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Chapter 12: 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
Answer: The limit of the function as (x, y) approaches (0, 蟺) is $$\frac{1}{2 \pi}$$.

Step by step solution

01

Analyse the limit

We are given the function, $$f(x, y) = \frac{\cos x y+\sin x y}{2 y}$$, and we want to find the limit as (x, y) approaches (0, 蟺). Since this is a multivariable limit, we can rewrite this as: $$\lim _{x \rightarrow 0}(\lim_{y \rightarrow \pi} f(x, y))$$
02

Calculate the limit as y approaches 蟺

First, let us calculate the limit as y approaches 蟺: $$\lim_{y \rightarrow \pi}f(x, y) = \lim_{y \rightarrow \pi} \frac{\cos x y+\sin x y}{2 y}$$. In this limit, we keep x constant and let y approach 蟺. We can notice that the denominator $$2 y$$ approaches $$2 \pi$$ as $$y \to \pi$$, which is nonzero. Now let us investigate what happens with the numerator: $$\lim_{y \rightarrow \pi} (\cos x y+\sin x y)$$ Since y approaches 蟺 while x is kept constant, we will have in the limit $$\cos x \pi + \sin x \pi$$. Thus, we have: $$\lim_{y \rightarrow \pi} f(x, y) =\frac{\cos x \pi + \sin x \pi}{2 \pi}$$
03

Calculate the limit as x approaches 0

Now, we want to evaluate the limit as x approaches 0, keeping in mind our answer from step 2: $$\lim_{x \rightarrow 0} (\frac{\cos x \pi + \sin x \pi}{2 \pi})$$ As x approaches 0, we will have $$\cos (0) + \sin (0)$$ in the numerator. Therefore, the limit becomes: $$\lim_{x \rightarrow 0} (\frac{\cos x \pi + \sin x \pi}{2 \pi}) = \frac{1 + 0}{2 \pi} = \frac{1}{2 \pi}$$
04

Final answer

So the limit of our given function as (x, y) approaches (0, 蟺) is: $$\lim _{(x, y) \rightarrow(0, \pi)} \frac{\cos x y+\sin x y}{2 y} = \frac{1}{2 \pi}$$

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