10
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Let $f$ be a continously differentiable function on $\mathbb R$. Suppose that

$$L=\lim_{x\to\infty}(f(x)+f'(x))$$ exists. If $0<L<\infty$, then which of the following statements is\are correct?

  1. If $\lim_{x\to\infty} f'(x)$ exists, then it is $0$.

  2. If $\lim_{x\to\infty} f(x)$ exists, then it is $L$.

  3. If $\lim_{x\to\infty} f'(x)$ exists, then $\lim_{x\to\infty}f(x)=0$.

  4. If $\lim_{x\to\infty} f(x)$ exists, then $\lim_{x\to\infty}f'(x)=0$.

My Guess

I could not conclude the answer and prove that properly. But, I guess that it must be 1 and 2. help me.

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  • 2
    $\begingroup$ Note that under the assumption, $\lim_{x\to\infty} f(x)$ exists if and only if $\lim_{x\to\infty} f'(x)$ exists. That makes several of the options equivalent. $\endgroup$ – Daniel Fischer Dec 22 '14 at 17:18
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    $\begingroup$ Show that $f'\to L > 0$ implies that $f$ is unbounded because $f(x) = f(0) + \int_0^x f'(t)dt$ can be arbitrarily large. $\endgroup$ – Myself Dec 22 '14 at 17:36
  • $\begingroup$ You said that you could not "prove that properly". How did you come to your conclusion, then? Is there anyway we can take your intuitive notion of the problem and formalize it? $\endgroup$ – Omnomnomnom Dec 22 '14 at 18:13
  • $\begingroup$ In 3 do you mean $\lim_{x\to \infty}f'(x)$? $\endgroup$ – Noah Olander Jun 5 '15 at 2:32
  • $\begingroup$ @ Noah : Limit of the derivative of $f$ as $x \to \infty$ $\endgroup$ – Struggler Jun 5 '15 at 2:39
2
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You are correct about 1 and 2.

Note also that 2 implies 4, since $$ \lim_{x \to \infty} f'(x) = L - \lim_{x \to \infty} f(x) $$ (assuming the latter limit exists).


Note that for arbitrary functions $g,h$: if $\lim_{x \to \infty} g(x)$ and $\lim_{x \to \infty} h(x)$ both exist, then $$ \lim_{x \to \infty} [g(x) \pm h(x)] = \lim_{x \to \infty} g(x) \pm \lim_{x \to \infty} h(x) $$

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  • $\begingroup$ is it true??@omnomnomnom $\endgroup$ – David Dec 22 '14 at 18:20
  • $\begingroup$ Is what true? I told you that 1 and 2 are true, and yes: it is true that 2 implies 4. $\endgroup$ – Omnomnomnom Dec 22 '14 at 18:22
  • $\begingroup$ i think it will not hold always..@Omnmnomnom $\endgroup$ – David Dec 22 '14 at 18:23
  • $\begingroup$ what is it that you do not think will hold always? Do you think that it will not always hold that 2 implies 4? $\endgroup$ – Omnomnomnom Dec 22 '14 at 18:24
  • $\begingroup$ yes...kindly clarify me...i am having a doubt in spliting the limit. $\endgroup$ – David Dec 22 '14 at 18:25
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Hint: Check this: $$\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty}\frac{e^xf(x)}{e^x} \color{red}{=} \lim_{x \to +\infty}\frac{e^x(f(x)+f'(x))}{e^x} = \lim_{x \to +\infty}f(x)+f'(x) = L,$$ by $\color{red}{\text{L'Hospital's rule}}$.

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  • $\begingroup$ I learned that here! $\endgroup$ – Ivo Terek Jun 5 '15 at 2:38
  • $\begingroup$ thanks your prompt reply and very tricky answer $\endgroup$ – Struggler Jun 5 '15 at 2:40
  • $\begingroup$ @IvoTerek Seems there is a tacit assumption that $e^xf(x) \to \infty$ as $x\to \infty$. Perhaps it is obvious, but the possibility (or not) that $e^xf(x)$ does not approach $\infty$ warrants a bit of discussion. Otherwise +1 $\endgroup$ – Mark Viola Jun 5 '15 at 2:57
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    $\begingroup$ @Dr.MV, though it's not usually stated as such, L'Hopital's rule for $f(x)/g(x)$ only requires that $g(x) \to \pm \infty$, not necessarily $f(x)$ as well. $\endgroup$ – Antonio Vargas Jun 5 '15 at 3:08
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    $\begingroup$ @Dr.MV It's mentioned very briefly in the wikipedia article for the rule (see the end of the second-to-last paragraph in the General Form section), and I have seen a proof for that case somewhere, but unfortunately I don't know an official reference for it. It seems that the proof on Wikipedia shows it, though. $\endgroup$ – Antonio Vargas Jun 5 '15 at 17:34
5
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Hint:

If $\lim f'(x) = M$, then $\lim f(x) = L-M$

Use the MVT: $f(x+1) - f(x) = f'(\xi)$ with $x < \xi < x+1$.

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  • $\begingroup$ kindly give details.. I couldnt understand.@rrl $\endgroup$ – David Dec 22 '14 at 17:47
  • $\begingroup$ @gloom: Notice $\lim [ f(x+1) - f(x)] = L-M - (L-M) = 0$ So $\lim f'(\xi) = 0$ where $\xi$ is carried along with $x$. $\endgroup$ – RRL Dec 22 '14 at 18:06
  • $\begingroup$ oh.. thank you.. so much..@rrl $\endgroup$ – David Dec 22 '14 at 18:17
  • $\begingroup$ so.. 4 also follows right??@rrl $\endgroup$ – David Dec 22 '14 at 18:18
3
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  1. If $\lim\limits_{x\to\infty}f'(x)$ exists, then $\lim\limits_{x\to\infty}f(x)=L-\lim\limits_{x\to\infty}f'(x)$ also exists. Then $$ \begin{align} \lim\limits_{x\to\infty}f'(x) &=\lim\limits_{x\to\infty}(f(x+1)-f(x))\\ &=\lim\limits_{x\to\infty}f(x+1)-\lim\limits_{x\to\infty}f(x)\\ &=0\tag{1} \end{align} $$
  2. If $\lim\limits_{x\to\infty}f(x)$ exists, then $(1)$ implies that $\lim\limits_{x\to\infty}f'(x)=0$ and therefore, $$ \begin{align} \lim\limits_{x\to\infty}f(x) &=L-\lim\limits_{x\to\infty}f'(x)\\ &=L\tag{2} \end{align} $$
  3. If $\lim\limits_{x\to\infty}f'(x)$ exists, then $(1)$ and $(2)$ say that $\lim\limits_{x\to\infty}f'(x)=0$ and $\lim\limits_{x\to\infty}f(x)=L$.

  4. If $\lim\limits_{x\to\infty}f(x)$ exists, then $(1)$ implies that $\lim\limits_{x\to\infty}f'(x)=0$.

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