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

We wish to evaluate the limit $$\displaystyle\lim_{x \to \frac{\pi}{4}} \dfrac{8\sqrt{2} - (\cos x + \sin x)^7}{\sqrt{2} - \sqrt{2}\sin 2x}\,. $$

Setting $$t = x - \dfrac{\pi}{4}$$ so that $$t \to 0$$, we observe that $$ \cos x + \sin x = \sqrt{2}\cos\!\Bigl(x - \tfrac{\pi}{4}\Bigr) = \sqrt{2}\cos t\,, $$ and hence $$ (\cos x + \sin x)^7 = (\sqrt{2})^7 \cos^7 t = 8\sqrt{2}\cos^7 t\,. $$

It follows that the numerator becomes $$ 8\sqrt{2} - 8\sqrt{2}\cos^7 t = 8\sqrt{2}\bigl(1 - \cos^7 t\bigr)\,. $$

Meanwhile, since $$ \sin 2x = \sin\!\Bigl(2t + \tfrac{\pi}{2}\Bigr) = \cos 2t\,, $$ the denominator simplifies as $$ \sqrt{2} - \sqrt{2}\sin 2x = \sqrt{2}\bigl(1 - \cos 2t\bigr) = \sqrt{2}\cdot 2\sin^2 t = 2\sqrt{2}\sin^2 t\,. $$

Combining these results gives $$ L = \lim_{t \to 0} \dfrac{8\sqrt{2}(1 - \cos^7 t)}{2\sqrt{2}\sin^2 t} = 4 \lim_{t \to 0} \dfrac{1 - \cos^7 t}{\sin^2 t}\,. $$

We factor the numerator using $$1 - \cos^7 t = (1 - \cos t)\bigl(1 + \cos t + \cos^2 t + \cdots + \cos^6 t\bigr)\,, $$ so that $$ L = 4 \lim_{t \to 0} \dfrac{(1 - \cos t)\,(1 + \cos t + \cos^2 t + \cdots + \cos^6 t)}{\sin^2 t}\,. $$

As $$t \to 0$$, we know $$\lim_{t \to 0} \dfrac{1 - \cos t}{\sin^2 t} = \lim_{t \to 0} \dfrac{1 - \cos t}{1 - \cos^2 t} = \lim_{t \to 0} \dfrac{1}{1 + \cos t} = \tfrac12$$ and $$\lim_{t \to 0} \bigl(1 + \cos t + \cos^2 t + \cdots + \cos^6 t\bigr) = 7\,. $$

Therefore, $$ L = 4 \times \tfrac12 \times 7 = 14\,. $$

The correct answer is Option A: $$14$$.