Multivariable Function Limits: UNIZOR.COM - Math4Teens - Calculus - Limi...

Notes to a video lecture on http://www.unizor.com

Multivariable Limits

Sometimes we have to deal with functions of two or more arguments and have to analyze the behavior of such functions as their arguments approach certain values.

The usual way to analyze this situation is to fix all arguments except one and see what happens with the function if that single argument approach the value we are interested in.
The result of this process is the reduction of variables by one, and we can repeat the same thing for the next argument, then the next etc.

For example, consider a function
F(x,y,z) =
= arctan(x) + 2^−y + z/(z+1)
and its behavior when all arguments increase without restriction.
1. Fix x and y, let z→+∞.
lim_z→+∞ z/(z+1) = 1
2. Our function now can be written as
F(x,y,+∞) = arctan(x)+2 ^−y+1
Fix x, let y→+∞.
lim_y→+∞ 2^−y = 0
3. Our function now is
F(x,+∞,+∞) = arctan(x)+0+1
Let x→+∞.
lim_x→+∞ arctan(x) = π/2
The limit of our function when z→+∞, y→+∞, x→+∞ is
lim F(x,y,z) = π/2 + 1

This is great, but we have a result that seems to depend on the order of arguments we analyze.
Can the result be different if we choose a different order, say, fix z and y getting a limit by x, then fix z getting a limit by y and finish with a limit by z?

The answer is:
Under certain relatively broad conditions, the final result is not dependent on the order of taking limits by different arguments.

We consider only a case of a function of two arguments (in a general case of many arguments the proof is similar but a bit more tedious) and prove the theorem below.

Prior to that, let's talk about a particular type of taking a multivariable function to a limit - a uniform convergence.
Assume, we have to analyze the limit of function F(x,y) as x→a and y→b.
We say that function F(x,y) is converging uniformly by y to Y(y) as x→a if for any positive ε there is positive δ such that
|F(x,y)−Y(y)| < ε
as long as both |x−a| < δ and |y−b| < δ.

In other words, as x→a, not only F(x,y)→Y(y) for any specific value of y (so-called point convergence of F(x,y) to Y(y)), but F(x,y) gets equally close to Y(y) (distance less than small ε) for all values of y within certain neighborhood of its limit point y=b.

Theorem

Assume that all limits below exists and satisfy the conditions listed.
IF
lim_y→b F(x,y) = X(x)
         uniformly by x
lim_x→a X(x) = A
lim_x→a F(x,y) = Y(y)
         uniformly by y
lim_y→b Y(y) = B
THEN
A = B

Proof

Assume, A ≠ B.
Let |A−B|=d.
Let ε=d/4.

Then the following logic holds.

lim_x→a F(x,y) = Y(y) (uniformly by y) ⇒
⇒ ∃δ_a: |Y(y)−F(x,y)| < ε ∀x∈[a−δ_a,a+δ_a] and
∀y∈[b−δ_a,b+δ_a]

lim_y→b Y(y) = B ⇒
⇒ ∃δ_b: |B−Y(y)| < ε
∀y∈[b−δ_b,b+δ_b]

Let δ = min(δ_a,δ_b)
Then from both above statements follows that within a small neighborhood of limit points x=a and y=b
∀x∈[a−δ,a+δ] and
∀y∈[b−δ,b+δ]
the following is true
|B−Y(y)|<ε and
|Y(y)−F(x,y)|<ε
from which, in turn, follows
|B−F(x,y)|<2ε=d/2

That is, within a small distance of arguments x and y around their limit points a and b the value of F(x,y) deviates from B by less than d/2.

Now repeat the same from the position of another limit of function F(x,y).

lim_y→b F(x,y) = X(x) (uniformly by x) ⇒
⇒ ∃γ_b: |X(x)−F(x,y)| < ε ∀x∈[a−γ_b,a+γ_b] and
∀y∈[b−γ_b,b+γ_b]

lim_x→a X(x) = A ⇒
⇒ ∃γ_a: |A−X(x)| < ε
∀x∈[a−γ_a,a+γ_a]

Let γ = min(γ_a,γ_b)
Then from both above statements follows that within a small neighborhood of limit points x=a and y=b
∀x∈[a−γ,a+γ] and
∀y∈[b−γ,b+γ]
the following is true
|A−X(x)|<ε and
|X(x)−F(x,y)|<ε
from which, in turn, follows
|A−F(x,y)|<2ε=d/2

That is, within a small distance of arguments x and y around their limit points a and b the value of F(x,y) deviates from A by less than d/2.

Make our neighborhood even smaller by choosing β=min(γ,δ) and consider
∀x∈[a−β,a+β] and
∀y∈[b−β,b+β]

Now we see that within this neighborhood F(x,y) is closer to A by less than d/2 and closer to B by less than d/2, where d is the distance between A and B, which is impossible.

