## Tuesday, August 30, 2016

### Unizor - Derivatives - Function Limit - Simple Problems

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

Function Limit - Simple Problems

Let's recall two definitions of a limit of function.

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 {xn} converging to r the sequence of function values {f(xn)} converges to a.
Symbolically:
∀{xn}→r ⇒ {f(xn}→a

Definition 2
For any positive ε there should be positive δ such that, if x is withinδ-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

Solving the problems below, you can use any of these definitions to prove the existence of a limit and to find its concrete value.

Problem 1

Consider a function defined for all real arguments x:
f(x) = Ax
(where A ≠ 0).
Prove that this function has a limit for x→0 and that this limit equals to 0.

Solution

Let's use Definition 1 above. Consider ANY sequence {xn}converging to 0.
We will prove that corresponding sequence of function values {f(xn)=Axn}converges to 0 as well.
From {xn}→0 follows that
∀ε>0 ∃Nn ≥ N ⇒ |xn| ≤ ε
The above statement is given.
What we have to prove is that
∀ε>0 ∃Mn ≥ M ⇒ |Axn| ≤ ε
We can prove the existence of such M for any ε by direct constructing it following these steps:
(a) choose ANY positive ε;
(b) calculate new ε' = ε/|A|;
(c) since {xn} converging to 0, for this ε' find M such that, ifn ≥ M, then |xn| ≤ ε';
From |xn| ≤ ε' follows that|Axn| ≤ |A|ε' = ε.
So, for ANY sequence {xn}converging to 0 and ANY positive ε we have found Msuch that, if n ≥ M, then|Axn| ≤ ε.
That means that functionf(x)=Ax converges to 0 as its argument x converges to 0.
End of proof.

Problem 2

Consider a function defined for all real arguments x:
f(x) = x²
Prove that this function has a limit for x→0 and that this limit equals to 0.

Solution

Let's use Definition 1 above. Consider ANY sequence {xn}converging to 0.
We will prove that corresponding sequence of function values {f(xn)=x²n}converges to 0 as well.
From {xn}→0 follows that
∀ε>0 ∃Nn ≥ N ⇒ |xn| ≤ ε
The above statement is given.
What we have to prove is that
∀ε>0 ∃Mn ≥ M ⇒ |n| ≤ ε
We can prove the existence of such M for any ε by direct constructing it following these steps:
(a) choose ANY positive ε;
(b) calculate new ε' = ε1/2;
(c) since {xn} converging to 0, for this ε' find M such that, ifn ≥ M, then |xn| ≤ ε';
From |xn| ≤ ε' follows that|n| ≤ (ε')² = ε.
So, for ANY sequence {xn}converging to 0 and ANY positive ε we have found Msuch that, if n ≥ M, then|n| ≤ ε.
That means that functionf(x)=x² converges to 0 as its argument x converges to 0.
End of proof.

Problem 3

Consider a function defined for all positive arguments x:
f(x) = log2(x)
Prove that this function has a limit for x→1 and that this limit equals to 0.

Solution

Let's use Definition 2 above. Consider ANY positive ε.
We will prove that there is suchδ that, as long as x is withinδ-neighborhood of value 1, function values f(x)=log2(x)will be within ε-neighborhoodof value 0.
Basically, we have to prove that the inequality |log2(x)| ≤ εfollows from |x−1| ≤ δ for someδ.
The inequality |log2(x)| ≤ ε can be represented as
−ε ≤ log2(x) ≤ ε
If this inequality is true, raising2 to the power of its components will produce an equivalent inequality, since power function 2x is monotonically increasing.
So, an equivalent inequality is
2−ε ≤ 2log2(x) ≤ 2ε
or, considering that 2log2(x) = x,2−ε ≤ x ≤ 2ε.
We can say now that, as long as2−ε ≤ x ≤ 2ε, it is true that−ε ≤ log2(x) ≤ ε and, equivalently, |log2(x)| ≤ ε.
Let's examine this inequality forx2−ε ≤ x ≤ 2ε.
We are considering positive ε to be as small as possible. That results in 2ε to be slightly greater than 1, while 2−ε to be slightly less then 1. So, an interval (2−ε,2ε) encompasses the value of 1 from both sides.
Let's chooseδ=min(1−2−ε, 2ε−1).
Now interval (1−δ, 1+δ) is inside a bigger interval (2−ε,2ε), while still encompassing the value 1.
Therefore, for anyx(1−δ, 1+δ) (this isδ-neighborhood of 1) it is true that −ε ≤ log2(x) ≤ ε (that isε-neighborhood of 0).
Our problem is solved, for any given ε we have found corresponding δ such that, if xis in δ-neighborhood of 1, it is true that f(x)=log2(x) is inε-neighborhood of 0.
That means that functionf(x)=log2(x) converges to 0 as its argument x converges to 1.
End of proof.