Find this limit without using L'Hospital's rule I have to find this limit without using l'Hôspital's rule:
$$\lim_{x\to 0} \frac{\alpha \sin \beta x - \beta \sin \alpha x}{x^2 \sin \alpha x}$$
Using L'Hôspital's rule gives:
$$\frac{\beta}{6(\alpha^2 - \beta^2)}$$
I am stuck where to begin without using the rule.

Using the Taylor series
$$\sin t = t - \frac{t^3}{3!} + \frac{t^5}{5!} - \dots$$
the numerator is
\begin{align*}
\alpha \left(\beta x - \frac{(\beta x)^3}{3!} + O(x^5)\right) - \beta \left(\alpha x - \frac{(\alpha x)^3}{3!} + O(x^5)\right) = \frac{\beta \alpha^3 - \alpha \beta^3}{6} x^3 + O(x^5)
\end{align*}
Then the fraction can be written as
\begin{align*}
\frac{\dfrac{\beta \alpha^3 - \alpha \beta^3}{6} x^3 + O(x^5)}{\alpha x^3 + O(x^5)} \to \frac{\beta \alpha^3 - \alpha \beta^3}{6\alpha}
\end{align*}
as $x \to 0$.
