Question

$\displaystyle\lim _{x \rightarrow \frac{\pi}{4}} \frac{8 \sqrt{2}-(\cos x+\sin x)^7}{\sqrt{2}-\sqrt{2} \sin 2 x}$ 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 2 x=t^2$
$ \sin 2 x=t^2-1 $
$ \displaystyle\lim _{t \rightarrow \sqrt{2}} \frac{8 \sqrt{2}-t^7}{\sqrt{2}-\sqrt{2}\left(t^2-1\right)} $
$ \displaystyle\lim _{t \rightarrow \sqrt{2}} \frac{8 \sqrt{2}-t^7}{2 \sqrt{2}-\sqrt{2} t^2} \text { (L-Hospital Rule) }$
$ \displaystyle\lim _{t \rightarrow \sqrt{2}} \frac{-7 t^6}{-2 \sqrt{2} t}=\displaystyle\lim _{t \rightarrow \sqrt{2}} \frac{7}{2 \sqrt{2}} \times t^5 $
$ =\frac{7}{2 \sqrt{2}} \times(\sqrt{2})^5=14$