Question

Evaluate the following limits:

$\underset{x\to 0}{\lim }\frac{\cos ax-\cos bx}{\cos cx-1}$

Answer 1

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{\cos ax-\cos bx}{\cos cx-1} \\ =\underset{x\to 0}{\lim }\frac{-2\sin \left({\displaystyle \frac{ax+bx}{2}}\right)\sin \left({\displaystyle \frac{ax-bx}{2}}\right)}{-2{\sin }^2{\displaystyle \frac{cx}{2}}}\left(1-\cos 2\theta =2{\sin }^2\theta \right) \\ =\underset{x\to 0}{\lim }\frac{\left({\displaystyle \frac{a+b}{2}}\right)x\times {\displaystyle \frac{\sin \left({\displaystyle \frac{a+b}{2}}\right)x}{\left({\displaystyle \frac{a+b}{2}}\right)x}}\times \left({\displaystyle \frac{a-b}{2}}\right)x\times {\displaystyle \frac{\sin \left({\displaystyle \frac{a-b}{2}}\right)x}{\left({\displaystyle \frac{a-b}{2}}\right)x}}}{{\displaystyle \frac{c^2{x}^2}{4}}\times {\displaystyle \frac{{\sin }^2{\displaystyle \frac{cx}{2}}}{{\displaystyle \frac{c^2{x}^2}{4}}}}} \\ =\frac{\left(a+b\right)\left(a-b\right)}{c^2}\times \frac{\underset{x\to 0}{\lim }{\displaystyle \frac{\sin \left({\displaystyle \frac{a+b}{2}}\right)x}{\left({\displaystyle \frac{a+b}{2}}\right)x}}\times \underset{x\to 0}{\lim }{\displaystyle \frac{\sin \left({\displaystyle \frac{a-b}{2}}\right)x}{\left({\displaystyle \frac{a-b}{2}}\right)x}}}{{\left(\underset{x\to 0}{\lim }{\displaystyle \frac{\sin {\displaystyle \frac{cx}{2}}}{{\displaystyle \frac{cx}{2}}}}\right)}^2} \end{gathered}$$
$$\begin{gathered} =\frac{a^2-b^2}{c^2}\times \frac{1\times 1}{1}\left(\underset{\theta \to 0}{\lim }\frac{\sin \theta }{\theta }=1\right) \\ =\frac{a^2-b^2}{c^2} \end{gathered}$$