2.1: Limits
( \newcommand{\kernel}{\mathrm{null}\,}\)
Before we really start, let's recall some useful notation.
- N is the set {1,2,3,⋯} of all natural numbers.
- R is the set of all real numbers.
- ∈ is read “is an element of”.
- ∉ is read “is not an element of”.
- {A|B} is read “the set of all A such that B”
- If S is a set and T is a subset of S, then S∖T is {x∈S|x∉T}, the set S with the elements of T removed. In particular, if S is a set and a is an element of S, then S∖{a}={x∈S|x≠ a} is the set S with the element a removed.
- If n is a natural number, Rn is used for both the set of n-component vectors ⟨x1,x2,⋯,xn⟩ and the set of points (x1,x2,⋯,xn) with n coordinates.
- If S and T are sets, then f:S→T means that f is a function which assigns to each element of S an element of T. The set S is called the domain of f.
-
[a,b]={x∈R|a≤x≤b}(a,b]={x∈R|a<x≤b}[a,b)={x∈R|a≤x<b}(a,b)={x∈R|a<x<b}
The definition of the limit of a function of more than one variable looks just like the definition 1 of the limit of a function of one variable. Very roughly speaking
lim→x→→af(→x)=L
if f(→x) approaches L whenever →x approaches →a. Here is a more careful definition of limit.
Let
- m and n be natural numbers 2
- →a∈Rm
- the function f(→x) be defined for all →x near 3 →a and take values in Rn
- L∈Rn
We write
lim→x→→af(→x)=L
if 4 the value of the function f(→x) is sure to be arbitrarily close to L whenever the value of →x is close enough to →a, without 5 being exactly →a.
Now that we have extended the definition of limit, we can extend the definition of continuity.
Let
- m and n be natural numbers
- →a∈Rm
- the function f(→x) be defined for all →x near →a and take values in Rn
- The function f is continuous at a point →a if
lim→x→→af(→x)=f(→a)
- The function f is continuous on a set D if it is continuous at every point of D.
Here are a few very simple examples. There will be some more substantial examples later — after, as we did in the CLP-1 text, we build some tools that can be used to build complicated limits from simpler ones.
- If f(x,y) is the constant function which always takes the value L, then
lim(x,y)→(a,b)f(x,y)=L
- If f:R2→R2 is defined by f(x,y)=(x,y), then
lim(x,y)→(a,b)f(x,y)=(a,b)
- By definition, as (x,y) approaches (a,b), x approaches a and y approaches b, so that if f:R2→R is defined by f(x,y)=x, then
lim(x,y)→(a,b)f(x,y)=a
Similarly, if g:R2→R is defined by g(x,y)=y, thenlim(x,y)→(a,b)g(x,y)=b
Limits of multivariable functions have much the same computational properties as limits of functions of one variable. The following theorem summarizes a bunch of them. For simplicity, it concerns primarily real valued functions. That is, functions that output real numbers as opposed to vectors. However it does contain one vector valued function. The function X in the theorem takes as input an n-component vector and returns an m-component vector. We will not deal with many vector valued functions here in CLP-3, but we will see a lot in CLP-4.
Let
- m and n be natural numbers
- →a∈Rm and →b∈Rn
- D be a subset of Rm that contains all →x∈Rm that are near →a
- c,F,G∈R
and
f,g:D∖{→a}→RX:Rm∖{→b}→D∖{→a}γ:R→R
Assume that
lim→x→→af(→x)=Flim→x→→ag(→x)=Glim→y→→bX(→y)=→alimt→Fγ(t)=γ(F)
Then
- lim→x→→a[f(→x)+g(→x)]=F+G
lim→x→→a[f(→x)−g(→x)]=F−G
- lim→x→→af(→x)g(→x)=FG
lim→x→→acf(→x)=cF
- lim→x→→af(→x)g(→x)=FG if G≠0
- lim→y→→bf(X(→y))=F
- lim→x→→aγ(f(→x))=γ(F)
This shows that multivariable limits interact very nicely with arithmetic, just as single variable limits did. Also recall, from Theorem 1.6.8 in the CLP-1 text,
The following functions are continuous everywhere in their domains
- polynomials, rational functions
- roots and powers
- trig functions and their inverses
- exponential and the logarithm
In this example we evaluate
lim(x,y)→(2,3)x+sinyx2y2+1
as a typical application of Theorem 2.1.5. Here “a=” means that part (a) of Theorem 2.1.5 justifies that equality. Start by computing separately the limits of the numerator and denominator.
