Show that $\lim_{x\to 0^+}\int_x^{2x}\frac{\sin t}{t^2}dt=\ln 2.$ 
Question: Show that $$\lim_{x\to 0^+}\int_x^{2x}\frac{\sin t}{t^2}dt=\ln 2.$$

Solution: Note that $$\lim_{x\to 0^+}\int_x^{2x}\frac{\sin t}{t^2}dt=\ln 2=\lim_{x\to 0^+}\int_x^{2x}\frac{dt}{t}\\\iff\lim_{x\to 0^+}\int_x^{2x}\frac{\sin t-t}{t^2}dt=0.$$ 
Thus it is sufficient to prove that $$\lim_{x\to 0^+}\int_x^{2x}\frac{\sin t-t}{t^2}dt=0.$$ 
Now, expanding $\sin t$ using Maclaurin's formula having remainder in Lagrange's form, we have $$\sin t=t-\frac{\cos\xi}{3!}t^3, \text{ where } \xi=\theta t\text{ and } 0<\theta<1.$$ 
Now $$\cos \xi\le 1\implies \frac{\cos \xi}{3!}t^3\le \frac{t^3}{3!}(\because t>0)\\\implies t-\frac{\cos \xi}{3!}t^3\ge t-\frac{t^3}{3!}\\\implies \sin t\ge t-\frac{t^3}{3!}.$$ 
Again since $t>0$, we have $\sin t<t$. 
Thus $\forall t\in[x,2x]$ and $\forall x>0,$ we have $$t-\frac{t^3}{3!}\le \sin t\le t\\\implies -\frac{t}{3!}\le \frac{\sin t-t}{t^2}\le 0\\\implies \int_x^{2x} -\frac{t}{3!}dt\le \int_x^{2x}\frac{\sin t-t}{t^2}dt\le 0\\\implies -\frac{x^2}{4}\le \int_x^{2x}\frac{\sin t-t}{t^2}dt\le 0.$$
Now since $\lim_{x\to 0^+} -\frac{x^2}{4}=0$ and $\lim_{x\to 0^+} 0=0$, thus by Sandwich Theorem we can conclude that $$\lim_{x\to 0^+} \int_x^{2x}\frac{\sin t-t}{t^2}dt=0.$$ Hence, we are done. 
Is this solution correct and rigorous enough? Are there any alternative solutions?

Let $ x\in\left(0,1\right] : $
\begin{aligned}\int_{x}^{2x}{\frac{\sin{t}}{t^{2}}\,\mathrm{d}t}&=\ln{2}+\int_{x}^{2x}{\frac{t-\sin{t}}{t^{2}}\,\mathrm{d}t}\\ &=\ln{2}+\int_{0}^{2x}{\frac{t-\sin{t}}{t^{2}}\,\mathrm{d}t}-\int_{0}^{x}{\frac{t-\sin{t}}{t^{2}}\,\mathrm{d}t}\\ &=\ln{2}+\frac{1}{2}\int_{0}^{x}{\frac{2u-\sin{\left(2u\right)}}{u^{2}}\,\mathrm{d}u}-\int_{0}^{x}{\frac{t-\sin{t}}{t^{2}}\,\mathrm{d}t}\\ \int_{x}^{2x}{\frac{\sin{t}}{t^{2}}\,\mathrm{d}t}&=\ln{2}+\int_{0}^{x}{\frac{\sin{t}\left(1-\cos{t}\right)}{t^{2}}\,\mathrm{d}t}\end{aligned}
Since $ x\mapsto\frac{x\left(1-\cos{x}\right)}{x^{2}} $ is continuous on $ \left(0,1\right] $, and can be extended to a continuous function $ \left[0,1\right] $, it can be upper-bounded on $ \left(0,1\right] $ by some constant $ M>0 :$ $$ \left(\exists M>0\right)\left(\forall x\in\left(0,1\right]\right),\ \frac{\sin{x}\left(1-\cos{x}\right)}{x^{2}}\leq M $$
Thus $ \int_{0}^{x}{\frac{\sin{t}\left(1-\cos{t}\right)}{t^{2}}\,\mathrm{d}t}\leq Mx\underset{x\to 0}{\longrightarrow}0 $, and hence : $$ \lim_{x\to 0}{\int_{x}^{2x}{\frac{\sin{t}}{t^{2}}\,\mathrm{d}t}}=\ln{2} $$
