Show that the function $f: x \mapsto \frac{e^{-x}}{x}|\sin{2x}|$ is integrable on $[0,+\infty[$.
My approach: $|\frac{e^{-x}}{x}|\sin{2x}|| = |\frac{2e^{-x}}{2x}|\sin{2x}|| = |2e^{-x}|\frac{\sin{2x}}{2x}|| \leq 2 e^{-x}$ since $\frac{\sin{2x}}{2x} \leq 1$. Thus $x \mapsto 2 e^{-x}$ is integrable on $[0,+\infty[$, and therefore $f$ is integrable.
Is there an error in my reasoning?
$$\lim_{x\to0^+}\frac{e^{-x}}{x}|\sin(2x)|=\left(\lim_{x\to0^+}e^{-x}\right)\left(\lim_{x\to0^+}\frac{\sin(2x)}{x}\right)$$ $$=\lim_{x\to0^+}2\frac{\sin(2x)}{2x}=2$$ and there are no other problematic points on the domain, now you can use the fact that: $$\int_1^\infty\frac{e^{-x}}{x}|\sin(2x)|\,dx<\int_1^\infty\frac{e^{-x}}{x}dx<\infty$$ then just justify that the integral from $0$ to $1$ is finite, which should be easy as we know its range over this domain