I need to prove that this limit exists and calculate it. I tried integration by parts, resulting with an integral which is called "The Cosine Integral", denoted $Ci(x)$. However, I need to deal with this question in a qualitative way, pretending as if I do not know the cosine integral or any other fancy function. I also tried to apply the so-called "Differentiability of the indefinite integral with variable endpoints", and I actually find the derivative function of the function for which I need to find the limit. But the derivative function is not defined at $x=0$, so I couldn't go that way either. Any suggestions?
Too long for comments.
For the limit itself, the problem is simple, as @Aaron Hendrickson did show in comments.
But we can go much further to get asymptotics and even more. Just write $$\frac{\sin (t)}{t^2}=\frac 1t+\sum_{n=1}^\infty(-1)^n\frac{ t^{2 n-1}}{(2 n+1)!} $$ $$\int_x^{3x}\frac{\sin (t)}{t^2}\,dt=\log(3)+\sum_{n=1}^\infty(-1)^n \frac{ \left(9^n-1\right) x^{2 n}}{2 n (2 n+1)!}$$
Computing the partial sums for $x=\frac \pi 6$ $$\left( \begin{array}{cc} p & \log(3)+\sum_{n=1}^p\\ 1 & 0.9158418368 \\ 2 & 0.9283687261 \\ 3 & 0.9278726579 \\ 4 & 0.9278854235 \\ 5 & 0.9278851944 \\ 6 & 0.9278851944 \end{array} \right)$$