# Study the convergence of $\int_{3}^{+\infty} \sin\left(\frac{1}{x}\right)- \tan\left(\frac{1}{x}\right)\, dx$

I have to study the convergence of $$\int_{3}^{+\infty} \sin\left(\frac{1}{x}\right)- \tan\left(\frac{1}{x}\right)\, dx .$$ A friend of mine suggest me to use Taylor development, but I don't understand how this can show that the integral converge. Can someone help me?

For large $x$ or equivalently small $\frac{1}{x}$ we have $\sin\frac{1}{x}\approx\frac{1}{x}-\frac{1}{6x^3}$ and $\tan\frac{1}{x}\approx\frac{1}{x}+\frac{1}{3x^3}$ so $\sin\frac{1}{x}-\tan\frac{1}{x}\approx-\frac{1}{2x^3}$, which has a finite integral on $\int_3^\infty dx$.
Both functions are continuous on the interval. Let $u=1/x$ and $du = -u^2dx$.
$$-\int_{1/3}^{0^+} \frac{\sin\left(u\right)- \tan\left(u\right)}{u^2}\, du$$
The question is the improper limit at $u=0$. Expand the functions in their Taylor series, keeping terms up to third order.
$$\int_{0^+}^{1/3} \frac{u - u + \mathcal{O}(u^3)}{u^2}\, du = \int_{0^+}^{1/3} \mathcal{O}(u)\,du < \infty$$
The integrand is of order $u$ and higher, so the integral is finite.