<s>[INST] 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? 
 [/INST] 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 $$\displaystyle y = \lim_{x\rightarrow \frac{\pi}{2}}\frac{(4\cos^3 x-3\cos x)^2}{\cos x} = \lim_{x\rightarrow \frac{\pi}{2}}\frac{\cos^2 x\cdot(4\cos^2 x-3)}{\cos x} = 0$$
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