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The problem is to prove that if $$\lim_{x\to\infty} f(x)$$ exists and $$\lim_{x\to\infty} f''(x)$$ exists, then $$\lim_{x\to\infty} f''(x) = 0$$My first attemp is that I divide it into two cases. First,I let $$l>0$$then according to the $\epsilon-\delta$ defition of limit,$$\forall\epsilon > 0,\exists N,$$ such that $\forall x$, if $x>N$, then $$|f''(x) - l| < \epsilon$$ Then I let $$\frac{l}{2} = \epsilon$$ and I then get $$\frac{-l}{2} < f''(x) - l < \frac{l}{2}$$ Then I know that $$f''(x) > \frac{l}{2}$$ Then according to the Mean Value Theorem, I get $$f'(x) \ge f'(N) + \frac{l}{2}(x-N)$$ then $$\lim_{x\to\infty} f'(x)$$ doesn't exist. Then it means that $$f'(x) > 0, \forall x > N$$ then since if $$\lim_{x\to\infty} f(x)$$ and $$\lim_{x\to\infty} f'(x)$$ both exists, $$\lim_{x\to\infty} f'(x) = 0$$ then in this case$$\lim_{x\to\infty} f(x)$$ doesn't exist.Then if $$l < 0$$ we do similar thing using $$-f$$Could anyone please tell me if my idea is correct or not? Thanks!

RobPratt
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Wendy Lyu
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  • Duplicate found with Approach0 – Martin R Oct 30 '22 at 12:29
  • You need to reserve displays (double dollar signs) for the rare occasions they are warranted. This is so distracting. – Ted Shifrin Oct 31 '22 at 01:02
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    Your proof that f' is unlimited looks right to me, but I don't understand the rest of your proof. On the other hand, if f' is positive and unlimited does not it prove that f(x) must be unlimited too? – blamocur Oct 31 '22 at 01:08
  • Yeah, blamocur, what you commented makes sense to me! So basically i proved that f' is unlimited and since it's positive and unlimited, then f must be unlimited too, then this leads to a contradiction, since we're given that limit of f(x) as x goes to infinity exists. Then the limit of f''(x) can't be greater than 0. Do i understand you idea correctly? – Wendy Lyu Oct 31 '22 at 01:12

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