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Chapter 9: Problem 752

\(\lim _{\mathrm{x} \rightarrow \pi}\left[\left\\{\sqrt{\left.(17+\operatorname{Cos} \mathrm{x})-4\\} /(\pi-\mathrm{x})^{2}\right]}=?\right.\right.\) (a) \((1 / 8)\) (b) \((1 / 16)\) (c) \((1 / 24)\) (d) \((1 / 64)\)

Short Answer

Expert verified
The answer is \(\frac{1}{64}\) (option (d)).

Step by step solution

01

Apply L'Hopital's Rule once

To apply L'Hopital's rule, we need to differentiate both the numerator and the denominator of the function with respect to x. Notice that we need to use Chain Rule to differentiate the numerator: Numerator: \[ \frac{d}{dx}(\sqrt{17 + \cos x} - 4) = \frac{1}{2} \cdot \frac{1}{\sqrt{17 + \cos x}} \cdot (-\sin x), \] Denominator: \[ \frac{d}{dx}(\pi - x)^2 = -2(\pi - x). \] Applying L'Hopital's Rule, we have: \[ \lim_{x \rightarrow \pi}\left[\frac{\frac{1}{2} \cdot \frac{1}{\sqrt{17 + \cos x}} \cdot (-\sin x)}{-2(\pi - x)}\right]. \]
02

Simplify and apply L'Hopital's Rule again

Now, we simplify the result from Step 1: \[ \lim_{x \rightarrow \pi}\frac{\sin x}{4(\pi - x)\sqrt{17 + \cos x}}. \] The form is still indeterminate \(\frac{0}{0}\) as x approaches \(\pi\). We can apply L'Hopital's Rule again: Numerator: \[ \frac{d}{dx}\sin x = \cos x, \] Denominator: \[ \frac{d}{dx} [4(\pi - x)\sqrt{17 + \cos x}] = -4\sqrt{17+\cos x} - \frac{4(\pi - x)(-\sin x)}{2\sqrt{17 + \cos x}}. \] Applying L'Hopital's rule again, we have: \[ \lim_{x \rightarrow \pi}\frac{\cos x}{-4\sqrt{17+\cos x} - \frac{4(\pi - x)(-\sin x)}{2\sqrt{17 + \cos x}}}. \]
03

Calculate the limit

Now, we evaluate the limit as x approaches \(\pi\): \[ \frac{-1}{-4\sqrt{16} - 0} = \frac{1}{64}. \] The answer to the given limit is \(\frac{1}{64}\), which corresponds to option (d).

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