Question

$\underset{x\to 0}{\lim }\frac{\cos ax-\cos bx}{\cos cx-\cos dx}$

Answer 1

$\underset{x\to 0}{\lim }\left[\frac{\cos ax-\cos bx}{\cos cx-\cos dx}\right]$

It is of the form $\left(\frac{0}{0}\right)$.

$$\begin{gathered} \Rightarrow \underset{x\to 0}{\lim }\left[\frac{-2\sin \left({\displaystyle \frac{ax+bx}{2}}\right)\sin \left({\displaystyle \frac{ax-bx}{2}}\right)}{-2\sin \left({\displaystyle \frac{cx+dx}{a2}}\right)\sin \left({\displaystyle \frac{cx-dx}{2}}\right)}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{{\displaystyle \frac{\sin \left({\displaystyle \frac{ax+bx}{2}}\right)}{\left({\displaystyle \frac{ax+b}{x}}\right)}}\times \left({\displaystyle \frac{ax+bx}{2}}\right)\times {\displaystyle \frac{\sin \left({\displaystyle \frac{ax-bx}{2}}\right)}{\left({\displaystyle \frac{ax-bx}{2}}\right)}}\times \left({\displaystyle \frac{ax-bx}{2}}\right)}{{\displaystyle \frac{\sin \left({\displaystyle \frac{cx+dx}{2}}\right)}{\left({\displaystyle \frac{cx+dx}{2}}\right)}}\times \left({\displaystyle \frac{cx+dx}{2}}\right)\times {\displaystyle \frac{\sin \left({\displaystyle \frac{cx-dx}{2}}\right)}{\left({\displaystyle \frac{cx-dx}{2}}\right)}}\times \left({\displaystyle \frac{cx-dx}{2}}\right)}\right] \\ =1\times \underset{x\to 0}{\lim }\left[\frac{\left({\displaystyle \frac{ax+bx}{2}}\right)\left({\displaystyle \frac{ax-bx}{2}}\right)}{\left({\displaystyle \frac{cx+dx}{2}}\right)\left({\displaystyle \frac{cx-dx}{2}}\right)}\right] \\ =\underset{x\to 0}{\lim }\frac{x^2\left(a+b\right)\left(a-b\right)}{x^2\left(c+d\right)\left(c-d\right)} \\ =\frac{a^2-b^2}{c^2-d^2} \end{gathered}$$