Question

$\underset{x\to \frac{\pi }{4}}{\lim }\frac{8\sqrt{2}-(\cos x+\sin x{)}^7}{\sqrt{2}-\sqrt{2}\sin 2x}$ is equal to

• A. 14
• B. 7
• C. $14\sqrt{2}$
• D. $7\sqrt{2}$

Answer 1

Answer: (A)

$\sin x+\cos x=t$
$1+\sin 2x=t^2$
$\sin 2x=t^2-1$
$\underset{t\to \sqrt{2}}{\lim }\frac{8\sqrt{2}-t^7}{\sqrt{2}-\sqrt{2}\left(t^2-1\right)}$
$\underset{t\to \sqrt{2}}{\lim }\frac{8\sqrt{2}-t^7}{2\sqrt{2}-\sqrt{2}{t}^2}\text{ (L-Hospital Rule) }$
$\underset{t\to \sqrt{2}}{\lim }\frac{-7t^6}{-2\sqrt{2}t}=\underset{t\to \sqrt{2}}{\lim }\frac{7}{2\sqrt{2}}\times t^5$
$=\frac{7}{2\sqrt{2}}\times (\sqrt{2}{)}^5=14$