Convergence of $\int_0^\infty \frac{\sin x}{x}dx$

Question

$$ \begin{align} \int_0^\infty \frac{\sin x}{x}dx \end{align} $$ Convergence at $x=0.$

Since, $\lim_{x\rightarrow 0}\frac{\sin x}{x} = 1$ so $0$ it is not a point of infinite discontinuity, hence the integrand is convergent at $0$.

Convergence at $x=\infty.$

Since, $\lim_{x\rightarrow \infty}\frac{\sin x}{x} = 0$ so $\infty$ it is not a point of infinite discontinuity, hence the integrand is convergent at $\infty$.

Will this be sufficient to prove that the integrand is convergent at $\infty$?

Answer 1

No, because you could use the same argument when you replace $\sin(x)$ with $\frac{x}{x+1}$, but the statement is obviously false in this case.