Does $\int_{1}^{\infty} g(x)\ dx$ imply $\lim\limits_{x\to \infty}g(x)=0$?

Question

Let $g:[1, \infty)\rightarrow \mathbb R$ a continuous non-negative function, such that $\int_{1}^{\infty} g(x)\ dx$ converges. Is it true that $\lim\limits_{x\to \infty}g(x)=0$ ?

I tried to find a counter-example but I can't figure a trivial one. I also tried to prove it by the definition of the convergence of $g(x)$ but couldn't show that limit is really $0$.

I remember this one. And I remember I didn't really find an expression but got it right anyway. Hope it's really identical :D — user76568, May 03 '17 at 09:26

score 10 · Accepted Answer · answered May 03 '17 at 09:16

10

Take a function made up of triangles (i.e. it starts at $0$, it goes to $1$, then comes back at $0$). Make it so that the area of each triangle is $1/n^2$ (i.e. the base of each triangle is $2/n^2$).

This means that $$\int_1^\infty g(x) = \sum_{n=1}^\infty \frac{1}{n^2}$$ converges.

But the limit of $g(x)$ as $x \to \infty$ is not $0$ as $g(x)$ alternates between $0$ and $1$

answered May 03 '17 at 09:16

Ant

21,522

Is this continous? – user76568 May 03 '17 at 09:20
1

@Dror It clearly is. All the triangles are "connected". – Zubzub May 03 '17 at 09:22
Whoops my bad.. – user76568 May 03 '17 at 09:24
So is it true to say that if $\lim\limits_{x\to \infty}g(x)$ exists, then it must be zero? – atefsawaed May 03 '17 at 09:49
@z00x Yes; if the limit is $l > 0$ then you can bound the integral with something like $\int_1^R g(x)dx \ge l/2\cdot R$, which diverges as $R \to \infty$, which implies that the integral $\int_1^\infty g(x)dx$ does not converge. So if the limit exists then it must be $0$ – Ant May 03 '17 at 10:36

Does $\int_{1}^{\infty} g(x)\ dx$ imply $\lim\limits_{x\to \infty}g(x)=0$?

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