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Suppose $f:\mathbb{R} \rightarrow \mathbb{R}$ is continuous on $[0,\infty)$ and $\lim\limits_{x \rightarrow \infty} (f(x) +\int^x_0 f(t) dt)$ exists. Show that $\lim\limits_{x \to \infty} f(x) = 0$.

Any hints to get me started would be appreciated.

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    If you assume that $\lim_{x\to \infty}$ exists, but does not equal zero, then you can show that $\lim_{x\to \infty} \int_0^x f(t)dt)$ necessarily does not exist. – Jbag1212 Feb 20 '19 at 05:09
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    @Jbag1212 what if the limit of $f$ doesn’t exist? – shalin Feb 20 '19 at 05:45

1 Answers1

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Let $F(x)=\int_0^{x} f(t) \, dt$ then the assumption is
$$\lim\limits_{x \rightarrow \infty} (F'(x) +F(x))=L.$$ Now, by If $\lim\limits_{x\rightarrow\infty} (f'(x)+f(x)) =L<\infty$, does $\lim\limits_{x\rightarrow\infty} f(x) $ exist? , we have that $$\lim\limits_{x \rightarrow \infty} F(x)=L.$$ and it follows straightforwardly that $$\lim\limits_{x \rightarrow \infty} f(x)=\lim\limits_{x \rightarrow \infty} F'(x)=\lim\limits_{x \rightarrow \infty} \left((F'(x)+F(x))-F(x)\right)=L-L=0.$$

Robert Z
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  • I am curious about the proof you linked. If $\lim_{x\to\infty}f(x) = 0$ isn't true that $\lim_{x\to\infty} e^xf(x) \ne \infty$ ? then L'Hopital cannot be used. I am just reading the comments :) – mate89 Feb 20 '19 at 05:56
  • No. In order to use L'Hopital to the ratio $g(x)/e^x$ all you need is that the denominator goes to infinity. See the remark here: https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule#General_form – Robert Z Feb 20 '19 at 06:33