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?

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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Medfield, MA

I don't think you can really prove this.

lim _{x->0} 1/x^{2 } does not exist (DNE) as a finite number;

although we can suspect that it diverges to infinity.

lim _{x->0} 1/x^{2} = ∞ really comes from a definition. And many courses say that all limits -> ∞ DNE.

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.

1st attempt - Better Way Below

So using the properties of limits we could also say

lim _{x->0} 1/x^{2} = (lim_{ x->0} 1) / (lim _{
x->0} x^{2})

given lim c = c and lim _{x->a} x = a and lim
_{x->0} f(x^{2}) = (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) = x^{2}

Then show that x^{2} -> 0 (by substitution, or more formally.

Now since 1/f(x) = x^{2} -> 0 Then f(x) = 1/x^{2} -> ∞ or DNE

Willowbrook, IL

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/x^{2}>M. Pick δ=1/√M, then if |x|<1/√M, 1/x^{2}>1/(1/√M)^{2}>M. Proof is complete.

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.

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 {a_{n}}=(-1)^{n}, for example.

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/x^{2} for me, at the time, would have been to show that if f(x) = 1/x^{2} then 1/f(x) = x^{2} and then to show that x^{2} 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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