Therefore, A = B.

Double Sequence Limits: UNIZOR.COM - Math4Teens - Calculus - Limit of Se...

Notes to a video lecture on http://www.unizor.com

Double Limits

Sometimes we are interested in sequences that depend on two natural numbers like {a_m,n}.
For example,
a_m,n=arctan(m)+2⁻ⁿ.

Now, if m→∞ and n→∞, how to calculate the limit of a_m,n?
It might be
lim_m→∞ [lim_n→∞ (a_m,n)]
or
lim_n→∞ [lim_m→∞ (a_m,n)]
depending on which index, m or n, we will use to go to a limit first, and there is no guarantee that the answers will be the same.

Let's check both methods.
lim_n→∞ (a_m,n) =
= lim_n→∞ (arctan(m) + 2⁻ⁿ) =
= arctan(m)
lim_m→∞ (arctan(m)) = π/2

If we change the order of limits,
lim_m→∞ (a_m,n) =
= lim_m→∞ (arctan(m) + 2⁻ⁿ) =
= π/2 + 2⁻ⁿ
lim_n→∞ (π/2 + 2⁻ⁿ) = π/2

As you see, the results are the same.

In some way, the equality of these two limits might be considered similar to a standard accounting procedure of checking the calculations.
Imagine an M⨯N matrix with numbers.
If you summarize the numbers within each of M rows with row #i having a sum R_i and summarize all these row totals by i from 1 to M, you will get a total of all numbers in an original table.
If you summarize the numbers within each of N columns with column #j having a sum S_j and summarize all these column totals by j from 1 to N, you should get exactly the same total of all numbers.
If these two calculated totals are not equal, that is if
Σ_i∈[1,M] R_i ≠ Σ_j∈[1,N] S_j
then your calculations are wrong.

The equality of two limits in the example above is not coincidental. We shall prove it as the theorem below.
But, to make our proof rigorous, we have to make a short comment about uniform convergence.
We say that sequence {a_m,n} is converging uniformly by m to b_m as n→∞ if for any positive ε there is number K such that
|b_m−a_m,n| < ε
as long as both n and m are greater than K.
Analogously, we say that sequence {a_m,n} is converging uniformly by n to c_n as m→∞ if for any positive ε there is number K such that
|c_n−a_m,n| < ε
as long as both n and m are greater than K.

Theorem

IF
lim_n→∞ (a_m,n) = b_m
(uniformly by m)
lim_m→∞ (b_m) = B
lim_m→∞ (a_m,n) = c_n
(uniformly by n)
lim_n→∞ (c_n) = C
THEN
B = C

Proof

Assume, B≠C and |B−C|=d.

Since lim_m→∞(b_m)=B, there exists some natural number M₁ such that
|B−b_m| < d/4 for all m>M₁

Since
lim_n→∞(a_m,n)=uniformly=b_m
there exists some natural number N₁ greater than M₁ such that
|a_m,n−b_m| < d/4 for all m,n>N₁

Then, for all pairs m,n greater than N₁ both following inequalities are true:
|B − b_m| < d/4 and
|b_m − a_m,n| < d/4
Therefore,
|B−a_m,n| < d/2
That is, our double sequence a_m,n with both indices m and n greater than N₁ is closer to number B than d/2.

Let's use the same approach, but start from index n.
Since lim_n→∞(c_n)=C, there exists some natural number N₂ such that
|C−c_n| < d/4 for all n>N₂

Since
lim_m→∞(a_m,n)=uniformly=c_n
there exists some natural number M₂ greater than N₂ such that
|a_m,n−c_n| < d/4 for all m,n>M₂

Then, for all pairs m,n greater than M₂ both following inequalities are true:
|C − c_n| < d/4 and
|c_n − a_m,n| < d/4
Therefore,
|C−a_m,n| < d/2
That is, our double sequence a_m,n with both indices m and n greater than M₂ is closer to number C than d/2.

Choosing K=max(N₁,M₂) we see that all a_m,n with indices greater than K should be closer to B than d/2 and at the same time should be closer to C than d/2.

With the distance between B and C equal to d it's impossible.
We came to contradiction that signifies that our initial assumption that B≠C is wrong.

Therefore, B=C.

For sequences that satisfy the conditions of the theorem above we can define their limit regardless of how we calculate it, starting from the first index or the second, and can combine indices in the notation
lim_m,n→∞ a_m,n = B

Not every sequence satisfies the conditions of this theorem.
For example, a_m,n=m/n has no unconditional limit.
If m is fixed and we vary n→∞, each b_m is zero.
Hence, lim_m→∞ b_m = 0
But if we fix n and vary m→∞, the resulting sequence {1/n,2/n,3/n...} has no limit for any n.