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

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Limits

lim_{x \to \frac{π}{4}} \frac{8 2 - ( c o s x + s i n x ) ^{7}}{2 - 2 s i n 2 x}

is equal to

Answer: (A)

x = \frac{π}{4}

cos x + sin x = 2

and

sin 2 x = 1

, giving

\frac{0}{0}

form. Let

t = cos x + sin x

. Then

t^{2} = 1 + sin 2 x

, so

sin 2 x = t^{2} - 1

. As

x \to \frac{π}{4}

t \to 2

. The limit becomes

lim_{t \to 2} \frac{8 2 - t ^{7}}{2 - 2 ( t ^{2} - 1 )} = lim_{t \to 2} \frac{8 2 - t ^{7}}{2 ( 2 - t ^{2} )}

. Using L'Hôpital's rule or factoring: the numerator factors as

(2)^{7} - t^{7}

with derivative

- 7 t^{6}

t = 2

, and denominator derivative is

- 22 t

. The limit is

\frac{- 7 ( 2 ) ^{6}}{- 2 2 \cdot 2} = \frac{7 \times 8}{4} = 14

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Single correct option

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