Solve L'Hopitals problem

Question

$$\lim_{x\rightarrow \frac{\pi}{2}} \frac{\sec x}{{\sec^2 3x}} $$

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?

Answer 1

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 $$\displaystyle y = \lim_{x\rightarrow \frac{\pi}{2}}\frac{(4\cos^3 x-3\cos x)^2}{\cos x} = \lim_{x\rightarrow \frac{\pi}{2}}\frac{\cos^2 x\cdot(4\cos^2 x-3)}{\cos x} = 0$$

Answer 2

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.