10
$\begingroup$

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.

$\endgroup$
6
  • 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
    Commented Dec 22, 2014 at 17:18
  • 1
    $\begingroup$ 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
    Commented Dec 22, 2014 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? $\endgroup$
    – Ben Grossmann
    Commented Dec 22, 2014 at 18:13
  • $\begingroup$ In 3 do you mean $\lim_{x\to \infty}f'(x)$? $\endgroup$
    – Noah Olander
    Commented Jun 5, 2015 at 2:32
  • $\begingroup$ @ Noah : Limit of the derivative of $f$ as $x \to \infty$ $\endgroup$
    – Struggler
    Commented Jun 5, 2015 at 2:39

4 Answers 4

Reset to default
12
$\begingroup$

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 $\color{red}{\text{L'Hospital's rule}}$.

$\endgroup$
10
  • $\begingroup$ I learned that here! $\endgroup$
    – Ivo Terek
    Commented Jun 5, 2015 at 2:38
  • $\begingroup$ thanks your prompt reply and very tricky answer $\endgroup$
    – Struggler
    Commented Jun 5, 2015 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
    Commented Jun 5, 2015 at 2:57
  • 1
    $\begingroup$ @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
    Commented Jun 5, 2015 at 3:08
  • 1
    $\begingroup$ @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
    Commented Jun 5, 2015 at 17:34
5
$\begingroup$

Hint:

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

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

$\endgroup$
4
  • $\begingroup$ kindly give details.. I couldnt understand.@rrl $\endgroup$
    – David
    Commented Dec 22, 2014 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
    Commented Dec 22, 2014 at 18:06
  • $\begingroup$ oh.. thank you.. so much..@rrl $\endgroup$
    – David
    Commented Dec 22, 2014 at 18:17
  • $\begingroup$ so.. 4 also follows right??@rrl $\endgroup$
    – David
    Commented Dec 22, 2014 at 18:18
3
$\begingroup$
  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 $\lim\limits_{x\to\infty}f'(x)=0$.

$\endgroup$
2
$\begingroup$

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 $$ \lim_{x \to \infty} [g(x) \pm h(x)] = \lim_{x \to \infty} g(x) \pm \lim_{x \to \infty} h(x) $$

$\endgroup$
8
  • $\begingroup$ is it true??@omnomnomnom $\endgroup$
    – David
    Commented Dec 22, 2014 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. $\endgroup$
    – Ben Grossmann
    Commented Dec 22, 2014 at 18:22
  • $\begingroup$ i think it will not hold always..@Omnmnomnom $\endgroup$
    – David
    Commented Dec 22, 2014 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? $\endgroup$
    – Ben Grossmann
    Commented Dec 22, 2014 at 18:24
  • $\begingroup$ yes...kindly clarify me...i am having a doubt in spliting the limit. $\endgroup$
    – David
    Commented Dec 22, 2014 at 18:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .