- #1
Buri
- 273
- 0
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!
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!
Last edited: