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 within
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, if
From |x_{n}| ≤ ε' follows that
So, for ANY sequence {x_{n}}converging to 0 and ANY positive ε we have found Msuch that, if
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, if
From |x_{n}| ≤ ε' follows that
So, for ANY sequence {x_{n}}converging to 0 and ANY positive ε we have found Msuch that, if
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) = log_{2}(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 within
Basically, we have to prove that the inequality
The inequality
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
or, considering that 2^{log2(x)} = x,
We can say now that, as long as
Let's examine this inequality forx:
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
Let's choose
Now interval (1−δ, 1+δ) is inside a bigger interval
Therefore, for any
Our problem is solved, for any given ε we have found corresponding δ such that, if xis in
That means that functionf(x)=log_{2}(x) converges to 0 as its argument x converges to 1.
End of proof.
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