Use L'Hopital's rule to find $ \lim \limits_{x \to 0} \left( \frac{ \tan\beta x - \beta \tan x}{\sin \beta x - \beta \sin x} \right) $ where $\beta $ is a non-zero constant and $\beta \ne \pm 1$.
I have applied L'Hospital's rule (twice) but it doesn't seem to be going anywhere.
I have tried writing in terms of sine and cosine but that didn't appear fruitful either.
I have verified numerically that the limit of this expression appears to be $-2$.
What am I missing?
$$\lim_{x\to 0}\frac{\tan(\beta x)-\beta\tan(x)}{\sin(\beta x)-\beta\sin(x)}$$ yields an indeterminate $\lim_{x\to 0}\frac{0}{0}$ case. Hence $$\lim_{x\to 0}\frac{\tan(\beta x)-\beta\tan(x)}{\sin(\beta x)-\beta\sin(x)}=\lim_{x\to 0}\frac{\frac{d}{dx}(\tan(\beta x)-\beta\tan(x))}{\frac{d}{dx}(\sin(\beta x)-\beta\sin(x))}=\lim_{x\to 0}\frac{\beta \sec^2(\beta x)-\beta \sec^2(x)}{\beta\cos(\beta x)-\beta\cos(x)}=\lim_{x\to0}\frac{\frac{1}{\cos^2(\beta x)}-\frac{1}{\cos^2(x)}}{\cos(\beta x)-\cos(x)}=\lim_{x\to 0}\frac{\cos^2(x)-\cos^2(\beta x)}{\cos^2(x)\cos^2(\beta x)}\cdot\frac{1}{\cos(\beta x)-\cos(x)}$$ $$=\lim_{x\to0}-\frac{\cos(\beta x)+\cos(x)}{\cos^2(x)\cos^2(\beta x)}=-\frac{\cos(0)+\cos(0)}{\cos^2(0)\cos^2(0)}=-2$$
Hint: Using the rules of L'Hospital one times and simplifying we get $$\lim_{x\to 0}{\frac {-\cos \left( \beta\,x \right) -\cos \left( x \right) }{ \left( \cos \left( \beta\,x \right) \right) ^{2} \left( \cos \left( x \right) \right) ^{2}}} $$
If the application of L'Hospital's Rule is not mandatory,
$$F=\lim \limits_{x \to 0} \left( \frac{ \tan\beta x - \beta \tan x}{\sin \beta x - \beta \sin x} \right) =\dfrac{\lim_{x\to0}\dfrac{\tan\beta x-\beta x}{x^3}-\beta\cdot\lim_{x\to0}\dfrac{\tan x-x}{x^3}}{\lim_{x\to0}\dfrac{\sin\beta x-\beta x}{x^3}-\beta\cdot\lim_{x\to0}\dfrac{\sin x-x}{x^3}}$$
Using Are all limits solvable without L'Hôpital Rule or Series Expansion,
$$F=\dfrac{\dfrac{\beta^3-\beta}3}{-\dfrac{\beta^3-\beta}6}=?\text{ if }\beta^3-\beta\ne0$$
Observe the condition in the question!
