Notes to a video lecture on http://www.unizor.com
Function Limit - Why Two Definitions?
Recall two definitions of function limit presented in the previous lecture.
Definition 1
Value a is a limit of functionf(x) when its argument xconverges to real number r, if for ANY sequence of argument values {x_{n}} converging to r the sequence of function values {f(x_{n})} converges to a.
Symbolically:
∀{x_{n}}→r ⇒ {f(x_{n}}→a
Definition 2
Value a is a limit of functionf(x) when its argument xconverges to real number r, if for any positive ε there should be positive δ such that, if x is within δ-neighborhood of r(that is, |x−r| ≤ δ), then f(x)will be within ε-neighborhood of a (that is, |f(x)−a| ≤ ε).
Symbolically:
∀ ε>0 ∃ δ>0:
|x−r| ≤ δ ⇒ |f(x)−a| ≤ ε
Sometimes the last two inequalities in the above definition are specified as "less" instead of "less or equal". It makes no difference.
First of all, let's answer the question of the title of this lecture: Why two definitions?
Obviously, there were historical reasons. Mathematicians of 18th and early 19th centuries suggested different approaches to function limits and functioncontinuity that led to both definitions. Cauchy, Bolzano Weierstrass and others contributed to these definitions.
The Definition 1 seems to sound "more human", it seems more natural, though difficult to deal with if we want to prove the existence of a limit.
The Definition 2, seemingly "less human", is easier to use when proving the existence of a limit. It is more constructive.
In this lecture we will prove the equivalence of both definitions. That is, if function has a limit according to Definition 1, it is the same limit according to Definition 2 and, inversely, from existence of a limit by Definition 2 follows that this same limit complies with Definition 1.
Theorem 1
IF
for any sequence of argument values {x_{n}} converging to r the sequence of function values{f(x_{n})} converges to a
[that is, if f(x)→a as x→raccording to Definition 1],
THEN
for any positive ε there should be positive δ such that, if x is within
[that is, it follows that f(x)→awhile x→r, according to Definition 2].
Proof
Choose any positive ε, however small, thereby fixing some
Let's prove an existence of δsuch that, if x is closer to r thanδ, then f(x) will be closer to athan ε.
Assume the opposite, that is, no matter what δ we choose, there is some value of argument x in the δ-neighborhood of r such that f(x) is outside of
Let's choose δ_{1}=1 and find the argument value x_{1} such that|x_{1}−r| ≤ δ_{1}, while |f(x) is outside of ε-neighborhood of a.
Next choose δ_{2}=1/2 and find the argument value x_{2} such that|x_{2}−r| ≤ δ_{2}, while |f(x) is outside of ε-neighborhood of a.
Next choose δ_{3}=1/3 and find the argument value x_{3} such that|x_{3}−r| ≤ δ_{3}, while |f(x) is outside of ε-neighborhood of a.
etc.
Generally, on the n^{th} step choose δ_{n}=1/n and find the argument value x_{n} such that
Continue this process of building sequence {x_{n}}.
This sequence, obviously, converges to r since
So, our assumption that, no matter what δ we choose, there is some value of argument x in the
End of proof.
Theorem 2
IF
for any positive ε there is positive δ such that, if x is within δ-neighborhood of r(symbolically, |x−r| ≤ δ), thenf(x) will be within
[that is, if f(x)→a as x→raccording to Definition 2],
THEN
for any sequence of argument values {x_{n}} converging to r the sequence of function values{f(x_{n})} converges to a
[that is, it follows that f(x)→awhile x→r, according to Definition 1].
Proof
Let's consider any sequence
In other words, for any positiveε we will find such number Nthat for all n ≥ N the inequality
Based on the premise of this theorem, there exists positive δsuch that, if
Since
So, for this particular N it is true that
End of proof.