Question

$\underset{x\to 0}{lim}\frac{\cos ax-\cos bx}{x^2}$

Answer 1

$\underset{x\to 0}{lim}\frac{\cos ax-\cos bx}{x^2}\left(\frac{0}{0}form\right)$

using L'Hospital Rule

$=\underset{x\to 0}{lim}\frac{-a\sin ax+b\sin bx}{2x}$

$=\underset{x\to 0}{lim}\frac{-a^2}{2}\left(\frac{\sin ax}{ax}\right)+\frac{b^2}{2}\left(\frac{\sin bx}{bx}\right)$

$=\frac{-a^2}{2}+\frac{b^2}{2}\left[therefore\underset{x\to 0}{lim}\frac{\sin x}{x}=1\right]$

$=\frac{b^2-a^2}{2}$