Key Concepts and Formulas

• L'Hôpital's Rule: If a limit of a quotient of two functions, $\frac{f(x)}{g(x)}$, results in an indeterminate form ($\frac{0}{0}$ or $\frac{\infty}{\infty}$) as $x$ approaches a certain value $a$, then the limit is equal to the limit of the quotient of their derivatives, i.e., $\lim\limits_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim\limits_{x \rightarrow a} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.
• Trigonometric Identities:
  • $\sin 2x = 2 \sin x \cos x$
  • $\cos 2x = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x = 2 \cos^2 x - 1$
• Differentiation Rules:
  • Chain Rule: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$
  • Derivative of a power: $\frac{d}{dx}[u^n] = n u^{n-1} \frac{du}{dx}$
  • Derivative of $\sin x$: $\cos x$
  • Derivative of $\cos x$: $-\sin x$

Step-by-Step Solution

Step 1: Evaluate the form of the limit. We are asked to find the limit: $L = \lim\limits_{x \rightarrow \frac{\pi}{4}} \frac{8 \sqrt{2}-(\cos x+\sin x)^{7}}{\sqrt{2}-\sqrt{2} \sin 2 x}$ Let's substitute $x = \frac{\pi}{4}$ into the numerator and the denominator. Numerator: $8 \sqrt{2}-(\cos \frac{\pi}{4}+\sin \frac{\pi}{4})^{7} = 8 \sqrt{2}-(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})^{7} = 8 \sqrt{2}-(\frac{2}{\sqrt{2}})^{7} = 8 \sqrt{2}-(\sqrt{2})^{7} = 8 \sqrt{2}-8 \sqrt{2} = 0$. Denominator: $\sqrt{2}-\sqrt{2} \sin (2 \cdot \frac{\pi}{4}) = \sqrt{2}-\sqrt{2} \sin (\frac{\pi}{2}) = \sqrt{2}-\sqrt{2}(1) = 0$. Since we have the indeterminate form $\frac{0}{0}$, we can apply L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule for the first time. Let $f(x) = 8 \sqrt{2}-(\cos x+\sin x)^{7}$ and $g(x) = \sqrt{2}-\sqrt{2} \sin 2 x$. We need to find the derivatives $f'(x)$ and $g'(x)$. $f'(x) = \frac{d}{dx}[8 \sqrt{2}-(\cos x+\sin x)^{7}]$ $f'(x) = 0 - 7(\cos x+\sin x)^{6} \cdot \frac{d}{dx}(\cos x+\sin x)$ $f'(x) = -7(\cos x+\sin x)^{6} \cdot (-\sin x + \cos x)$

$g'(x) = \frac{d}{dx}[\sqrt{2}-\sqrt{2} \sin 2 x]$ $g'(x) = 0 - \sqrt{2} \cdot \frac{d}{dx}(\sin 2 x)$ $g'(x) = -\sqrt{2} \cdot (\cos 2x \cdot 2)$ $g'(x) = -2\sqrt{2} \cos 2x$

Applying L'Hôpital's Rule: $L = \lim\limits_{x \rightarrow \frac{\pi}{4}} \frac{-7(\cos x+\sin x)^{6} (\cos x - \sin x)}{-2\sqrt{2} \cos 2x}$

Step 3: Evaluate the form of the limit after the first application of L'Hôpital's Rule. Substitute $x = \frac{\pi}{4}$ into the new expression: Numerator: $-7(\cos \frac{\pi}{4}+\sin \frac{\pi}{4})^{6} (\cos \frac{\pi}{4} - \sin \frac{\pi}{4}) = -7(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})^{6} (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) = -7(\sqrt{2})^{6} (0) = 0$. Denominator: $-2\sqrt{2} \cos (2 \cdot \frac{\pi}{4}) = -2\sqrt{2} \cos (\frac{\pi}{2}) = -2\sqrt{2} (0) = 0$. We still have the indeterminate form $\frac{0}{0}$, so we need to apply L'Hôpital's Rule again.

Step 4: Apply L'Hôpital's Rule for the second time. Let $p(x) = -7(\cos x+\sin x)^{6} (\cos x - \sin x)$ and $q(x) = -2\sqrt{2} \cos 2x$. We need to find the derivatives $p'(x)$ and $q'(x)$. For $p'(x)$, we use the product rule: $\frac{d}{dx}(uv) = u'v + uv'$. Let $u = -7(\cos x+\sin x)^{6}$ and $v = (\cos x - \sin x)$. $u' = -7 \cdot 6(\cos x+\sin x)^{5} \cdot (-\sin x + \cos x) = -42(\cos x+\sin x)^{5} (\cos x - \sin x)$. $v' = -\sin x - \cos x = -(\sin x + \cos x)$.

