$\displaystyle \begin{array}{cc}\lim\\x\rightarrow0\end{array} \frac{\sin2x}{\sin3x}$
How can I solve that without using L'Hôpital's rule (which is considered illegal in my calculus class)?
For small x (which we have here, since we're letting x tend to 0) you can use the first order Taylor expansion for sin(x) which is x. So for small x:
$\displaystyle \frac{{\sin \left( {2x} \right)}}{{\sin \left( {3x} \right)}} \approx \frac{{2x}}{{3x}} \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {2x} \right)}}{{\sin \left( {3x} \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{{2x}}{{3x}} = \frac{2}{3}$
It's also possible that you have to use the 'standard limit':
$\displaystyle \mathop {\lim }\limits_{\alpha \to 0} \frac{{\sin \alpha }}{\alpha }=1$
In that case, rewrite as follows:
$\displaystyle \mathop {\lim }\limits_{x \to 0} \frac{{2\sin \left( {2x} \right)}}{{2x}}\frac{{3x}}{{3\sin \left( {3x} \right)}} = 2\mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {2x} \right)}}{{2x}} \cdot \frac{1}{3}\mathop {\lim }\limits_{x \to 0} \frac{{3x}}{{\sin \left( {3x} \right)}} = \frac{2}{3}$