Question

Calc III Tutorial: Limits and Continuities of Multi-Variable Functions

Answer 1

\({\bf{Definitions:}}\)
\(\lim_{(x,y) \rightarrow (x0,y0)}f(x,y) = L\)
if, for ε > 0 there exists δ < 0 such that for all (x,y)
|f(x,y) - L| < ε wherever 0 < sqrt((x-x0)^2 + (y-y0)^2) < δ

Limit Properties: basically the same as the limit properties from calc i
1. adding or subtracting two functions --> add or subtract the limits
2. multiply or divide two functions --> multiply or divide the limits
3. multiply a function by a constant --> multiply the limit by the same constant
4. raise a function to a constant (including roots) --> perform the same operation on the limit

if you are asked to find a limit, you would start by just plugging in the given x,y values into the function, but of course this is an issue when you get undefined/an indeterminate form. afaik L'hopital's doesn't work for multi-variable functions so you will have to manipulate the equation algebraically and apply the delta-epsilon proof

Answer 2

ex. limit of 4xy^2 / (x^2 + y^2)

if x = 0, then the limit of the function is 0 for nonzero y (just plugging in x = 0)
likewise if y = 0 then the limit of the function is 0 for nonzero x

so the limit is likely to be 0, we can test this using a delta-epsilon proof (x0,y0) --> (0,0)

|4xy^2 / (x^2 + y^2) - L| < ε for 0 < sqrt(x^2 + y^2) < δ

y^2 <= y^2 + x^2, assuming real x, which means y^2 / (y^2 + x^2) will always be less than 1, making 4xy^2 / (x^2 + y^2) <= 4x = 4sqrt(x^2) <= 4sqrt(x^2 + y^2)

plugging this into the epsilon-portion of the proof gives us

4sqrt(x^2 + y^2) < ε
0 < sqrt(x^2 + y^2) < δ
making δ = ε/4 and fulfilling both conditions of the proof
so the limit is 0

Answer 3

\({\bf{Continuities:}}\) function f(x,y) is continuous at (x0,y0) if
1. f(x0,y0) is defined
2. the limit of f exists as (x,y) --> (x0,y0)
3. limit of f(x,y) = f(x0,y0) as (x,y) --> (x0,y0)
the function is continuous if it satisfies these three conditions across its domain

Two-path test: if the function has diff. limits across two different paths in the domain of f as (x,y) approaches (x0,y0) then the limit f does not exist

Answer 4

Source material is section 14.2 of Thomas' Calculus, Early Transcendentals, 14th edition by Hass, Heil, Weir, et. al.