Limit of $$\frac{\cos(ax)-\cos(bx)}{\cos(cx)-\cos(dx)}$$ as $x$ approaches to $0$.
I have calculated this limit using L Hospital's rule to be $$ \frac{-a^2+b^2}{-c^2+d^2}$$ But on the book the limit is given to be $$ \frac{a^2-b^2}{c^2-d^2}$$ Please help me understand what is the actual limit.
On
As noticed your solution is fine, note also that we can avoid l'Hospital by standard limit $\frac{1-\cos x}{x^2} \to \frac12$
$$\frac{\cos(ax)-\cos(bx)}{\cos(cx)-\cos(dx)}=\frac{\cos(ax)-1+1-\cos(bx)}{\cos(cx)-1+1-\cos(dx)}=$$
$$=\frac{a^2\frac{\cos(ax)-1}{a^2x^2}+b^2\frac{1-\cos(bx)}{b^2x^2}}{c^2\frac{\cos(cx)-1}{c^2x^2}+d^2\frac{1-\cos(dx)}{d^2x^2}} \to\frac{-\frac12 a^2+\frac12 b^2}{-\frac12 c^2+\frac12d^2}=\frac{- a^2+b^2}{- c^2+d^2}=\frac{a^2-b^2}{c^2-d^2}$$
As an alternative using that $\cos (ax) =1-2\sin^2(ax/2)$ etc.
$$\ldots=\frac{-\frac{a^2}4\frac{\sin^2(ax/2)}{a^2x^2/4}+\frac{b^2}4\frac{\sin^2(bx/2)}{b^2x^2/4}}{-\frac{c^2}4\frac{\sin^2(cx/2)}{c^2x^2/4}+\frac{d^2}4\frac{\sin^2(dx/2)}{d^2x^2/4}} \to \frac{a^2-b^2}{c^2-d^2}$$
The two expressions given are exactly the same. Observe that \begin{align*} \frac{-a^2 + b^2}{-c^2 + d^2} &= \frac{-a^2 + b^2}{-c^2 + d^2} \times \frac{-1}{-1} \\ &= \frac{-(-a^2+b^2)}{-(-c^2+d^2)} \\ &= \frac{a^2 - b^2}{c^2 - d^2}.\end{align*}