Let $f:[1,\infty)$ a continuious and non-negative function s.t. $\int_1^\infty f(x)$ converges.
I don't understand why the existence of $lim_{x\rightarrow\infty} f($x) neccessairly means that $lim_{x\rightarrow\infty} f($x) = 0.(why?)
Plus, Does $lim_{x\rightarrow\infty} f(x) = 0$ even if the assumption of the existence of is not given?(why?)
Lastly, Is the function $f$ bounded over the line $[1,\infty)$?(why?)
Thanks in adnvance!
Suppose that $l:= \lim_{x\rightarrow\infty} f(x)>0$. Then there is $c_1>1$ such that $f(x) \ge l/2$ for $x \ge c_1$. For $u$ with $u>c_1$ we then have
$\int_u^{u+1}f(x) dx \ge l/2$.
If $ 0< \epsilon < l/2$ we have by Cauchy: there is $c_2 \ge c_1$ such that
$\int_u^{u+1}f(x) dx < \epsilon$ for $u>c_2$, a contradiction.
Yes, $f$ is bounded: there is $r>1$ such that $f(x) \le 2l$ for $x>r$. Since $f$ is continuous, $f$ is also bounded on $[1,r]$