Proving integral diverges limits assume $f(x)$ is continuous on $[0,\infty)$ and that $\lim_{x \to \infty}f(x)=L > 0$ How would you prove that $\int_0^\infty f(x) \, dx$ diverges?
I know that since $\lim_{x \to \infty}f(x)=L$ then $f(x) \leq L$ However from here I have no idea how to show that $\lim_{b \to \infty}\int_0^b f(x) \, dx$ does not exist.
I think I've got it
by definition of the limit f $L-\frac L 2 < f(x)<L+\frac L 2$ so from there $0\leq L-\frac{L}{2} \leq f(x)$ and so by the integral comparison test since $\int_0^\infty L-\frac L 2 \, dx$ diverges then $\int_0^\infty f(x) \, dx$ diverges
 A: Hint prove that there exists $x_o$ such that $x>x_0$ implies that $f(x)>c>0$ and $\int_{x_0}^{+\infty} f>\int_{x_0}^{+\infty} c$. Now remark that $\int_0^{+\infty}f(x)=\int_0^{x_0}f(x)+\int_{x_0}^{+\infty}f(x)$. Since $\int_0^{x_0}f(x)$ is finite and $\int_{x_0}^{+\infty}f(x)$ infinite, you can conclude that $\int_0^{+\infty}f(x)$ is infinite.
A: $$
\int_0^\infty f(x)\, dx \ge \int_N^\infty f(x)\,dx \ge \int_N^M f(x)\,dx \ge\int_N^M \frac L 2 \,dx = \frac{(M-N)L} 2.
$$
Since $f(x) \to L$ as $x\to\infty,$ you can find $N$ so large that whenever $x\ge N$ then $f(x) >\dfrac L2.$ Then choose $M$ to be anything bigger than $N.$ Since the inequality holds no matter how big $M$ gets, the value of the integral must be $+\infty.$
It is not correct to say, as you did, that $f(x) <L.$ You may have, for example, $f(x) = L + \dfrac 1 x.$
A: If $L >0$ it exists $y$ such as for $x>y$  $f(x)>0$ and
$$
 f(x) \underset{(+\infty)}{\sim}L
$$
Furthermore $\displaystyle \int_{0}^{+\infty}L \text{ d}x$ diverges hence it is the same for $\displaystyle \int_{0}^{+\infty}f(x) \text{ d}x$
A: By definition of limit there exist $x_M$ such that $\forall x>x_M \quad f(x)>L-\epsilon=L_M>0$ thus
$$\displaystyle \int_{0}^{+\infty}f(x) \text{ d}x>\displaystyle \int_{x_M}^{+\infty}L_M \text{ d}x=+\infty$$
