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 {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
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 {x_n}converging to 0.
We will prove that corresponding sequence of function values {f(x_n)=Ax_n}converges to 0 as well.
From {x_n}→0 follows that
∀ε>0 ∃N: n ≥ N ⇒ |x_n| ≤ ε
The above statement is given.
What we have to prove is that
∀ε>0 ∃M: n ≥ M ⇒ |Ax_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' = ε/|A|;
(c) since {x_n} converging to 0, for this ε' find M such that, ifn ≥ M, then |x_n| ≤ ε';
From |x_n| ≤ ε' follows that|Ax_n| ≤ |A|ε' = ε.
So, for ANY sequence {x_n}converging to 0 and ANY positive ε we have found Msuch that, if n ≥ M, then|Ax_n| ≤ 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 {x_n}converging to 0.
We will prove that corresponding sequence of function values {f(x_n)=x²_n}converges to 0 as well.
From {x_n}→0 follows that
∀ε>0 ∃N: n ≥ N ⇒ |x_n| ≤ ε
The above statement is given.
What we have to prove is that
∀ε>0 ∃M: n ≥ M ⇒ |x²_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' = ε^1/2;
(c) since {x_n} converging to 0, for this ε' find M such that, ifn ≥ M, then |x_n| ≤ ε';
From |x_n| ≤ ε' follows that|x²_n| ≤ (ε')² = ε.
So, for ANY sequence {x_n}converging to 0 and ANY positive ε we have found Msuch that, if n ≥ M, then|x²_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 2^x is monotonically increasing.
So, an equivalent inequality is
2^−ε ≤ 2^log₂(x) ≤ 2^ε
or, considering that 2^log₂(x) = x,2^−ε ≤ x ≤ 2^ε.
We can say now that, as long as2^−ε ≤ x ≤ 2^ε, it is true that−ε ≤ log₂(x) ≤ ε and, equivalently, |log₂(x)| ≤ ε.
Let's examine this inequality forx: 2^−ε ≤ x ≤ 2^ε.
We are considering positive ε to be as small as possible. That results in 2^ε to be slightly greater than 1, while 2^−ε to be slightly less then 1. So, an interval (2^−ε,2^ε) encompasses the value of 1 from both sides.
Let's chooseδ=min(1−2^−ε, 2^ε−1).
Now interval (1−δ, 1+δ) is inside a bigger interval (2^−ε,2^ε), 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.