Question

$\underset{x\to 0}{\lim }\frac{1-\cos 5x}{1-\cos 6x}$

Answer 1

$$\begin{gathered} \underset{x\to 0}{\lim }\left[\frac{1-\cos \left(5x\right)}{1-\cos \left(6x\right)}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{2{\sin }^2\left({\displaystyle \frac{5x}{2}}\right)}{2{\sin }^{2}3x}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{{\sin }^2\left({\displaystyle \frac{5x}{2}}\right)}{{\left({\displaystyle \frac{5x}{2}}\right)}^2}\times \frac{{\left({\displaystyle \frac{5x}{2}}\right)}^2}{{\displaystyle \frac{{\sin }^{2}3x}{{\left(3x\right)}^2}}\times {\left(3x\right)}^2}\right] \\ =\underset{x\to 0}{\lim }\left[{\left\{\frac{\sin \left({\displaystyle \frac{5x}{2}}\right)}{{\displaystyle \frac{5x}{2}}}\right\}}^2\frac{{\displaystyle \frac{25}{4}}{x}^2}{{\left\{{\displaystyle \frac{\sin \left(3x\right)}{3x}}\right\}}^2\times 9x^2}\right] \\ =\frac{25}{9\times 4} \\ =\frac{25}{36} \end{gathered}$$