since it is not defined there. However, that appears to be a removable discontinuity, so define f(0, 0) = 1 in addition to what you have.