QuestionFind the limit: $\lim _{x \rightarrow \pi} \frac{\cos x+\sin (2 x)+1}{x^{2}-\pi^{2}}$ . Options: (A) $\frac{1}{2 \pi}$ (B) $\frac{1}{\pi}$ (C) 1 (D) nonexistent.

Studdy Solution

STEP 1

Question: What is the form of the limit as $x$ approaches $\pi$ for the function $\frac{\cos x+\sin (2 x)+1}{x^{2}-\pi^{2}}$ ?
A) $\frac{0}{0}$ B) $\frac{\infty}{\infty}$ C) $0$ D) $\infty$
Answer: A) $\frac{0}{0}$

STEP 2

Question: What is the derivative of $\cos x + \sin (2x) + 1$ ?
A) $-\sin x + 2\cos (2x)$ B) $-\sin x + \cos (2x)$ C) $\sin x + 2\cos (2x)$ D) $-\sin x + 2\sin (2x)$
Answer: A) $-\sin x + 2\cos (2x)$

STEP 3

Question: What is the derivative of $x^2 - \pi^2$ ?
A) $2x$ B) $2x - 2\pi$ C) $2\pi$ D) $2\pi x$
Answer: A) $2x$

STEP 4

Question: After applying L'Hopital's Rule, what is the new limit expression?
A) $\lim _{x \rightarrow \pi} \frac{-\sin x + 2\cos (2x)}{2\pi}$ B) $\lim _{x \rightarrow \pi} \frac{-\sin x + 2\cos (2x)}{2x}$ C) $\lim _{x \rightarrow \pi} \frac{\sin x + 2\cos (2x)}{2x}$ D) $\lim _{x \rightarrow \pi} \frac{\sin x + \cos (2x)}{2x}$
Answer: B) $\lim _{x \rightarrow \pi} \frac{-\sin x + 2\cos (2x)}{2x}$

STEP 5

Question: What is the value of $\frac{-\sin \pi + 2\cos (2\pi)}{2\pi}$ ?
A) $0$ B) $1$ C) $\frac{1}{\pi}$ D) $2$
Answer: C) $\frac{1}{\pi}$

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