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?

Put $t=ux$ to reduce the expression under limit to $$\int_{1}^{2}\frac{\sin ux} {u^2x}\,du$$ The integrand $$f(x, u) =\frac{\sin (ux)} {u^2x},f(0,u)=\frac{1}{u}$$ is continuous in $[0,h]\times[1,2]$ and hence we can switch limit with integral to get the desired limit as $$\int_{1}^{2}\frac{du}{u}=\log 2$$
