# Finding the value of a limit of trigonometric functions

The question $$\lim_{x\to\pi/2} \frac{2^{1/2} - (1 + \sin x)^2}{\cos^2x}$$

My attempt

I figured that I needed to remove the $\cos^2x$ in the denominator, but I was unable to do so. I used a lot of trigonometric identities to try and simplify the $\cos^2x$, but to no avail. Please help.

Try and avoid using L'Hopital's rule, because I'm not permitted to use that to solve this question (although I couldn't solve it even when I tried to use it; The denominator just refused to be anything but zero). Thanks in advance.

• As lab bhattacharjee supposed too, I think that there is a typo in the formula. Should it be $$\lim_{x\to\pi/2} \frac{2^{1/2} - (1 + \sin x)^{1/2}}{\cos^2x}$$ instead ? Or one more typo in a textbook ? – Claude Leibovici Oct 27 '15 at 14:29

HINT:

$$\lim_{x\to\frac{\pi}{2}} \frac{2^{\frac{1}{2}} - (1 + \sin(x))^2}{\cos^2(x)}=$$ $$\lim_{x\to\frac{\pi}{2}} \frac{\sqrt{2} - (1 + \sin(x))^2}{\cos^2(x)}=$$ $$\lim_{x\to\frac{\pi}{2}} \left(\sqrt{2}-(1+\sin(x))^2\right)\sec^2(x)=$$ $$\left(\lim_{x\to\frac{\pi}{2}}\sqrt{2}-(1+\sin(x))^2\right)\left(\lim_{x\to\frac{\pi}{2}}\sec^2(x)\right)=$$ $$\left(\lim_{x\to\frac{\pi}{2}}\sqrt{2}-(1+\sin(x))^2\right)\left(\lim_{x\to\frac{\pi}{2}}\frac{1}{\cos^2(x)}\right)=$$ $$\left(\sqrt{2}-(1+\sin\left(\frac{\pi}{2}\right))^2\right)\left(\lim_{x\to\frac{\pi}{2}}\frac{1}{\cos^2(x)}\right)=$$ $$\left(\sqrt{2}-4\right)\left(\lim_{x\to\frac{\pi}{2}}\frac{1}{\cos^2(x)}\right)=$$ $$\left(\sqrt{2}-4\right)\left(\lim_{x\to\frac{\pi}{2}}\sec^2(x)\right)$$

• But won't 1/cos^2(x) evaluate to a 1/0 form when x tends to (pi/2)? (In the last step) – EuclidAteMyBreakfast Oct 27 '15 at 13:37
• @EuclidAteMyBreakfast What does $\sec^2(x)$ do, when x approaches $\frac{\pi}{2}$? – Jan Oct 27 '15 at 13:43

Assuming $$\lim_{y\to0}\dfrac{\sqrt2-\sqrt{1+\sin x}}{\cos^2x}$$

Set $\dfrac\pi2-x=y$

$$\lim_{y\to0}\dfrac{\sqrt2-\sqrt{1+\cos y}}{\sin^2y}=\lim_{y\to0}\dfrac{2-(1+\cos y)}{\sin^2y}\cdot\dfrac1{\lim_{y\to0}(\sqrt2+\sqrt{1+\cos y})}$$

Now $$\lim_{y\to0}\dfrac{2-(1+\cos y)}{\sin^2y}=\lim_{y\to0}\dfrac{1-\cos y}{(1-\cos y)(1+\cos y)}=?$$