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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.

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If you haven't learned taylor series yet (tbongers answer), I think you can use the squeeze theorem to do this one, although I suspect the solution will be far more difficult. – MHH Feb 2 '14 at 1:03
Possible duplicate of this. – user 170039 Oct 19 '14 at 17:07

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 $x \to 0$.

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I think that you had a mistake somewhere when you applied L'Hospital's rule and arrived to $$\frac{\beta}{6(\alpha^2 - \beta^2)}$$ as reported in your post.

To get rid of problems with $x$ in the denominator, you must apply L'Hospital's rule three times and arrive to $$\frac{\beta \alpha^3 - \alpha \beta^3}{6\alpha}$$ which, fortunately (!), matches what T. Bongers obtained using Taylor series.

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