lim(x,y)→(2,3)(x+siny) a=lim(x,y)→(2,3)x+lim(x,y)→(2,3)sinye=lim(x,y)→(2,3)x+sin(lim(x,y)→(2,3)y)= 2+sin3lim(x,y)→(2,3)(x2y2+1) a= lim(x,y)→(2,3)x2y2+lim(x,y)→(2,3)1b= (lim(x,y)→(2,3)x)(lim(x,y)→(2,3)x)(lim(x,y)→(2,3)y)(lim(x,y)→(2,3)y)+1= 2232+1
Since the limit of the denominator is nonzero, we can simply divide.
lim(x,y)→(2,3)x+sinyx2y2+1 c= lim(x,y)→(2,3)(x+siny)lim(x,y)→(2,3)(x2y2+1)= 2+sin337
Here we have used that sinx is a continuous function.
While the CLP-1 text's Definition 1.3.3 of the limit of a function of one variable, and our Definition 2.1.2 of the limit of a multivariable function look virtually identical, there is a substantial practical difference between the two. In dimension one, you can approach a point from the left or from the right and that's it. There are only two possible directions of approach. In two or more dimensions there is “much more room” and there are infinitely many possible types of approach. One can even spiral in to a point. See the middle and right hand figures below.
The next few examples illustrate the impact that the“extra room” in dimensions greater than one has on limits.
As a second example, we consider lim(x,y)→(0,0)x2yx2+y2. In this example, both the numerator, x2y, and the denominator, x2+y2, tend to zero as (x,y) approaches (0,0), so we have to be more careful.
A good way to see the behaviour of a function f(x,y) when (x,y) is close to (0,0) is to switch to the polar coordinates, r,θ, that are defined by
x=rcosθy=rsinθ
The points (x,y) that are close to (0,0) are those with small r, regardless of what θ is. Recall that lim(x,y)→(0,0)f(x,y)=L when f(x,y) approaches L as (x,y) approaches (0,0). Substituting x=rcosθ, y=rsinθ into that statement turns it into the statement that lim(x,y)→(0,0)f(x,y)=L when f(rcosθ,rsinθ) approaches L as r approaches 0. For our current example
x2yx2+y2=(rcosθ)2(rsinθ)r2=rcos2θsinθ
As |rcos2θsinθ|≤r tends to 0 as r tends to 0 (regardless of what θ does as r tends to 0) we have
lim(x,y)→(0,0)x2yx2+y2=0
As a third example, we consider lim(x,y)→(0,0)x2−y2x2+y2. Once again, the best way to see the behaviour of f(x,y)=x2−y2x2+y2 for (x,y) close to (0,0) is to switch to polar coordinates.
f(x,y)=x2−y2x2+y2=(rcosθ)2−(rsinθ)2r2=cos2θ−sin2θ=cos(2θ)
Note that, this time, f is independent of r but does depend on θ. Here is a greatly magnified sketch of a number of level curves for f(x,y).
Observe that
- as (x,y) approaches (0,0) along the ray with 2θ=30∘, f(x,y) approaches the value √32 (and in fact f(x,y) takes the value cos(30∘)=√32 at every point of that ray)
- as (x,y) approaches (0,0) along the ray with 2θ=60∘, f(x,y) approaches the value 12 (and in fact f(x,y) takes the value cos(60∘)=12 at every point of that ray)
- as (x,y) approaches (0,0) along the ray with 2θ=90∘, f(x,y) approaches the value 0 (and in fact f(x,y) takes the value cos(90∘)=0 at every point of that ray)
- and so on
So there is not single number L such that f(x,y) approaches L as r=|(x,y)|→0, no matter what the direction of approach is. The limit lim(x,y)→(0,0)x2−y2x2+y2 does not exist.
Here is another way to come to the same conclusion.
- Pick any really small positive number. We'll use 10−137 as an example.
