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I'm trying to prove that if the limit of continuous function $f$ is $0$ then the improper integral that goes from $0$ to infinity converges.

What I've done is say that for every $x > x_0$ then $\vert f(x) - 0\vert < E$ with $E>0$, then for all $x > x_0$, $f(x) < E$.

So if the integral of $E$ from $0$ to infinity converges then the integral of $f$ also does.

But is this true? Because now I'm thinking it isn't...

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    That is not true. $1/x$ – 2017-02-18

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That is absolutely not true.

Take for instance (among so many other examples):

$$f(x)=\frac 1{x+1}.$$

Then

$$\int_0^\infty f=+\infty$$

so it is not convergent, but

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

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    Probably $f(x)=1/(x+1)$ is better, in order to avoid problems at $0$. – 2017-02-18
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    Thanks for answering! so glad I came here to ask!! – 2017-02-18
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I don't think you would try and prove that

if $\lim\limits_{n\to\infty}a_n=0$, then the series $\sum\limits_{n\ge0}a_n$ converges

Well, this is essentially the same. The harmonic series provides a counterexample that shows the connection. Indeed, $$ \int_0^k\frac{1}{x+1}\,dx\ge\sum_{n=2}^{k}\frac{1}{n} $$ so, if $\displaystyle\int_0^{\infty}\frac{1}{x+1}\,dx$ converges, also $\displaystyle\sum_{n\ge1}\frac{1}{n}$ would converge.