Verifying Proof: lim [x→0] f(x)g(x) Does Not Exist

[Buri]

Is this proof correct?

Prove that if lim [x→0] |f(x)| = ∞ and lim [x→0] g(x) does not exist, then lim [x→0] f(x)g(x) also does not exist.

Proof:

(1) lim [x→0] |f(x)| = ∞ means that ∀N > 0 ∃δ > 0 such that if 0 < |x| < δ ⇒ |f(x)| > N

And

(2) lim [x→0] g(x) does not exist means that ∃ε > 0 such that ∀δ > 0 if 0 < |x| < δ ⇒ |g(x) - M| > ε.

So this will be a direct proof in which I'll show that ∃ε > 0 such that ∀δ > 0, if 0 < |x| < δ ⇒ |f(x)g(x) - L| > ε.

So if 0 < |x| < δ ⇒ |f(x)g(x) - L| = |f(x)||g(x) - L/f(x)| > N|g(x) - L/f(x)| > Nε > ε

Hence, the limit doesn't exist.

I'm using (1) when I divide by f(x) as it guarantees that f(x) > N > 0 on (-δ,δ) and I'm using (2) to get the last inequality as using the ε from (2) guarantees that |g(x) - L/f(x)| isn't small.

I'd appreciate if someone could look this over.

Thanks!

 

[SammyS]

Is this proof correct?

Prove that if lim [x→0] |f(x)| = ∞ and lim [x→0] g(x) does not exist, then lim [x→0] f(x)g(x) also does not exist.

Proof:

(1) lim [x→0] |f(x)| = ∞ means that ∀N > 0 ∃δ > 0 such that if 0 < |x| < δ ⇒ |f(x)| > N

And

(2) lim [x→0] g(x) does not exist means that ∃ε > 0 such that ∀δ > 0 if 0 < |x| < δ ⇒ |g(x) - M| > ε.

This looks like what it means for lim [x→0] g(x) ≠ M. You need to generalize M to make this statement say that the limit does not exist no matter what M is.
Also you just need |g(x) - M| > ε for some x such that 0 < |x| < δ .

So this will be a direct proof in which I'll show that ∃ε > 0 such that ∀δ > 0, if 0 < |x| < δ ⇒ |f(x)g(x) - L| > ε.

So if 0 < |x| < δ ⇒ |f(x)g(x) - L| = |f(x)||g(x) - L/f(x)| > N|g(x) - L/f(x)| > Nε > ε

Hence, the limit doesn't exist.

I'm using (1) when I divide by f(x) as it guarantees that f(x) > N > 0 on (-δ,δ) and I'm using (2) to get the last inequality as using the ε from (2) guarantees that |g(x) - L/f(x)| isn't small.

I'd appreciate if someone could look this over.

Thanks!

 

[Buri]

Thanks for those points. For your first remark I am technically using the fact that it isn't for any M, as I concluded it didn't also exist for L/f(x). I guess I should have been explicit about it. And about the 'some x', yeah that's true. It seems like it doesn't actually mess up the proof now does it? Seems like it still works with these changes?

 

[Buri]

Anyone?

 

[alomari2010]

i think it is correct 100%

 

[SammyS]

Is this proof correct?

Prove that if lim [x→0] |f(x)| = ∞ and lim [x→0] g(x) does not exist, then lim [x→0] f(x)g(x) also does not exist.

Proof:

(1) lim [x→0] |f(x)| = ∞ means that ∀N > 0 ∃δ > 0 such that if 0 < |x| < δ ⇒ |f(x)| > N

And

(2) lim [x→0] g(x) does not exist means that ∃ε > 0 such that ∀δ > 0 if 0 < |x| < δ ⇒ |g(x) - M| > ε.

So this will be a direct proof in which I'll show that ∃ε > 0 such that ∀δ > 0, if 0 < |x| < δ ⇒ |f(x)g(x) - L| > ε.

So if 0 < |x| < δ ⇒ |f(x)g(x) - L| = |f(x)||g(x) - L/f(x)| > N|g(x) - L/f(x)| > Nε > ε

Hence, the limit doesn't exist.

I'm using (1) when I divide by f(x) as it guarantees that f(x) > N > 0 on (-δ,δ) and I'm using (2) to get the last inequality as using the ε from (2) guarantees that |g(x) - L/f(x)| isn't small.

I'd appreciate if someone could look this over.

Thanks!

It doesn't look all that convincing to me. Take some specific example and see if there are any holes in your assertions.

I think you need to have an N and a δ₁ from (1) and an ε and a δ₂ and perhaps M from (2). Relate M & L. Then define δ = min( δ₁, δ₂). You also need to be sure that you choose an x from (-δ,0)U(0,δ) such that |g(x) - L/f(x)| > ε .

With your present proof, it looks as if for some values of L and x that f(x)g(x)=L, so that |f(x)g(x)-L|=0 .