Solve L'Hopitals problem $$\lim_{x\rightarrow \frac{\pi}{2}} \frac{\sec x}{{\sec^2 3x}} $$
I used LH: $$\lim_{x\rightarrow \frac{\pi}{2}} \frac{\sec x \tan x}{6\sec 3x \sec 3x \tan 3x}$$
then: $$\lim_{x\rightarrow \frac{\pi}{2}} \frac{\sec x\tan x}{ 6 \sec^2 3x \tan 3x}$$
Now I'm stuck there.
What should I do next? 
 A: Solution With out Using $\bf{D-Lhopital \; Rule}$
Let $$\displaystyle y = \lim_{x\rightarrow \frac{\pi}{2}}\frac{\sec x}{(\sec 3x)^2} = \lim_{x\rightarrow \frac{\pi}{2}}\frac{(\cos 3x)^2}{\cos x}.$$
Now Using $$\bullet\; \cos 3x = 4\cos^3 x-3\cos x$$
We get $$\displaystyle y = \lim_{x\rightarrow \frac{\pi}{2}}\frac{(4\cos^3 x-3\cos x)^2}{\cos x} = \lim_{x\rightarrow \frac{\pi}{2}}\frac{\cos^2 x\cdot(4\cos^2 x-3)}{\cos x} = 0$$
A: Its much better to change $\sec$ into $\cos$ and then we get
\begin{align}
L &= \lim_{x \to \pi/2}\frac{\sec x}{\sec^{2}3x}\notag\\
&= \lim_{x \to \pi/2}\frac{\cos^{2}3x}{\cos x}\notag\\
&= \lim_{h \to 0}\dfrac{\cos^{2}3\left(\dfrac{\pi}{2} + h\right)}{\cos \left(\dfrac{\pi}{2} + h\right)}\text{ (by putting }x = (\pi/2) + h)\notag\\
&= \lim_{h \to 0}\frac{\sin^{2}3h}{-\sin h}\notag\\
&= -\lim_{h \to 0}\sin 3h\cdot\frac{\sin 3h}{\sin h}\notag\\
&= -\lim_{h \to 0}\sin 3h\cdot(3 - 4\sin^{2}h)\notag\\
&= -0\cdot (3 - 4\cdot 0) = 0
\end{align}
We have used the formula $\sin 3h = 3\sin h - 4\sin^{3}h$ in the above evaluation.
