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This propped up while I was going through some old question papers. If $f \in C^1[0,1]$ so that $ \lim \limits_{x\to \infty} \dfrac{x.f(x)}{f'(x)}=2$.Then :

1) Show that for $s<2$, $\lim\limits_{x \to \infty} x^{-s}f(x) \longrightarrow \infty$

2) Find $\dfrac{\int \limits_{0}^{1}f(x)\,\mathrm dx}{x.f'(x)}$

The first I tried employing L'Hospital's rule, thought I am still not getting the answer. The second I thought may be approached by writing $f(x) = f(0) + x\int_{0}^{1}f'(tx)\mathrm dt$, though again not sure where this leads.

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    $\begingroup$ If $\operatorname{dom}(f)=[0,1]$ how can you even talk about the limit at $+\infty$? $\endgroup$ – Git Gud Oct 12 '13 at 10:36
  • $\begingroup$ Yeah, I thought about that myself.Maybe a misprint in the question.I apologise, but thats how the question has been given. $\endgroup$ – Vishesh Oct 12 '13 at 10:38
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    $\begingroup$ Thanks for clarifying. Perhaps they mean that $f$ is defined on $\Bbb R$ and it is differentiable on $\Bbb R$, furthermore $f'$ is continuous on $[0,1]$. $\endgroup$ – Git Gud Oct 12 '13 at 10:39
  • $\begingroup$ @GitGud. Thanks for the edit. How did you enlarge the size?? $\endgroup$ – Vishesh Oct 12 '13 at 10:46
  • $\begingroup$ \dfrac instead of \frac $\endgroup$ – Git Gud Oct 12 '13 at 10:47
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First problem :

We want to show that $$\lim_{x\rightarrow \infty}f=\infty$$

If not, $\lim_{x\rightarrow \infty}f=C>0$ Then $\lim_{x\rightarrow \infty}f'=0$ It is a contradiction. If $ \lim_{x\rightarrow \infty} f=0$, then $$ 0> \lim_{x\rightarrow \infty} \frac{x^2}{\ln\ f} = \lim_{x\rightarrow \infty}\frac{2x}{\frac{f'}{f}}=4$$ It is also a contradiction.

So
$$ \lim_{x\rightarrow \infty} \frac{f}{x^s }= \lim_{x\rightarrow \infty} \frac{f'}{sx^{s-1}}= \lim_{x\rightarrow \infty} \frac{f'}{sx^{s-1}}\frac{xf}{2 f'}= \lim_{x\rightarrow \infty} \frac{x^{2-s}f}{2s}=\infty$$ where $s<2$.

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  • $\begingroup$ Thanks. But I am unable to grasp a few things here. First is that limit of 4 you deduced and the other is that multiplication by $\dfrac{xf}{2f'}$ $\endgroup$ – Vishesh Oct 12 '13 at 11:12
  • $\begingroup$ Aah thanks.I will wait a bit to see if someone comes up with an answer to the second, else I will accept your answer $\endgroup$ – Vishesh Oct 12 '13 at 11:20
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In my opinion this is a trick question to test your capability of detecting disguised nonsense. The part between "If" and "then" is absurd, because for a limit of a function$~f$ to be defined at a point$~p$, that point must be a limit point of the domain of$~f$, which for $p=\infty$ and $f\in C^1[0,1]$ is always false (if the domain of $f$ were not $[0,1]$, then one should have said the restriction of $f$ to$~[0,1]$ is in $C^1[0,1]$, but it clearly does not say that). So with a hypothesis that is never satisfied, there is nothing to prove, whatever the desired conclusion is. The first question resembles something that might be true for some functions not in $C^1[0,1]$, so it gives you an opportunity to spend some time uselessly on a proof, but the second question should drive home that this really was not meant to be done: in the numerator $x$ is a bound variable, but in the denominator it is a free variable; moreover there is nothing in the hypothesis that could give you any information about the value of the integral in the numerator, nor about the value of the denominator even if the value of $x$ were specified (and the denominator might then well be$~0$ too).

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  • $\begingroup$ Wow, ok. I never realised that. I get your argument. But the above solution made me feel this was solvable too.Thanks. $\endgroup$ – Vishesh Oct 13 '13 at 2:43

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