Convergence of a special integral Determine the value of the integral below:
$\lim_{x\rightarrow 0^+} \int_{x}^{2x} \frac{\sin{(t)}}{t^2} \,dt\ $
My attempt to solution:
I could not solve the problem. And I do not know how to continue.
$1)$ I started analysing just  integral, and after that I applied the integration by parts:
$\int_{x}^{2x} \frac{\sin{(t)}}{t^2} \,dt\ = -\frac{\sin{(2x)}}{2x} +\frac{\sin{(x)}}{x} - \int_{x}^{2x} \frac{-\cos{(t)}}{t} \,dt\ $
$\int_{x}^{2x} \frac{\sin{(t)}}{t^2} \,dt\ = -\frac{\sin{(2x)}}{2x} +\frac{\sin{(x)}}{x} + \int_{0}^{2x} \frac{\cos{(t)}}{t} \,dt\ - \int_{0}^{x} \frac{\cos{(t)}}{t} \,dt\ $
$2)$ Now, applying the limit:
$\lim_{x\rightarrow +0} \int_{x}^{2x} \frac{\sin{(t)}}{t^2} \,dt\ = \lim_{x\rightarrow +0} (-\frac{\sin{(2x)}}{2x} +\frac{\sin{(x)}}{x} + \int_{0}^{2x} \frac{\cos{(t)}}{t} \,dt\ - \int_{0}^{x} \frac{\cos{(t)}}{t} \,dt\ )$
$3)$ If we prove that $\exists\lim_{x\rightarrow +0} \frac{\sin{(x)}}{x}$, $\exists\lim_{x\rightarrow +0}\int_{0}^{x} \frac{\cos{(t)}}{t} \,dt\ $, we could be able to apply the sum of limits:
$\lim_{x\rightarrow +0} \int_{x}^{2x} \frac{\sin{(t)}}{t^2} \,dt\ = \lim_{x\rightarrow +0} -\frac{\sin{(2x)}}{2x} +\lim_{x\rightarrow +0} \frac{\sin{(x)}}{x} + \lim_{x\rightarrow +0}\int_{0}^{2x} \frac{\cos{(t)}}{t} \,dt\ - \lim_{x\rightarrow +0} \int_{0}^{x} \frac{\cos{(t)}}{t} \,dt\ $
$4)$ I know that $\lim_{x\rightarrow +0}\frac{\sin{(x)}}{x} = 1$, but I do not know if $ \exists\lim_{x\rightarrow +0} \int_{0}^{x} \frac{\cos{(t)}}{t} \,dt\ $, and how to find the value of it.
I tried for long time, but I could not find.

As the interval $[x, 2x]$ gets smaller and smaller the integrand will be $1/t + O(t)$, so the integral is $\log 2x - \log x + O(x^2)$, giving you $\log 2$ as the result as $x\to 0^+$.
