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Question Number 20138 by tammi last updated on 22/Aug/17

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

Answered by ajfour last updated on 22/Aug/17

$$=\underset{x\to 0}{\lim }\frac{2\sin {}^2\left(\frac{ax}{2}\right)}{2\sin {}^2\left(\frac{bx}{2}\right)}=\frac{a^2}{b^2}\underset{x\to 0}{\lim }{\left[\frac{\sin \left(\frac{ax}{2}\right)}{\left(\frac{ax}{2}\right)}\times \frac{\left(\frac{bx}{2}\right)}{\sin \left(\frac{bx}{2}\right)}\right]}^2$$$$=\frac{a^2}{b^2}\times \underset{x\to 0}{\lim }\left[\frac{\sin \left(ax/2\right)}{ax/2}\right]\mathbf{÷}\underset{x\to 0}{\lim }\left[\frac{\sin \left(bx/2\right)}{bx/2}\right]$$$$=\frac{a^2}{b^2}\times 1\mathbf{÷}1=\frac{{\mathit{a}}^2}{{\mathit{b}}^2}.$$

Commented by tammi last updated on 23/Aug/17

$$thanksalottt$$

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