Unizor - Function Limits - Properties

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Function Limit -
Properties of Limits

Theorem 1

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

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-δ definition of a function limit.
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 within δ-neighborhood of r, it's true that f(x) is within ε'-neighborhood of a.
Step 4 - we have proven that, as long as x is within δ-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 within ε-neighborhood of A·a.
End of proof.

Theorem 2

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

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-δ definition 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 within δ1-neighborhood of r, it's true that f(x) is within ε'-neighborhood of a.
Step 4 - using the convergence of g(x), for this ε' find δ2 such that, if x is within δ2-neighborhood of r, it's true that g(x) is within ε'-neighborhood of b.
Step 5 - calculate
δ = min(δ1, δ2);
for this δ both statements are true:
f(x) is within ε'-neighborhood of a and
g(x) is within ε'-neighborhood of b.
Step 6 - we have proven that, as long as x is within δ-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 x→r and
g(x)→b as x→r,
then
f(x)·g(x)→a·b as x→r.

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 to a·b.

(3b) |f(x)·g(x)−a·b| = |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|

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 within δ4-neighborhood of r, it's true that f(x) is within ε1-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 within δ5-neighborhood of r, it's true that g(x) is within ε2-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-neighborhood of a and
g(x) is within ε2-neighborhood of b.
Step 7 - we have proven that, as long as x is within δ-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|+1)·ε/[2(|b|+1)] +
+ |a|·ε/(2|a|) = ε
End of proof.

Theorem 4

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

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) 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 within δ2-neighborhood of r, it's true that f(x) is within ε'-neighborhood of a.
Step 4 - set δ = min(δ1, δ2).
Step 5 - we have proven that, as long as x is within δ-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| / a² ≤
≤ 2·(ε·a²/2) / a² = ε.

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

[ZorShekhtman]

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