- Pick any real number F between −1 and 1. We'll use F=√32 as an example.
- Looking at the sketch above, we see that f(x,y) takes the value F along an entire ray θ=const, r>0. In the case F=√32, the ray is 2θ=30∘, r>0. In particular, because the ray extends all the way to (0,0), f takes the value F for some (x,y) obeying |(x,y)|<10−137.
- That is true regardless of which really small number you picked. So f(x,y)=x2−y2x2+y2 does not approach any single value as r=|(x,y)| approaches 0 and we conclude that lim(x,y)→(0,0)x2−y2x2+y2 does not exist.
Optional — A Nasty Limit That Doesn't Exist
In this example we study the behaviour of the function
f(x,y)={(2x−y)2x−yif x≠y0if x=y
as (x,y)→(0,0). Here is a graph of the level curve, f(x,y)=−3, for this function.
Here is a larger graph of level curves, f(x,y)=c, for various values of the constant c.
As before, it helps to convert to polar coordinates — it is a good approach 6 . In polar coordinates
f(rcosθ,rsinθ)={r(2cosθ−sinθ)2cosθ−sinθif cosθ≠sinθ0if cosθ=sinθ
If we approach the origin along any fixed ray θ=const, then f(rcosθ,rsinθ) is the constant (2cosθ−sinθ)2cosθ−sinθ (or 0 if cosθ=sinθ) times r and so approaches zero as r approaches zero. You can see this in the figure below, which shows the level curves again, with the rays θ=18π and θ=316π superimposed.
If you move towards the origin on either of those rays, you first cross the f=3 level curve, then the f=2 level curve, then the f=1 level curve, then the f=12 level curve, and so on.
That f(x,y)→0 as (x,y)→(0,0) along any fixed ray is suggestive, but does not imply that the limit exists and is zero. Recall that to have lim(x,y)→(0,0)f(x,y)=0, we need f(x,y)→0 no matter how (x,y)→(0,0). It is not sufficient to check only straight line approaches.
In fact, the limit of f(x,y) as (x,y)→(0,0) does not exist. A good way to see this is to observe that if you fix any r>0, no matter how small, f(x,y) takes all values from −∞ to +∞ on the circle x2+y2=r2. You can see this in the figure below, which shows the level curves yet again, with a circle x2+y2=r2 superimposed. For every single −∞<c<∞, the level curve f(x,y)=c crosses the circle.
Consequently there is no one number L such that f(x,y) is close to L whenever (x,y) is sufficiently close to (0,0). The limit lim(x,y)→(0,0)f(x,y) does not exist.
Another way to see that f(x,y) does not have any limit as (x,y)→(0,0) is to show that f(x,y) does not have a limit as (x,y) approaches (0,0) along some specific curve. This can be done by picking a curve that makes the denominator, x−y, tend to zero very quickly. One such curve is x−y=x3 or, equivalently, y=x−x3. Along this curve, for x≠0,
f(x,x−x3)=(2x−x+x3)2x−x+x3=(x+x3)2x3=(1+x2)2x⟶{+∞as x→0 with x>0−∞as x→0 with x<0
The choice of the specific power x3 is not important. Any power xp with p>2 will have the same effect.
If we send (x,y) to (0,0) along the curve x−y=ax2 or, equivalently, y=x−ax2, where a is a nonzero constant,
limx→0f(x,x−ax2)=limx→0(2x−x+ax2)2x−x+ax2=limx→0(x+ax2)2ax2=limx→0(1+ax)2a=1a
This limit depends on the choice of the constant a. Once again, this proves that f(x,y) does not have a limit as (x,y)→(0,0).
Exercises
Stage 1
Suppose f(x,y) is a function such that lim(x,y)→(0,0)f(x,y)=10.
True or false: |f(0.1,0.1)−10|<|f(0.2,0.2)−10|
A millstone pounds wheat into flour. The wheat sits in a basin, and the millstone pounds up and down.
Samples of wheat are taken from various places along the basin. Their diameters are measured and their position on the basin is recorded.
Consider this claim: “As the particles get very close to the millstone, the diameters of the particles approach 50 μm.” In this context, describe the variables below from Definition 2.1.2.
- x
- a
- L
Let f(x,y)=x2x2+y2.
