Originally Posted by
Also sprach Zarathustra Here something useful...
$\displaystyle lim_{x\to 0} \frac{sin x}{x}=1$
in your case (q. a):
$\displaystyle lim_{x\to 0} \frac{sin 4x}{9x}=\frac{4}{9}lim_{x\to 0} \frac{sin 4x}{4x}$
Try apply this also for b.