Question

$$\lim_{x \to \pi/4} \frac{(1 - \cos x)^2}{\tan^2 x - \sin^2 x} \text{ is equal to:}$$

• A. 0
• B. $(\sqrt{2} - 1)^2$
• C. 1
• D. $\infty$

Answer 1

The Correct Option is B

Let $x = \pi/4$, then: $$\cos(\pi/4) = \frac{1}{\sqrt{2}} \Rightarrow 1 - \cos x = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}},\quad \text{so: } (1 - \cos x)^2 = \left(\frac{\sqrt{2} - 1}{\sqrt{2}}\right)^2$$ $$\tan^2 x = 1,\quad \sin^2 x = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \Rightarrow \tan^2 x - \sin^2 x = 1 - \frac{1}{2} = \frac{1}{2}$$ $$\Rightarrow \text{Limit} = \frac{(\sqrt{2} - 1)^2/\ 2}{1/2} = (\sqrt{2} - 1)^2$$