Sun K.
asked • 05/10/14Can you prove this?
Prove that lim x-->0 1/x^2=∞.
Please show all your work. And tell me what level of math this is. Is it Calculus I or Calculus II or Calculus III or other?
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2 Answers By Expert Tutors
Inactive Tutor answered • 05/11/14
Tutor
New to Wyzant
Proving that f(x)→∞ does not mean to prove that the limit does not exist. Discontinuous function with a finite jump at x=a also does not have a limit as x→a. In this case, we need to prove that:
∀M>0 ∃δ: ∀x such that |x-a|<δ, f(x)>M.
This means, in layman words, that no matter how big M is, we can always find an interval containing point x=a such that the value of our function is greater than M inside this interval. This obviously excludes cases of finite discontinuities, since if M is greater than max(f(a+),f(a-)), maximum between left- and right-hand limits, there is no interval around x=a where our function exceeds that value.
In your case, let us take some M. Then, since a=0, we need to find δ such that for any |x|<δ 1/x2>M. Pick δ=1/√M, then if |x|<1/√M, 1/x2>1/(1/√M)2>M. Proof is complete.
Inactive Tutor
I do not mean to say that to prove it it goes to infinity you prove it DNE;
but because it goes to infinity many texts, courses, professors say it DNE since it does not go to a finite number. That is what was taught in the courses I took, way back when.
(or maybe we are interpreting each other backwards)
But then again the δ notation for limits was not taught back when I took calculus, at least not in the calculus courses I took. I only encountered that notation last night so I still need to study that notation more to make sure I read and write it properly. I do not know if that is because it is a new notation or because the calculus I took was an engineering series. I suspect it is a new notation since math majors were in the same class as me.
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05/11/14
Inactive Tutor
I do not know how it was in US, but in Russia this ε-δ notation was already used at least 25 years ago, maybe more.
I would say it is fine to say limit DNE when a sequence does not converge to a finite number, only this case shall then be treated separately from bounded but divergent sequences, like {an}=(-1)n, for example.
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05/11/14
Inactive Tutor
The notation may be very old but none of the texts or courses I ever had used it, but that was 50 years ago.
You are right, saying for bounded but divergent is infinity because the limit does not exist is not right.
There are plenty of limits that do not exist that are not infinity.
I am not saying that if a limit does not exist it is infinity. I am saying the opposite. That if a limit is infinity it does not exist. That a function increasing (or decreasing) forever without bound does not have a limit - even thought we might commonly say or notate it as the limit is infinity.
The proposition that: The function f(x) -> ∞ as x -> a if and only if the function 1/f(x) -> 0 as x -> a is what was shown/proven in class and that we learned.
So the way to prove lim x->0 1/x2 for me, at the time, would have been to show that if f(x) = 1/x2 then 1/f(x) = x2 and then to show that x2 goes to zero. Then by the proposition lim f(x) = ∞.
Maybe I should have erased my 1st attempt using the properties of limits. I still think it is valid, but it is very clumsy and I see in hind-site causes confusion.
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05/11/14
Inactive Tutor answered • 05/11/14
Tutor
New to Wyzant
I don't think you can really prove this.
lim x->0 1/x2 does not exist (DNE) as a finite number;
although we can suspect that it diverges to infinity.
lim x->0 1/x2 = ∞ really comes from a definition. And many courses say that all limits -> ∞ DNE.
Definition We say that a function f(x) diverges to infinity, denoted by f(x) -> ∞ as x -> a,
if, for every ε > 0, there exists δ > 0 such that |f(x)| > ε whenever 0 < |x - a| < δ.
Consider the function f(x) = 1/x. We suspect that f(x) -> ∞ as x -> 0.
Here a = 0. Let ε > 0 be chosen. Then
|f(x)| = | 1/x | = 1/|x| > ε whenever | x - 0 | = |x| < δ = 1/ε.
|f(x)| = | 1/x | = 1/|x| > ε whenever | x - 0 | = |x| < δ = 1/ε.
Proposition The function f(x) -> ∞ as x -> a if and only if the function 1/f(x) -> 0 as x -> a.
1st attempt - Better Way Below
So using the properties of limits we could also say
lim x->0 1/x2 = (lim x->0 1) / (lim x->0 x2)
given lim c = c and lim x->a x = a and lim x->0 f(x2) = (lim x->0 f(x))2
= 1 / (lim x->0 x)2 = 1 / (0)2 = undefined
Better Way
But now that I think about it you could use the proposition to show this in a better way.
f(x) -> ∞ if 1/f(x) - > 0
So show that 1/f(x) = x2
Then show that x2 -> 0 (by substitution, or more formally.
Now since 1/f(x) = x2 -> 0 Then f(x) = 1/x2 -> ∞ or DNE
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Inactive Tutor
05/10/14