- Find a ray approaching the origin along which f(x,y)=1.
- Find a ray approaching the origin along which f(x,y)=0.
- What does the above work show about a limit of f(x,y)?
Let f(x,y)=x2−y2
- Express the function in terms of the polar coordinates r and θ, and simplify.
- Suppose (x,y) is a distance of 1 from the origin. What are the largest and smallest values of f(x,y)?
- Let r>0. Suppose (x,y) is a distance of r from the origin. What are the largest and smallest values of f(x,y)?
- Let ϵ>0. Find a positive value of r that guarantees |f(x,y)|<ϵ whenever (x,y) is at most r units from the origin.
- What did you just show?
Suppose f(x,y) is a polynomial. Evaluate lim(x,y)→(a,b)f(x,y), where (a,b)∈R2.
Stage 2
Evaluate, if possible,
- lim(x,y)→(2,−1) (xy+x2)
- lim(x,y)→(0,0) xx2+y2
- lim(x,y)→(0,0) x2x2+y2
- lim(x,y)→(0,0) x3x2+y2
- lim(x,y)→(0,0) x2y2x2+y4
- lim(x,y)→(0,0) (sinx)(ey−1)xy
- Find the limit: \displaystyle \lim_{(x,y)\to(0,0)}\frac{x^8+y^8}{x^4+y^4}\text{.}
- Prove that the following limit does not exist: \displaystyle \lim_{(x,y)\to(0,0)}\frac{xy^5}{x^8+y^{10}}\text{.}
Evaluate each of the following limits or show that it does not exist.
- \displaystyle \lim_{(x,y)\rightarrow (0,0)}\frac{x^3-y^3}{x^2+y^2}
- \displaystyle \lim_{(x,y)\rightarrow (0,0)}\frac{x^2-y^4}{x^2+y^4}
Stage 3
Evaluate each of the following limits or show that it does not exist.
- \displaystyle \lim_{(x,y)\rightarrow (0,0)}\frac{2x^2 + x^2y - y^2x + 2y^2}{x^2 + y^2}
- \displaystyle \lim_{(x,y)\rightarrow(0,1)} \frac{x^2y^2 -2 x^2y + x^2} {(x^2 + y^2-2y+1)^2}
Define, for all (x,y)\ne(0,0)\text{,} f(x,y)=\frac{x^2y}{x^4+y^2}\text{.}
- Let 0\le \theta \lt 2\pi\text{.} Compute \displaystyle\lim_{r\rightarrow 0^+}f(r\cos\theta,r\sin\theta)\text{.}
- Compute \displaystyle\lim_{x\rightarrow 0}f(x,x^2)\text{.}
- Does \displaystyle\lim_{(x,y)\rightarrow (0,0)}f(x,y) exist?
Compute the following limits or explain why they do not exist.
- \displaystyle \lim_{(x,y)\rightarrow(0,0)}\frac{xy}{x^2+y^2}
- \displaystyle \lim_{(x,y)\rightarrow(0,0)}\frac{\sin(xy)}{x^2+y^2}
- \displaystyle \lim_{(x,y)\rightarrow(-1,1)}\frac{x^2+2xy^2+y^4}{1+y^4}
- \displaystyle \lim_{(x,y)\rightarrow(0,0)}|y|^x
- Definition 1.3.3 in the CLP-1 text.
- In this text, we will interested in m,n\in\big\{1,2,3\big\}\text{,} but the definition works for all natural numbers m,n\text{.}
- To be precise, there is a number r\gt 0 such that f(\vec{x}) is defined for all \vec{x} obeying |\vec{x}-\vec{a}|\lt r\text{.}
- There is a precise, formal version of this definition that looks just like Definition 1.7.1 of the CLP-1 text.
- You may find the condition “without being exactly \vec{a}” a little strange, but there is a good reason for it, which we have already seen in Calculus I. In the definition f'(x) = \lim\limits_{x\rightarrow a} \frac{f(x)-f(a)}{x-a}\text{,} the function whose limit is being taken, namely \frac{f(x)-f(a)}{x-a}\text{,} is not defined at all at x=a\text{.} This will again happen when we define derivatives of functions of more than one variable.
- Not just a pun.