# $L=\lim_{x\to\infty}(f(x)+f'(x))$ exists . Which of the following statements is\are correct?

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.

• 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. Dec 22 '14 at 17:18
• 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. Dec 22 '14 at 17:36
• 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? Dec 22 '14 at 18:13
• In 3 do you mean $\lim_{x\to \infty}f'(x)$? Jun 5 '15 at 2:32
• @ Noah : Limit of the derivative of $f$ as $x \to \infty$ Jun 5 '15 at 2:39

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

• I learned that here! Jun 5 '15 at 2:38
• @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 Jun 5 '15 at 2:57
• @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. Jun 5 '15 at 3:08
• @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. Jun 5 '15 at 17:34

Hint:

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

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

• kindly give details.. I couldnt understand.@rrl Dec 22 '14 at 17:47
• @gloom: Notice $\lim [ f(x+1) - f(x)] = L-M - (L-M) = 0$ So $\lim f'(\xi) = 0$ where $\xi$ is carried along with $x$.
– RRL
Dec 22 '14 at 18:06
• oh.. thank you.. so much..@rrl Dec 22 '14 at 18:17
• so.. 4 also follows right??@rrl Dec 22 '14 at 18:18
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 $\lim\limits_{x\to\infty}f'(x)=0$.

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

• is it true??@omnomnomnom Dec 22 '14 at 18:20
• Is what true? I told you that 1 and 2 are true, and yes: it is true that 2 implies 4. Dec 22 '14 at 18:22
• i think it will not hold always..@Omnmnomnom Dec 22 '14 at 18:23
• what is it that you do not think will hold always? Do you think that it will not always hold that 2 implies 4? Dec 22 '14 at 18:24
• yes...kindly clarify me...i am having a doubt in spliting the limit. Dec 22 '14 at 18:25