# Limit of $y$ if limit of $y+y'$ goes to $0$? [duplicate]

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 StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 15 '15 at 13:41

• 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. – Did Mar 15 '15 at 10:07

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

$$y'(x)+y(x)=f(x).$$

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 $\lim_{x \to \infty} f(x)=0$.

• The last limit can be evaluated to 0 using... the usual elementary N-epsilon definition of limit and the fact that $e^x\to\infty$. – Did Mar 15 '15 at 10:10
• @Did Can you please elaborate? – user1337 Mar 15 '15 at 10:13
• 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 $x\geqslant N$. – Did Mar 15 '15 at 10:21

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

• 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. – sayantankhan Mar 15 '15 at 9:55
• If the limit didn't exist, you wouldn't have $\lim_{x\to\infty}(y+y')=0$ in the first place. – Demosthene Mar 15 '15 at 9:57
• 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 $-\sin(x)$. – sayantankhan Mar 15 '15 at 10:05
• @Demosthene Sorry but why? Functions with no limit at infinity do exist, think about the sine. – Did Mar 15 '15 at 10:08
• Your last two comments seem to show that you still do not realize why your approach is seriously flawed. – Did Mar 15 '15 at 10:23