Limits of Multivariable Functions

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 2K views
CoolDude420
Messages
199
Reaction score
9

Homework Statement


Find the following limit:
502f2de48d.png


Homework Equations

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)
 
Physics news on Phys.org
CoolDude420 said:
But if you try and sub in 0 for x and y
When you sub in ##\epsilon_x## and ##\epsilon_y ## things might go better. But factoring shows that straight away.
 
CoolDude420 said:
rational functions which are a ratio of two polynomials are continuous on R^2.
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.
CoolDude420 said:
the limits of continuous functions can be computed by direct substitution.
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:
  • Like
Likes   Reactions: BvU
CoolDude420 said:

Homework Statement


Find the following limit:
502f2de48d.png


Homework Equations

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)
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.