Answer
$\dfrac{a}{b}$
Work Step by Step
Write the Taylor series for the function as: $\sin ax=ax-\dfrac{(ax)^3}{3!}+\dfrac{(ax)^5}{5!}+.......$
Now, we have:
$\lim\limits_{x \to 0} \dfrac{\sin (ax)}{\sin (bx)}=\lim\limits_{x \to 0} \dfrac{ax-\dfrac{(ax)^3}{3!}+\dfrac{(ax)^5}{5!}+.......}{bx-\dfrac{(bx)^3}{3!}+\dfrac{(bx)^5}{5!}+.......}\\=\dfrac{a-\dfrac{a^3x^2}{3!}+\dfrac{a^5x^4}{5!}-.......}{b-\dfrac{b^3x^2}{3!}+\dfrac{b^5x^4}{5!}-.......}\\=\dfrac{a}{b}$
Therefore, we have: $\lim\limits_{x \to 0} \dfrac{\sin (ax)}{\sin (bx)}=\dfrac{a}{b}$
