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Unizor - Function Limits - Exercise 2
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Unizor - Creative Minds through Art of Mathematics - Math4Teens
Function Limit - Exercise
(x→−∞)
Try to do these exercises yourself.
All function limits below are supposed to be calculated as argument x tends to negative infinity, that is infinitely decreasing without bounds, eventually getting less than any real number fixed beforehand and staying below that number ever since.
In other words, we say that
f(x)→L as x→−∞, if
∀ positive ε (however small)
∃A: (x ≤ A)⇒|f(x)−L| ≤ ε
In some cases it's useful to use the following
Theorem
If f(x)→L as x→−∞,
then f(−x)→L as x→+∞
and converse (former follows from latter).
Proof
Actually, this theorem is following from a more general theorem about limit of a compounded function.
Indeed, set g(x)=−x and consider limit of function f(g(x)) as x→+∞.
However, it might be useful to prove this theorem directly.
Given that
∀ positive ε (however small)
∃A: (x ≤ A) ⇒ |f(x)−L| ≤ ε
Choose positive ε (however small).
Let B = −A.
Let x' = −x.
For any x ≥ B
it is true that −x ≤ −B.
Therefore, x' ≤ A.
Therefore, |f(x')−L| ≤ ε
Since f(x') = f(−x),
it's true that |f(−x)−L| ≤ ε
So, we have proven that
∀ positive ε (however small) ∃B:
(x ≥ B) ⇒ |f(−x)−L| ≤ ε
That is, x→+∞ ⇒ f(−x)→L
End of proof.
Find the limits of the following functions as their argument x infinitely decreasing.
1. (2x²+3x+4)/(3x²+4x+5)
Answer: 2/3
2. (2x³+3x+4)/(3x²+4x+5)
Answer: Function will be infinitely decreasing or, non-rigorously, its limit is −∞
3. (2x²+3x+4)/(3x³+4x+5)
Answer: 0
4. x·sin(1/x)
Answer: 1
5. x·3^x
Hint: Switch to x→+∞ and the results of corresponding exercise from a previous topic.
Answer: 0
6. x^2·3^x
Hint: Switch to x→+∞ and the results of corresponding exercise from a previous topic.
Answer: 0
Function Limit - Exercise
(x→−∞)
Try to do these exercises yourself.
All function limits below are supposed to be calculated as argument x tends to negative infinity, that is infinitely decreasing without bounds, eventually getting less than any real number fixed beforehand and staying below that number ever since.
In other words, we say that
f(x)→L as x→−∞, if
∀ positive ε (however small)
∃A: (x ≤ A)⇒|f(x)−L| ≤ ε
In some cases it's useful to use the following
Theorem
If f(x)→L as x→−∞,
then f(−x)→L as x→+∞
and converse (former follows from latter).
Proof
Actually, this theorem is following from a more general theorem about limit of a compounded function.
Indeed, set g(x)=−x and consider limit of function f(g(x)) as x→+∞.
However, it might be useful to prove this theorem directly.
Given that
∀ positive ε (however small)
∃A: (x ≤ A) ⇒ |f(x)−L| ≤ ε
Choose positive ε (however small).
Let B = −A.
Let x' = −x.
For any x ≥ B
it is true that −x ≤ −B.
Therefore, x' ≤ A.
Therefore, |f(x')−L| ≤ ε
Since f(x') = f(−x),
it's true that |f(−x)−L| ≤ ε
So, we have proven that
∀ positive ε (however small) ∃B:
(x ≥ B) ⇒ |f(−x)−L| ≤ ε
That is, x→+∞ ⇒ f(−x)→L
End of proof.
Find the limits of the following functions as their argument x infinitely decreasing.
1. (2x²+3x+4)/(3x²+4x+5)
Answer: 2/3
2. (2x³+3x+4)/(3x²+4x+5)
Answer: Function will be infinitely decreasing or, non-rigorously, its limit is −∞
3. (2x²+3x+4)/(3x³+4x+5)
Answer: 0
4. x·sin(1/x)
Answer: 1
5. x·3^x
Hint: Switch to x→+∞ and the results of corresponding exercise from a previous topic.
Answer: 0
6. x^2·3^x
Hint: Switch to x→+∞ and the results of corresponding exercise from a previous topic.
Answer: 0
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