## Wednesday, August 31, 2016

### Unizor - Derivatives - Properties of Function Limits

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

Function Limit - Properties of Limits

Theorem 1

If f(x)a as xr, then for any positive A
A·f(x)A·a as xr.

Proof

(1a) Using a definition of function limit based on convergence of {f(xn)} to a for any {xn}r, the proof is in the corresponding property of a sequence limit, since {A·f(xn)}converges to A·a.
Indeed, take any sequence of arguments {xn} that converges to r.
Since {f(xn)}a,
{A·f(xn)}A·a by known property of the converging sequences.
So, we have just proven that
∀ {xn}r{A·f(xn)}A·a, which corresponds to the first definition of convergence of function A·f(x) to A·a.

(1b) Let's prove this using ε-δdefinition of a function limit.
It's given that
∀ ε>0 ∃ δ>0:
|x−r| ≤ δ ⇒ |f(x)−a| ≤ ε

We have to prove that
∀ ε>0 ∃ δ>0:
|x−r| ≤ δ ⇒ |A·f(x)−A·a| ≤ ε

Using the fact that f(x)a inε-δ sense, do the following steps.
Step 1 - choose any positive ε.
Step 2 - calculate ε' = ε/A.
Step 3 - using the convergence of f(x), for this ε' find δ such that, if x is withinδ-neighborhood of r, it's true that f(x) is withinε'-neighborhood of a.
Step 4 - we have proven that, as long as x is withinδ-neighborhood of r, it is true that
a−ε' ≤ f(x) ≤ a+ε';
since ε'=ε/A, represent it using original ε as
a−ε/A ≤ f(x) ≤ a+ε/A.
Step 5 - multiply all parts of this inequality by A (it's positive, so signs of inequality are preserved) getting
A·a−ε ≤ A·f(x) ≤ A·a+ε.
This means that A·f(x) is withinε-neighborhood of A·a.
End of proof.

Theorem 2

If f(x)a as xr and
g(x)b as xr,
then
f(x)+g(x)a+b as xr.

Proof

(2a) Using a definition of function limit based on convergence of {f(xn)} to a and{g(xn)} to b for any {xn}r, the proof is in the corresponding property of a sequence limits, since {f(xn)+g(xn)} converges to a+b.

(2b) Let's prove this using ε-δdefinition of a function limit.
Step 1 - choose any positive ε.
Step 2 - calculate ε' = ε/2.
Step 3 - using the convergence of f(x), for this ε' find δ1 such that, if x is withinδ1-neighborhood of r, it's true that f(x) is withinε'-neighborhood of a.
Step 4 - using the convergence of g(x), for this ε' find δ2 such that, if x is withinδ2-neighborhood of r, it's true that g(x) is withinε'-neighborhood of b.
Step 5 - calculate
δ = min(δ1, δ2);
for this δ both statements are true:
f(x) is within ε'-neighborhoodof a and
g(x) is within ε'-neighborhoodof b.
Step 6 - we have proven that, as long as x is withinδ-neighborhood of r, it is true that
a−ε' ≤ f(x) ≤ a+ε' and
b−ε' ≤ g(x) ≤ b+ε';
since ε'=ε/2, represent it using original ε as
a−ε/2 ≤ f(x) ≤ a+ε/2 and.
b−ε/2 ≤ g(x) ≤ b+ε/2;
Step 7 - add these inequalities getting
a+b−ε ≤ f(x)+g(x) ≤ a+b+ε.
This means that f(x)+g(x) is within ε-neighborhood of a+b.
End of proof.

Theorem 3

If f(x)a as xr and
g(x)b as xr,
then
f(x)·g(x)a·b as xr.

Proof

(3a) Using a definition of function limit based on convergence of {f(xn)} to a and{g(xn)} to b for any {xn}r, the proof is in the corresponding property of a sequence limits, since {f(xn)·g(xn)} converges toa·b.

(3b) Let's make some preparations before proving this theorem using ε-δ terminology.
We have to prove that
|f(x)·g(x)−a·b|0 as xr.
Recall a known inequality about absolute values:
|A+B| ≤ |A|+|B|.
We will use it in this proof.

Since both our functions have finite limits, their values are sufficiently close to their corresponding limits as long as an argument is sufficiently close to a limit point.
In particular, there is always such δ1 that |f(x)−a| ≤ 1 for|x−r| ≤ δ1 and there is always such δ2 that |g(x)−b| ≤ 1 for|x−r| ≤ δ2.
Therefore, for|x−r| ≤ δ3=min(δ12) we can be sure that |f(x)| ≤ |a|+1 and|g(x)| ≤ |b|+1.