So, $p'(x) = u'v + uv'$ $p'(x) = [-42(\cos x+\sin x)^{5} (\cos x - \sin x)](\cos x - \sin x) + [-7(\cos x+\sin x)^{6}][- (\sin x + \cos x)]$ $p'(x) = -42(\cos x+\sin x)^{5} (\cos x - \sin x)^2 + 7(\cos x+\sin x)^{7}$

For $q'(x)$: $q'(x) = \frac{d}{dx}[-2\sqrt{2} \cos 2x]$ $q'(x) = -2\sqrt{2} (-\sin 2x \cdot 2)$ $q'(x) = 4\sqrt{2} \sin 2x$

Applying L'Hôpital's Rule again: $L = \lim\limits_{x \rightarrow \frac{\pi}{4}} \frac{-42(\cos x+\sin x)^{5} (\cos x - \sin x)^2 + 7(\cos x+\sin x)^{7}}{4\sqrt{2} \sin 2x}$

Step 5: Evaluate the limit after the second application of L'Hôpital's Rule. Substitute $x = \frac{\pi}{4}$: Numerator: $-42(\cos \frac{\pi}{4}+\sin \frac{\pi}{4})^{5} (\cos \frac{\pi}{4} - \sin \frac{\pi}{4})^2 + 7(\cos \frac{\pi}{4}+\sin \frac{\pi}{4})^{7}$ $= -42(\sqrt{2})^{5} (0)^2 + 7(\sqrt{2})^{7} = 0 + 7 \cdot 8\sqrt{2} = 56\sqrt{2}$.

Denominator: $4\sqrt{2} \sin (2 \cdot \frac{\pi}{4}) = 4\sqrt{2} \sin (\frac{\pi}{2}) = 4\sqrt{2} (1) = 4\sqrt{2}$.

So the limit is: $L = \frac{56\sqrt{2}}{4\sqrt{2}} = \frac{56}{4} = 14$

Alternative Approach using Substitution and Taylor Series (for deeper understanding, not strictly required for this problem if L'Hopital is mastered):

Let $x = \frac{\pi}{4} + h$. As $x \rightarrow \frac{\pi}{4}$, $h \rightarrow 0$. $\cos x = \cos(\frac{\pi}{4} + h) = \cos \frac{\pi}{4} \cos h - \sin \frac{\pi}{4} \sin h = \frac{1}{\sqrt{2}}(\cos h - \sin h)$ $\sin x = \sin(\frac{\pi}{4} + h) = \sin \frac{\pi}{4} \cos h + \cos \frac{\pi}{4} \sin h = \frac{1}{\sqrt{2}}(\cos h + \sin h)$

$\cos x + \sin x = \frac{1}{\sqrt{2}}(\cos h - \sin h) + \frac{1}{\sqrt{2}}(\cos h + \sin h) = \frac{2}{\sqrt{2}} \cos h = \sqrt{2} \cos h$.

$\sin 2x = \sin(2(\frac{\pi}{4} + h)) = \sin(\frac{\pi}{2} + 2h) = \cos 2h$.

The limit becomes: $\lim\limits_{h \rightarrow 0} \frac{8 \sqrt{2}-(\sqrt{2} \cos h)^{7}}{\sqrt{2}-\sqrt{2} \cos 2h} = \lim\limits_{h \rightarrow 0} \frac{8 \sqrt{2}-2^{7/2} \cos^7 h}{\sqrt{2}(1 - \cos 2h)}$ $= \lim\limits_{h \rightarrow 0} \frac{8 \sqrt{2}-8\sqrt{2} \cos^7 h}{\sqrt{2}(1 - \cos 2h)} = \lim\limits_{h \rightarrow 0} \frac{8(1 - \cos^7 h)}{1 - \cos 2h}$ Using Taylor series expansions for small $h$: $\cos h \approx 1 - \frac{h^2}{2}$ $\cos^7 h \approx (1 - \frac{h^2}{2})^7 \approx 1 - 7 \frac{h^2}{2}$ (using binomial expansion $(1+u)^n \approx 1+nu$ for small $u$) $\cos 2h \approx 1 - \frac{(2h)^2}{2} = 1 - \frac{4h^2}{2} = 1 - 2h^2$.

Substituting these approximations: $\lim\limits_{h \rightarrow 0} \frac{8(1 - (1 - \frac{7h^2}{2}))}{(1 - (1 - 2h^2))} = \lim\limits_{h \rightarrow 0} \frac{8(\frac{7h^2}{2})}{2h^2} = \lim\limits_{h \rightarrow 0} \frac{28h^2}{2h^2} = 14$ This confirms the result obtained using L'Hôpital's Rule.

Common Mistakes & Tips

• Algebraic Errors: Be meticulous with signs and powers, especially when differentiating and simplifying expressions involving trigonometric functions.
• Incorrect Application of L'Hôpital's Rule: Ensure the limit is indeed in an indeterminate form ($\frac{0}{0}$ or $\frac{\infty}{\infty}$) before applying the rule. Also, remember to differentiate the numerator and denominator separately.
• Simplification: After applying L'Hôpital's Rule, always re-evaluate the form of the limit. If it's still indeterminate, apply the rule again. Simplification before substitution can sometimes make the evaluation easier. For instance, recognizing that $\cos x - \sin x = 0$ at $x = \frac{\pi}{4}$ is crucial.

Summary

The problem involves evaluating a limit that results in the indeterminate form $\frac{0}{0}$ at $x = \frac{\pi}{4}$. L'Hôpital's Rule is the most direct method to solve this. We apply the rule twice, differentiating the numerator and denominator each time, and simplifying the expressions. After the second application of L'Hôpital's Rule and careful evaluation, we arrive at the value 14. An alternative approach using substitution and Taylor series approximations also yields the same result, reinforcing the correctness of the answer.

The final answer is $\boxed{14}$.