Question

Evaluate the limit: \[ \lim_{x \to \frac{\pi}{4}} \frac{4\sqrt{2} - (\cos x + \sin x)^5}{1 - \sin 2x} = \]

• A. \( 5\sqrt{2} \)
• B. \( 3\sqrt{2} \)
• C. \( 2\sqrt{2} \)
• D. \( \sqrt{2} \)

Hint:

When dealing with indeterminate forms, using L'Hopital's rule or expanding the functions around the point of interest can help evaluate the limit.

Answer 1

The Correct Option is A

Step 1: Simplify the expression inside the limit. As \( x \) approaches \( \frac{\pi}{4} \), both \( \cos x \) and \( \sin x \) approach \( \frac{\sqrt{2}}{2} \). Hence, \[ \cos x + \sin x = \sqrt{2}. \] Step 2: Substitute \( \cos x + \sin x = \sqrt{2} \) into the numerator. \[ (\cos x + \sin x)^5 = (\sqrt{2})^5 = 4\sqrt{2}. \] Step 3: Now, substitute into the limit expression: \[ \lim_{x \to \frac{\pi}{4}} \frac{4\sqrt{2} - 4\sqrt{2}}{1 - \sin 2x}. \] We now handle the denominator. As \( x \to \frac{\pi}{4} \), \( \sin 2x = \sin \frac{\pi}{2} = 1 \). Hence, the denominator becomes: \[ 1 - 1 = 0. \] Step 4: At this stage, we have an indeterminate form \( \frac{0}{0} \). We can apply L'Hopital's rule or expand both the numerator and denominator using series expansions around \( x = \frac{\pi}{4} \). First, for the numerator \( 4\sqrt{2} - (\cos x + \sin x)^5 \), expand \( \cos x + \sin x \) around \( x = \frac{\pi}{4} \): \[ \cos x + \sin x = \sqrt{2} + (x - \frac{\pi}{4}) \cdot \text{(small term)}. \] Then, for the denominator \( 1 - \sin 2x \), expand \( \sin 2x \) around \( x = \frac{\pi}{4} \). By applying the expansion methods or L'Hopital's rule, the final limit evaluates to \( 5\sqrt{2} \).