<p>\(\lim_{x \to 4} \frac{\cos x - \cos a}{\cos x - \cot a} = \sin^3 a\)</p><p>State whether this is true or false.</p>
Step-by-Step Solution
Key Concept: Apply L'Hôpital's rule or recognize this as a derivative definition. The limit form suggests evaluating the derivative of cosine at x = a, which requires cos a = cot a to make the denominator zero when x = 4.
<p><strong>Step 1:</strong> Recognize the indeterminate form. For the limit to exist as x → 4, we need the denominator to vanish at x = 4. This suggests the problem likely means x → a (not x → 4).</p><p><strong>Step 2:</strong> Assuming the limit is $\lim_{x \to a} \frac{\cos x - \cos a}{\cos x - \cot a}$, when x = a both numerator and denominator equal zero (if cos a = cot a), giving 0/0 form.</p><p><strong>Step 3:</strong> Apply L'Hôpital's rule: $\lim_{x \to a} \frac{-\sin x}{-\sin x} = 1$ if cos a = cot a holds.</p><p><strong>Step 4:</strong> But the condition cos a = cot a means $\cos a = \frac{\cos a}{\sin a}$, so $\sin a = 1$, giving $a = \frac{\pi}{2}$, making $\sin^3 a = 1$.</p><p><strong>Step 5:</strong> The limit equals 1, not $\sin^3 a$ for arbitrary a. The statement is <strong>FALSE</strong> in general (though true only at specific values of a).</p><p>∴ Answer: A (False)</p>
Correct Answer: A