trigonometry - Evaluation of limit at infinity: $lim_{xtoinfty} x^2 sin(ln(cos(frac{pi}{x})^{1/2}))$

$$\lim_{x\to\infty} x^2 \sin(\ln(\cos(\frac{\pi}{x})^{1/2}))$$

What I tried was writing $1/x=t$ and making the limit tend to zero and writing the cos term in the form of sin