Question

Evaluate $\underset{\theta \to \frac{\pi }{4}}{lim}\frac{\sqrt{2}-\cos \theta -\sin \theta }{(4\theta -\pi {)}^2}$

• A. $\frac{1}{16\sqrt{2}}$
• B. $\frac{1}{4\sqrt{2}}$
• C. $-\frac{1}{4\sqrt{2}}$
• D. $-\frac{1}{16\sqrt{2}}$

Answer 1

The correct option is A $\frac{1}{16\sqrt{2}}$

$\underset{\theta \to \frac{\pi }{4}}{lim}\frac{\sqrt{2}-\cos \theta -\sin \theta }{(4\theta -\pi {)}^2}=\underset{\theta \to \frac{\pi }{4}}{lim}\frac{\sqrt{2}-\sqrt{2}\left[\cos [\theta -\frac{\pi }{4}]\right]}{(4\theta -\pi {)}^2}$

$=\underset{\theta \to \frac{\pi }{4}}{lim}\frac{\sqrt{2}[1-\cos [\theta -\frac{\pi }{4}]]}{16{[\theta -\frac{\pi }{4}]}^2}=\frac{1}{16\sqrt{2}}$

Since $\underset{x\to 0}{lim}\frac{1-\cos x}{x^2}=\frac{1}{2}$