Answer
$$\frac{-9}{50} $$
Work Step by Step
Given $$\lim_{h\to 0 }\frac{1-\cos 3h}{ \cos^25h-1}$$
Then
\begin{align*}
\lim_{h\to 0 }\frac{1-\cos 3h}{ \cos^25h-1}&=\lim_{h\to 0 }\frac{2\sin^2(3h/2)}{ -\sin^25h}\\
&=-2\lim_{h\to 0 }\frac{\frac{9h^2/4}{9h^2/4}\sin^2(3h/2)}{ \frac{25h^2}{25h^2} \sin^25h}\\
&=-2\frac{9/4\left(\lim_{h\to 0 } \frac{\sin 3h/2}{3h/2}\right)^2}{ \lim_{h\to 0 }25\left(\frac{\sin(5h)}{5h }\right)^2 }\\
&=\frac{-9}{50}
\end{align*}
