Limits of Multivariable Functions

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The discussion centers on the continuity of the rational function f(x,y) = (x^2+y^2-x^3-xy^2)/(x^2+y^2) and the computation of its limit at the point (0,0). While rational functions are generally continuous on R^2, this specific function is not continuous at (0,0) due to the 0/0 indeterminate form when substituting these values. The conversation highlights that additional constraints on the polynomials may be necessary for the continuity statement to hold true. It is suggested that defining f(0,0) as 1 could address the removable discontinuity at that point. The importance of understanding the conditions under which direct substitution applies is emphasized.
CoolDude420
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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)
 
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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.
 
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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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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