Let $y$ be a differentiable function on $\mathbb{R}$. If $$\lim_{x\rightarrow \infty}(y(x)+y'(x))=0$$ Then how does one show that $$\lim_{x\rightarrow \infty}y(x)=0$$ I'd appreciate some help on this problem. Thanks a lot.


marked as duplicate by Git Gud, Davide Giraudo real-analysis Mar 15 '15 at 13:41

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  • $\begingroup$ This was asked several times on the site before, with answers going from the wrong argument given below (assuming more or less implicitly that every function must have a limit) to the correct approach based on the explicit form of the solution of $y'+y=f$, also given below. $\endgroup$ – Did Mar 15 '15 at 10:07

Let $y'(x)+y(x)=:f(x)$, so that $y(x)$ trivially satisfies the ODE


The general solution of this equation is

$$ce^{-x}+e^{-x}\int_0^x e^t f(t) \mathrm{d}t.$$

Hence there exists some constant $c_0$ such that

$$y(x)=c_0 e^{-x}+e^{-x} \int_0^x e^t f(t) \mathrm{d} t, $$ and taking the limit as $x \to \infty$ gives

$$\lim_{x \to \infty} y(x)=c_0 \lim_{x \to \infty} e^{-x}+\lim_{x \to \infty} \frac{\int_0^x e^t f(t) \mathrm{d} t}{e^x}=\lim_{x \to \infty} \frac{\int_0^x e^t f(t) \mathrm{d} t}{e^x}. $$

The last limit can be evaluated to $0$ using l'Hospital's rule combined with the fact that $\lim_{x \to \infty} f(x)=0$.

  • $\begingroup$ The last limit can be evaluated to 0 using... the usual elementary N-epsilon definition of limit and the fact that $e^x\to\infty$. $\endgroup$ – Did Mar 15 '15 at 10:10
  • $\begingroup$ @Did Can you please elaborate? $\endgroup$ – user1337 Mar 15 '15 at 10:13
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    $\begingroup$ One knows that, for every positive $\varepsilon$, $|f(x)|\leqslant\varepsilon$ for every $x\geqslant N$ for some finite $N$, then one can split the integral from $0$ to $x$ at $N$ and use this upper bound on the part from $N$ to $x$ to deduce some upper bound on $|y(x)|$ for every $x\geqslant N$. $\endgroup$ – Did Mar 15 '15 at 10:21

HINT: think about what would happen to $\lim_{x\to\infty}y$ if $\lim_{x\to\infty}y'\neq 0$.

  • $\begingroup$ I see where you're going at. This shows that if the limit of $y$ exists, it has to be $0$. But it could very well turn out that the limit does not even exist for $y$. That's where I'm stuck at. $\endgroup$ – sayantankhan Mar 15 '15 at 9:55
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    $\begingroup$ If the limit didn't exist, you wouldn't have $\lim_{x\to\infty}(y+y')=0$ in the first place. $\endgroup$ – Demosthene Mar 15 '15 at 9:57
  • $\begingroup$ No, what I'm saying is you could have functions whose individual limits are not defined, but the limit of their sum is defined. One example is, $\sin(x)$ and $-\sin(x)$. $\endgroup$ – sayantankhan Mar 15 '15 at 10:05
  • $\begingroup$ @Demosthene Sorry but why? Functions with no limit at infinity do exist, think about the sine. $\endgroup$ – Did Mar 15 '15 at 10:08
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    $\begingroup$ Your last two comments seem to show that you still do not realize why your approach is seriously flawed. $\endgroup$ – Did Mar 15 '15 at 10:23

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