Now we get to actual calculus dealing with functions of two variables, and we naturally begin with limits. Recall from Calculus 1 the definition of a limit:
A function f(x) has limit L as x approaches c if ∣f(x)−L∣ (the distance between the function values and the limit) gets arbitrary small (∣f(x)−L∣<ϵ) for values of x that are sufficiently close to c (∣x−c∣<δ).
Let's extend this definition to functions of two variables in R3. Instead of x approaching c, we'll have (x,y) approaching (a,b). So here's the definition of what it means for
(x,y)→(a,b)limf(x,y)=L:
A function f(x,y) has limit L as (x,y)→(a,b) if ∣f(x,y)−L∣ can be made arbitrarily small for points sufficiently close to (a,b).
Limit Rules
The rules for limits with functions of two variables are essentially identical to the ones for limits with functions of one variable in Calculus 1.
First, suppose that lim(x,y)→(a,b)f(x,y)=L and lim(x,y)→(a,b)g(x,y)=M.
Constant Function: f(x,y)=c
(x,y)→(a,b)limc=c
Linear Function: f(x,y)=x or f(x,y)=y
(x,y)→(a,b)limx(x,y)→(a,b)limy=a=b
Sum
(x,y)→(a,b)lim[f(x,y)+g(x,y)]=L+M
Difference
(x,y)→(a,b)lim[f(x,y)−g(x,y)]=L−M
Constant Multiple
(x,y)→(a,b)limcf(x,y)=cL
Product
(x,y)→(a,b)lim[f(x,y)⋅g(x,y)]=L⋅M
Quotient
(x,y)→(a,b)lim[g(x,y)f(x,y)]=ML if M̸=0
Power
(x,y)→(a,b)lim[f(x,y)]n=Ln
Examples
Evaluate (x,y)→(5,−2)limx5+4x3y−5xy2.
Solution
Use the limit rules outlined above to break this down to basic forms:
(x,y)→(5,−2)limx5+4x3y−5xy2=(x,y)→(5,−2)limx5+(x,y)→(5,−2)lim4x3y−(x,y)→(5,−2)lim5xy2=((x,y)→(5,−2)limx)5+4((x,y)→(5,−2)limx)3((x,y)→(5,−2)limy)−5((x,y)→(5,−2)limx)((x,y)→(5,−2)limy)2=55+4(5)3(−2)−5(5)(−2)2=2025
Note
We could evaluate this limit by simply evaluating the function at (5,−2) because the function is continuous (a polynomial in this case).
Evaluate (x,y)→(0,0)limx2+y2x2.
Solution
If we try evaluating the function at (0,0), we'll get 0 in both the numerator and denominator. It turns out that this limit does not exist. How can we prove this? Think about the following picture in two dimensions:
If we approach from one side, the function values approach one value, and if we approach from another side, the function values approach a different value, which means that the two-sided limit doesn't exist.
We can do something similiar in three dimensions: if we approach (0,0) along two different lines and show that the limits are different in each case.
For this one, we'll approach along the x-axis and then the y-axis.
Along the x-axis: let y=0x→0limx2+02x2=x→0limx2x2=1
Along the y-axis: let x=0y→0lim02+y202=y→0lim0=0
Since these two limits differ, we've proven that the limit does not exist. We can graph this surface to observe this behavior. Notice how approaching the origin along the x-axis puts you up on the ridge, where the function values are 1, and approaching along the y-axis puts you in the valley, where the function values are 0.
Evaluate (x,y)→(0,0)limx2+2y2x2sin2y.
Solution
For this one, we'll use the Squeeze Theorem, noting that since y2≥0 regardless of what y is, that means that
x2+2y2x2≤1⟶x2+2y2x2sin2y≤sin2y
Since this is different for different values of m, we conclude that this limit does not exist.
Show that (x,y)→(0,0)limx2−y2y does not exist.
(work it out)
Continuity
Recall from Calculus I that we define continuity using limits; specifically, if the limit exists (approaching from both sides gives the same result) and equals the function value at that point, the function is continuous there. Therefore, there are three conditions (the limit exists, the function value is defined, and the two are equal):
A function f is continuous at c if x→climf(x)=f(c).
The same is true with functions of two variables: a function f(x,y) is continuous at a point (a,b) if
(x,y)→(a,b)limf(x,y)=f(a,b).
Is the following function continuous?
f(x,y)=⎩⎨⎧x2+y43xy20 if (x,y)=(0,0) if (x,y)=(0,0)
Solution
The limit of this function as (x,y) approaches (0,0) does not exist. To prove this, we need to substitute x=my2 (a linear substitution isn't good enough):
(x,y)→(0,0)limx2+y43xy2=x→0lim(m2+1)y43my4=m2+13m
Since this limit does not exist, the function is not continuous.
At what points of R2 is f(x,y)=x2−6x+7xy2 continuous?
Solution
Since this is a polynomial, it is continuous at all points in R2. There's no issue with dividing by 0 or taking the square root of a negative number.
At what points of R2 is f(x,y)=x2y2+1xy continuous?
All points in R2
At what points of R2 is f(x,y)=x2−y22xy continuous?
R2∖{(x,y):y2=x2}
At what points of R2 is f(x,y)=x(y2−1)x2+y2 continuous?
R2∖{(x,y):x=0 or y=1,−1}
Composite Functions
If we compose two continuous functions, we'll get another continuous function: f(g(x,y)) is continuous at (a,b) if g is continuous at (a,b) and f is continuous at g(a,b).
At what points of R2 is f(x,y)=x2+y2 continuous?
Solution
Since x2+y2 is continuous for all values of x and y, and x is continuous as long as x≥0, this compositive function is continuous on the following set:
{(x,y)∈R2:x2+y2≥0}
At what points of R2 is f(x,y)=sinxy continuous?
All points in R2
At what points of R2 is f(x,y)=4−x2−y2 continuous?
{(x,y)∈R2:x2+y2≤4}
At what points of R2 is f(x,y)=ln(x2+y2) continuous?