$f:\mathbb{R}\to\mathbb{R}$ is continuous and $\int_{0}^{\infty} f(x)dx$ exists

Question

$f:\mathbb{R}\to\mathbb{R}$ is continuous and $\int_{0}^{\infty} f(x)dx$ exists could any one tell me which of the following statements are correct?

$1. \text{if } \lim_{x\to\infty} f(x) \text{ exists, then it is 0}$

$2. \lim_{x\to\infty} f(x) \text{ must exists, and it is 0}$

$3. \text{ in case if f is non negative } \lim_{x\to\infty} f(x) \text{ exists, and it is 0}$

$4. \text{ in case if f is differentiable } \lim_{x\to\infty} f'(x) \text{ exists, and it is 0}$

I solved one problem in past which says: if $f$ is uniformly continuos and $\int_{0}^{\infty} f(x)dx$ exists then $\lim_{x\to\infty} f(x)=0$, so the condition in one says $f$ is uniformly continuous? and hence $1$ is true? well I have no idea about the other statements. will be pleased for your help.

Answer 1

The point is that the limit need not exist. You can construct something according to the comment by J.J. For instance, you can consider a function such that at any integer the height is one but the width is $1/n^2$. Then the area of each rectangle is $1/n^2$ and

$$ \int_0^\infty f(x)dx=\sum\frac{1}{n^2}<\infty. $$

But the limit does not exists. You get

$$ \lim_{n\rightarrow\infty} f(n)=1, $$

$$ \lim_{n\rightarrow\infty} f(n-\delta)=0,\;\; 0<\delta<<1 $$

Thus, 1 is true. If the limit exists, it should be zero (if it isn't then the function can't be integrable, right?). 2 is not true (see the hint by J.J. and the first part of my answer above). I think that 3, 4 are not true due to the same reason.

Edit: To prove that 4 is not true you can construct a (more decaying) example and then integrate it.

Answer 2

1. True and this a proof by contradiction if $\lim_{x\to\infty}f(x)=\ell\neq 0$ and suppose $\ell>0$ (the case $\ell<0$ is similar) then there's $A>0$ s.t $$\forall x\geq A\quad f(x)\geq \frac{\ell}{2}$$ so we have $$\int_A^\infty f(x)dx\geq \frac{\ell}{2}\int_A^\infty dx=+\infty \quad \Rightarrow\Leftarrow$$

2. and 3. aren't correct and we can construct a counterexample as suggested by J.J