Make an invariant transformation with our expression:
|f(x)·g(x)−a·b| =
|f(x)·g(x)−a·g(x)+a·g(x)−a·b|

The new expression, using the inequality with absolute values mentioned above, is bounded from above as follows:
|f(x)·g(x)−a·g(x)+a·g(x)−a·b|
|f(x)·g(x)−a·g(x)| +
|a·g(x)−a·b| =
|g(x)|·|f(x)−a|+|a|·|g(x)−b| ≤
≤ (|b|+1)·|f(x)−a|+|a|·|g(x)−b|

After these preparations we are ready to prove the theorem using ε-δ definition of a function limit.
Step 1 - choose any positive ε.
Step 2 - let ε1 = ε/[2(|b|+1)].
Step 3 - using the convergence of f(x), for this ε1 find δ4 such that, if x is withinδ4-neighborhood of r, it's true that f(x) is withinε1-neighborhood of a.
Step 4 - let ε2 = ε/(2|a|).
Step 5 - using the convergence of g(x), for this ε2 find δ5 such that, if x is withinδ5-neighborhood of r, it's true that g(x) is withinε2-neighborhood of b.
Step 6 - calculate
δ = min(δ1, δ2, δ3, δ4, δ5);
for this δ both statements are true:
f(x) is within ε1-neighborhoodof a and
g(x) is within ε2-neighborhoodof b.
Step 7 - we have proven that, as long as x is withinδ-neighborhood of r, it is true that
|f(x)−a| ≤ ε1 and
|g(x)−b| ≤ ε2;
since ε1=ε/(|b|+1) and ε2=ε/|a|,
represent this using original ε as
|f(x)−a| ≤ ε/(|b|+1) and
|g(x)−b| ≤ ε/|a|;
Step 7 - use the above inequalities in the transformed inequality that we have to prove:
|f(x)·g(x)−a·b| ≤
≤ (|b|+1ε/[2(|b|+1)] +
+ |aε/(2|a|) = ε

This means that f(x)·g(x) is within ε-neighborhood of a·b.
End of proof.

Theorem 4

If f(x)a ≠ 0 as xr
then
1/f(x)1/a as xr.

Proof

(4a) Using a definition of function limit based on convergence of {f(xn)} to a for any {xn}r, the proof is in the corresponding property of a sequence limits, since {1/f(xn)}converges to 1/a.

(4b) Let's prove this using ε-δdefinition of a function limit.
It's given that
∀ ε>0 ∃ δ>0:
|x−r| ≤ δ ⇒ |f(x)−a| ≤ ε

We have to prove that, assuming a ≠ 0,
∀ ε>0 ∃ δ>0:
|x−r| ≤ δ ⇒ |1/f(x)−1/a| ≤ ε

First of all, since a ≠ 0, function values of f(x) are separated from 0 as long as argument x is sufficiently close to a limit point r.
Indeed, take ε = a/2 and find δ1such that f(x) is inε=a/2-neighborhood of a (that is, a/2 ≤ f(x) ≤ 3a/2) as long as argument x is inδ1-neighborhood of the limit point r.

Notice now that
|1/f(x)−1/a|=|f(x)−a| / |f(x)|·|a|
Since a/2 ≤ f(x) ≤ 3a/2 in theδ1-neighborhood of the limit point r, the denominator of the last expression is greater thana²/2.
Therefore,
|1/f(x)−1/a| ≤ 2·|f(x)−a| /
Using the fact that f(x)a inε-δ sense, do the following steps.
Step 1 - choose any positive ε.
Step 2 - calculate ε' = ε·a²/2.
Step 3 - using the convergence of f(x), for this ε' find δ2 such that, if x is withinδ2-neighborhood of r, it's true that f(x) is withinε'-neighborhood of a.
Step 4 - set δ = min(δ1, δ2).
Step 5 - we have proven that, as long as x is withinδ-neighborhood of r, it is true that
|f(x) − a| ≤ ε' = ε·a²/2.
Step 6 - substitute this into inequality above:
|1/f(x)−1/a| ≤ 2·|f(x)−a| /  ≤
≤ 2·(ε·a²/2) / a² = ε.

This means that 1/f(x) is withinε-neighborhood of 1/a.
End of proof.