Find the limit as $(x,y)$ approaches $(0,0)$ of $\frac{\sin(x)(1-\cos(y))}{6xy^{2}}$ I used the path $y=x$ and got that the limit equals $12$. How do I manipulate the original expression to prove that the limit actually equals $12$ for all paths? Thank you!

@manooh
Computing the limit of $(1-\cos x)/x^2$ without l'Hopital's rule is easy:
$$
\lim_{x\to 0} \frac {1 - \cos x} {x^2} = \lim_{x\to 0} \frac {1 - (1 - 1/2x^2 + \mathcal o(x^3))} {x^2} = \frac 1 2 + \lim_{x\to 0} \frac {\mathcal o (x^3)}{x^2} = \frac 1 2
$$
