Let $f(x)$ be a differentiable function on $(0, \infty)$ with $\lim_{x\to \infty} f(x) - xf'(x) = L\in \mathbb{R}$. I'm trying to prove or disprove that $\lim_{x\to\infty} f'(x)$ exists as well. Here's what I have so far: if the limit does exist, it must also equal $\lim_{x\to\infty} \frac{f(x)}{x}$ (divide the limit condition by $x$). Then I rewrote the condition as $x^2 \frac{d}{dx} \frac{f(x)}{x} \to L$, and from this I expect (using some $1/x^2$ asymptotic argument) my statement to be true - but I'm not sure how to rigorously proceed with this argument.. Can someone provide some next steps, or a counter example?
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Try the case, whenever $\frac{f(x)}{x}=f'(x) $ , but $lim_{x\to\infty} \frac{f(x)}{x} $ doesn't exist! Is there any such case???? – A learner Aug 01 '20 at 04:30
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@FarazMasroor what do you precisely mean when you say $\lim_{x\to \infty} f'(x)$ exists,do you mean exists in $\mathbb R$ of exists in $\mathbb R\cup {\pm \infty}$ ?Please make it clear? – Kishalay Sarkar Aug 01 '20 at 05:39
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Yes, $\lim_{x \to\infty} f'(x)$ exists under those conditions. In fact it suffices to require that the function $$ r(x) = f(x) - xf'(x) $$ is bounded for $x \to \infty$, i.e. that $$ M = \sup \{ |r(x)| : x \ge 1 \} < \infty \, . $$
Proof: $$ f'(x) = \frac{f(x)}{x} - \frac{r(x)}{x} $$ so that the claim is equivalent to the existence of $\lim_{x \to\infty} \frac{f(x)}{x}$. Here we can apply the Cauchy criterion to $$ F(x) = \frac{f(x)}{x} \, . $$ For $x \ge 1$ is $$ |F'(x)| = \left|\frac{xf'(x) - f(x)}{x^2}\right| = \frac{|r(x)|}{x^2} \le \frac{M}{x^2} \, . $$ It follows that for $1 \le x < y$ $$ |F(x) - F(y)| \le \int_x^y \frac{M}{t^2} \, dt = M \left( \frac 1x - \frac 1y \right) < \frac Mx \, . $$ (This is a straight-forward estimate if the fundamental theorem of calculus can be applied to $F$, e.g. if $F'$ is continuous or Riemann integrable. For the general case see $|f'(x)| \le g(x)$ implies $|f(b) - f(a)| \le \int_a^b g(x) dx$, without assuming $f'$ to be integrable.).
So for every $\epsilon > 0$ $$ x, y > \frac{M}{\epsilon} \implies |F(x) - F(y) | < \epsilon $$ and that completes the proof.
Martin R
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Nice proof indeed! I am wondering if one can replace integrals with the use of mean value theorem. – Paramanand Singh Aug 01 '20 at 10:35
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@ParamanandSingh: I have added a proof using only the MVT which does not require the derivative to be continuous (or integrable). – Martin R Aug 01 '20 at 15:00
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Your proof via MVT is similar to the proof of Fundamental Theorem of Calculus (evaluation of integral via anti-derivative) and should be used as an example to illustrate its power. – Paramanand Singh Aug 01 '20 at 15:10
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1@ParamanandSingh: But the FTC requires the derivative to be integrable. What I needed here is essentially that $|F'(x)| \le h(x)$ implies $|F(y) - F(x)| \le \int_x^y h(t) dt$, and apparently that holds without additional assumptions on $F'$. I wonder if there is a simpler argument, without considering a sequence of partitions. – Perhaps I'll post a question about this. – Martin R Aug 01 '20 at 15:18
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Even I am unable to think of a simpler version (otherwise I would have posted something here). Regarding FTC and MVT I always find that MVT is much stronger and powerful. Yes your approach just assumes the existence of $f'$ and not its integrability. Let's wait for some answers in this thread before you ask your question. – Paramanand Singh Aug 01 '20 at 15:22
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@ParamanandSingh: Well, here it is https://math.stackexchange.com/q/3776825/42969 :) – Martin R Aug 01 '20 at 16:33
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Here is a path to a proof. Denote $f(x) = x\cdot g(x)$ Then equivalently what we are asking is that if the limit
$$\lim_{x\to\infty}-x^2g'(x) = L$$
exists, then does the limit
$$\lim_{x\to\infty}g(x)+xg'(x)$$
exist? Further we have by squeeze theorem that
$$\lim_{x\to\infty} xg'(x) = 0$$
so we can simplify this to the question does the limit
$$\lim_{x\to\infty}g(x)$$
exist given the condition above?
Ninad Munshi
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