Evaluate the following limit. Use lhôpital's Rule when it is convenient and applicable. \( \lim _{x \rightarrow-\pi} \frac{9 \cos x+9}{(x+\pi)^{2}} \)
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To evaluate the limit \[ \lim _{x \rightarrow-\pi} \frac{9 \cos x + 9}{(x+\pi)^{2}}, \] we start by plugging in \( x = -\pi \): - The numerator becomes \( 9 \cos(-\pi) + 9 = 9(-1) + 9 = -9 + 9 = 0 \). - The denominator becomes \( (-\pi + \pi)^{2} = 0^{2} = 0 \). Since both the numerator and denominator approach 0, we can use L'Hôpital's Rule. Applying L'Hôpital's Rule involves differentiating the numerator and the denominator: 1. Differentiate the numerator: \[ \frac{d}{dx}(9 \cos x + 9) = 9(-\sin x) = -9 \sin x. \] 2. Differentiate the denominator: \[ \frac{d}{dx}((x + \pi)^{2}) = 2(x + \pi). \] Now we apply L'Hôpital's Rule: \[ \lim_{x \rightarrow -\pi} \frac{9 \cos x + 9}{(x + \pi)^{2}} = \lim_{x \rightarrow -\pi} \frac{-9 \sin x}{2(x + \pi)}. \] Next, we evaluate this limit as \( x \) approaches \(-\pi\). - The numerator: \( -9 \sin(-\pi) = -9(0) = 0 \). - The denominator: \( 2(-\pi + \pi) = 2(0) = 0 \). Since we again have a \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule again. Differentiating again: 1. Differentiate the numerator: \[ \frac{d}{dx}(-9 \sin x) = -9 \cos x. \] 2. Differentiate the denominator: \[ \frac{d}{dx}(2(x + \pi)) = 2. \] Now we apply L'Hôpital's Rule a second time: \[ \lim_{x \rightarrow -\pi} \frac{-9 \sin x}{2(x + \pi)} = \lim_{x \rightarrow -\pi} \frac{-9 \cos x}{2}. \] Evaluating this limit: - The numerator: \( -9 \cos(-\pi) = -9(-1) = 9 \). - The denominator: \( 2 \). Thus, we have: \[ \lim_{x \rightarrow -\pi} \frac{-9 \cos x}{2} = \frac{9}{2}. \] The final answer is \[ \boxed{\frac{9}{2}}. \]
