Show $b_n=\frac{\int_{0}^\epsilon \cos^n(x) dx}{\int_{\epsilon}^{1/2} \cos^n(x) dx}\to\infty.$ I want to show the sequence $$b_n = \frac{\int_{0}^\epsilon \cos^n(x) dx}{\int_{\epsilon}^{1/2} \cos^n(x) dx} $$
tends to $\infty$ for every $ \epsilon : \frac{1}{2} > \epsilon > 0$
Solution:
\begin{align*} b_n &\geq \frac{1}{\cos^n(\epsilon) \left (0.5 - \epsilon \right ) }\int_{0}^\epsilon \cos^n(x) dx\\
&\geq  \frac{1}{\cos^n(\epsilon) }\int_{0}^\epsilon \cos^n(x) dx \\
& \geq \frac{1}{\cos^n(\epsilon) }\int_{0}^{\epsilon/2} \cos^n(x) dx \\
& \geq \frac{\cos^n(\epsilon/2)}{\cos^n(\epsilon) } \frac{\epsilon}{2} \\
 & = \left ( \frac{\cos(\epsilon/2)}{\cos(\epsilon)} \right )^n \frac{\epsilon}{2} \to \infty \text{ as } n \to \infty
\end{align*}
EDIT: Adapted from the answer given below. Credit to pre-kidney

Consider the ratio $$M_{\epsilon}=\frac{\cos(\epsilon/2)}{\cos(\epsilon)},$$
and observe that $M_\epsilon>1$ for all $0<\epsilon\leq \tfrac12$ since $\cos$ is strictly decreasing on the interval $[0,\tfrac 12]$. Since $\cos x$ is positive on the interval $[0,\tfrac 12]$, it follows that
$$
\int_0^{\epsilon}\cos^nx\ dx\geq \int_0^{\epsilon/2}\cos^nx\ dx\overset{\star}{\geq }\frac{\epsilon}{2}\cos^n\bigl(\frac{\epsilon}{2}\bigr)= \frac{\epsilon}{2}M_\epsilon^n\cos^n(\epsilon),
$$
where $\bigl(\overset{\star}{\geq}\bigr)$ follows since $\cos^n x$ is decreasing on $[0,\epsilon/2]$.
On the other hand,
$$
\int_{\epsilon}^{\tfrac12}\cos^n x\ dx\leq (\tfrac12-\epsilon)\cos^n(\epsilon),
$$
since $\cos^n x$ is decreasing on $[\epsilon,\tfrac12]$. Combining the inequalities gives you the result, since
$$
\frac{\int_0^{\epsilon}\cos^n x\ dx}{\int_{\epsilon}^{\tfrac12}\cos^n x\ dx}\geq \frac{\tfrac{\epsilon}{2} M_\epsilon^n}{\tfrac12-\epsilon}=C_\epsilon M_\epsilon^n,
$$
where $C_\epsilon=\epsilon/(1-2\epsilon)$ is a positive constant and $M_\epsilon>1$ implies that the ratio diverges to $\infty$ as $n\to\infty$.

From the proof, you can see that the only facts about $\cos x$ that were used is that it is strictly decreasing and positive on $[0,\tfrac 12]$, thus the same holds for any function $f(x)$ satisfying these two conditions.
