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Chapter 13: Problem 15

Evaluate the following limits. $$\lim _{(x, y) \rightarrow(0, \pi)} \frac{\cos x y+\sin x y}{2 y}$$

Short Answer

Expert verified
Question: Evaluate the limit of the following function as (x, y) approaches (0, 蟺): $$\lim _{(x, y) \rightarrow(0, \pi)} \frac{\cos x y+\sin x y}{2 y}$$ Answer: The limit exists and is equal to 0.

Step by step solution

01

Rewriting the expression as a single trigonometric function

First, we will rewrite the given function using the sine and cosine sum identity: $$\sin(x+y)=\cos x \sin y + \sin x \cos y$$ By multiplying x to both sides, we get: $$x \sin(x+y) = x \cos x \sin y + x \sin x \cos y$$ And so, the given expression becomes: $$\frac{x\sin(x+y)}{2y}$$
02

Apply the appropriate limit techniques

Now, we need to apply the appropriate limit techniques to evaluate the limit of this function. We will use the sandwich theorem to help us find the limit. We can find an absolute value expression such that: $$\left|\frac{x\sin(x+y)}{2y} \right| \leq \frac{|x|\cdot1}{2|y|}$$ Since \(-1 \leq \sin(x+y) \leq 1\). We have: $$\lim_{(x, y)\rightarrow(0, \pi)} \left|\frac{x\sin(x+y)}{2y} \right| \leq \lim_{(x, y)\rightarrow(0, \pi)} \frac{|x|\cdot1}{2|y|}$$
03

Evaluate the limit

Now, we will evaluate the limit of the right side of the inequality: $$\lim_{(x, y)\rightarrow(0, \pi)} \frac{|x|\cdot1}{2|y|} = \frac{0}{2|\pi|} = 0$$ According to the sandwich theorem: $$\lim_{(x, y)\rightarrow(0, \pi)} \left|\frac{x\sin(x+y)}{2y}\right| = 0$$ Thus, the limit of the function exists and is equal to 0. Therefore: $$\lim _{(x, y) \rightarrow(0, \pi)} \frac{\cos x y+\sin x y}{2 y} = 0$$

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