Assume that $\mathop {\lim }\limits_{x \to \infty } f(x) + f'(x) = 0$. Prove: $\mathop {\lim }\limits_{x \to \infty } f(x) = 0$

Question

Let $f:\mathbb{R}\rightarrow \mathbb{R}$ differentiable everywhere. Assume that $\mathop {\lim }\limits_{x \to \infty } f(x) + f'(x) = 0$.
Prove: $\mathop {\lim }\limits_{x \to \infty } f(x) = 0$

I've seen a suggestion for a solution defining $g(x)=e^xf(x)$ and taking its derivative. Personally, I don't like this trick. Can you suggest a more conservative way to solve it?

Thanks!

Answer 1

You can reason loosely (and yes, I know I'm violating all kinds of rules by doing this, but I did say "loosely") that $\lim (f' + f)=0$ means $\lim f' = -\lim f$, so $\lim f' / \lim f =-1$. Then $\lim (f'/f) = -1$; hence $\lim (\log f)' = -1$, so $\lim \log f = -\infty$ and thus $\lim f = 0$.

Now it's up to you to make this rigorous.

Answer 2

Only today I saw analogous questions 2 times. Here are a few links that I found searching on the site

If a function has a finite limit at infinity, does that imply its derivative goes to zero?

If a function has a finite limit at infinity, are there any additional conditions that could imply that its derivative converges to zero?

If $\lim_{x\to\infty}(f(x)+f'(x))=L$ show that $\lim_{x\to\infty} f(x) = L$ and $\lim_{x\to\infty} f'(x) = 0$

Maybe searching would help...

Answer 3

We show that for every $\varepsilon >0$, there exists $X$ such that $f(x) \le \varepsilon$ when $x>X$. This statement implies that $\limsup_{x\to\infty} f(x) \le 0$; then replacing $f(x)$ by $-f(x)$ immediately shows that $\liminf_{x\to\infty} f(x) \ge 0$, which finishes the proof that $\lim_{x\to\infty} f(x) = 0$.

By hypothesis, we can choose $W$ such that $f(x) + f'(x) < \frac\varepsilon2$ for all $x\ge W$. We claim there must be a point $X\ge W$ for which $f(X) \le \varepsilon$. Indeed, if $f(W) \le \varepsilon$ then set $X=W$; otherwise, note that $f'(x) < -f(x) + \frac\varepsilon2 < -\frac\varepsilon2$ for all $x\ge W$ for which $f(x) \ge \varepsilon$. In particular, $f(x) \le f(W) - \frac\varepsilon2(x-W)$ for all $x\ge W$ until $f(x) \le \varepsilon$; therefore there must be such an $X$ in the interval $[W,W+\frac2\varepsilon(f(W)-\varepsilon)]$.

Now we show that $f(x)\le\varepsilon$ for all $x\ge X$. Suppose not: choose $z>X$ such that $f(z) > \varepsilon$. The continuous function $f$ has a maximum on the interval $[X,z]$; it's not at $X$ since $f(z)>f(X)$, and it's not at $z$ because $f'(z) < -f(z) + \frac\varepsilon2 < -\frac\varepsilon2$ and so the values of $f$ just to the left of $z$ are larger than $f(z)$. Therefore there exists $y\in(X,z)$ that is a local maximum of $f$. Consequently $f'(y)=0$ and $f(y)\ge f(z) > \varepsilon$, so that $f(y)+f'(y) > \varepsilon$; but this $y>X\ge W$, contradicting the choice of $W$.

Answer 4

From the given condition we can see that both the limit $lim_{x\rightarrow \infty}f(x)$ and $lim_{x\rightarrow \infty}f'(x)$ exist.

Suppose that $lim_{x\rightarrow \infty}f'(x)=l$. Now , if $l>0$ The $f'(x)>\frac{l}{2}$ whenever $x>s$ for some real $s$. Therefore $f(x)>\frac{lx}{2}+C$ for some constant $C$, while $x>s$ Therefore $lim_{x\rightarrow\infty}f(x)$ tends to $\infty$.

In a similar way we can show for l<0 Therefore we get $l=0$

Hence $lim_{x\rightarrow \infty}f(x)=0$