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Given that $f(x)$ is a continuous function and satisfies $f'(x)>0$ on $(-\infty,\infty)$ and $f''(x)=2 \forall x \in(0,\infty)$.

We need to find the limit

$$\lim_{x\to\infty}\frac{3x^2-\frac{3}{x^2+1}-4f'(x)}{f(x)}$$

Now the numerator is tending to infinity so denominator must also go to infinity else limit won't exist.So I tried the L'Hospitals rule and it became$$\lim_{x\to\infty}\frac{6x+\frac{6x}{(x^2+1)^2}-4f''(x)}{f'(x)}$$The numerator is still infinity so once again applying L'Hospitals rule (assuming denominator must still be infinity) we get

$$\lim_{x\to\infty}\frac{6+\frac{6(x^2+1)^2-6x×2(x^2+1)×2x}{(x^2+1)^4}+0}{f''(x)}$$

Now putting $f''(x)=2$ we get

$$3+\lim_{x\to\infty}\frac{3(x^2+1)^2-12x^2(x^2+1)^2}{(x^2+1)^4}$$

Collecting the coefficients of $x^4$ from numerator and denominator we get the limit to be$3-9=-6$ but the answer is not -6.

Is applying LHospital wrong?Help.
Thanks.

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  • $\begingroup$ As far as I see it the limit is $\;3\;$ . Why didn't you write what the actual limit is? $\endgroup$ – DonAntonio Jan 16 '18 at 16:09
  • $\begingroup$ In your last limit, the numerator is $O(x^6)$ while the denominator is $O(x^8)$. Does that help? $\endgroup$ – Mark Viola Jan 16 '18 at 16:30
  • $\begingroup$ Nice question. +1 I have generalized the problem somewhat in my answer. $\endgroup$ – Paramanand Singh Jan 16 '18 at 19:14
  • $\begingroup$ @MarkViola Yeah my mistake.That was easy enough.I don't know for what mysterious reasons I thought the final limit has degree four in both numerator and denominator.A silly mistake.Btw thanks. $\endgroup$ – Prakhar Mishra Jan 17 '18 at 4:34
  • $\begingroup$ Thanks all of you +1 to all but I can accept only one. $\endgroup$ – Prakhar Mishra Jan 17 '18 at 4:42
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The two given conditions mean $\;\lim\limits_{x\to\infty}f(x)=\infty\;$ (why?), and then, from your work, we reached

$$\lim_{x\to\infty}\frac{6+\frac{6(x^2+1)^2-6x×2(x^2+1)×2x}{(x^2+1)^4}+0}{f''(x)}=\frac{6+0+0}2=3$$

I don't really understand what you did after the above part in your question...but the middle summand in the numerator above is

$$\frac{6(x^2+1)-24x^2}{(x^2+1)^3}\xrightarrow[x\to\infty]{}0$$

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    $\begingroup$ It was a silly mistake.I thought degree of both numerator and denominator is four so limit should be ratio of coefficients of $x^4$.But thanks I understood my mistake. $\endgroup$ – Prakhar Mishra Jan 17 '18 at 4:40
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Since $f''(x)=2~~for ~~x>0$, Therefore $f$ has the form $f(x)=x^2+bx +c~~~for~~~x>0$

Since the limit is at $+\infty$ it suffices to consider $x>0$ $$\lim_{x\to\infty}\frac{3x^2-\frac{3}{x^2+1}-4f'(x)}{f(x)} = \lim_{x\to\infty}\frac{3x^2-\frac{3}{x^2+1}-8x-4b}{x^2+bx +c} =3$$

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  • $\begingroup$ why a down vote here? $\endgroup$ – Guy Fsone Jan 16 '18 at 16:15
  • $\begingroup$ You didn't read carefully. $f'(x)=2e^x \text{ if }x\le 0, 2x+2 \text{ if }x\ge0$ can do. $\endgroup$ – Yves Daoust Jan 16 '18 at 16:17
  • $\begingroup$ @Guy Some people just really love to downvote...but I think you missed the fact that $\;f''(x)-2\;$ but only for $\;x\in (0,\infty)\;$ ... $\endgroup$ – DonAntonio Jan 16 '18 at 16:17
  • $\begingroup$ $f''=2$ only for $x>0$. So $f'(x)=2x+b$ is valid only for $x>0$ $\endgroup$ – Andrei Jan 16 '18 at 16:18
  • $\begingroup$ @Andrei thanks for the remark $\endgroup$ – Guy Fsone Jan 16 '18 at 16:25
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From $f''(x)=2$, we know that $f$ is a quadratic function with leading term $x^2$. There will be no cancellation of $3x^2$ at the numerator, and by ignoring low order terms, the expression simplifies to

$$\frac{3x^2}{x^2}.$$

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  • $\begingroup$ Note that the information on $f'$ is of no use. $\endgroup$ – Yves Daoust Jan 16 '18 at 16:23
  • $\begingroup$ I think you missed $\;f''(x)\;$ only for $\;x\ge 0\;$ ...so not necessarily a quadratic. $\endgroup$ – DonAntonio Jan 16 '18 at 16:24
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    $\begingroup$ @DonAntonio: I left implicit that $x>0$ (but you can guess it from the problem statement). The behavior in thenegatives is irrelevant. $\endgroup$ – Yves Daoust Jan 16 '18 at 16:25
  • $\begingroup$ The information of $\;f'\;$ is rather useful to conclude that $\;f(x)\xrightarrow[x\to\infty]{}\infty\;$ without having to use the function is a quadratic on the positive reals . $\endgroup$ – DonAntonio Jan 16 '18 at 16:25
  • $\begingroup$ @DonAntonio: which is perfectly of no use as you can't compare to $3x^2/4$. $\endgroup$ – Yves Daoust Jan 16 '18 at 16:29
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The condition $f''(x) =2$ makes the problem far simpler because then we can use integration to get $f(x) =x^2+bx+c$ and perform the limit evaluation easily. The answer however remains the same if we are given the weaker hypothesis that $f''(x) \to 2$ as $x\to\infty$ and I prove this below.

First we observe that derivative $f'$ is positive and hence $f$ is strictly increasing. Therefore $f(x) $ tends to a limit or to $\infty$ as $x\to\infty$. By similar argument $f'(x) $ also tends to a limit or to $\infty$. If $f(x) $ tends to a limit then $f(x+1)-f(x)=f'(c)$ (via mean value theorem) tends to $0$. This is a contradiction as derivative $f'$ is positive and strictly increasing for large $x$ so it can't tend to $0$. Thus it follows that $f(x) \to\infty$ as $x\to\infty$. Similarly $f'(x) \to \infty$ because $f''(x) \to 2$.

Thus the desired limit is equal to the limit of $$\frac{3x^2-4f'(x)}{f(x)}$$ And since denominator tends to $\infty $ we can try L'Hospital's Rule to get the ratio $$\frac{6x-4f''(x)}{f'(x)}$$ whose limit is same as that of $6x/f'(x)$. Applying L'Hospital's Rule once again we get the ratio $6/f''(x)$ which tends to $3$ and therefore the original limit is also $3$.

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  • $\begingroup$ Nice discussion and generalization! $\endgroup$ – gimusi Jan 16 '18 at 19:15
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The issue is with the largest power of $x$. At the numerator is $4$, at the denominator is $8$ (because $(x^2+1)^4$). So the last limit is $0$

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  • $\begingroup$ The only thing that they say is that the second derivative is 2 for $x>0$. It can be $-2$ for $x<0$ $\endgroup$ – Andrei Jan 16 '18 at 16:16

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