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Chapter 12: Problem 18

Evaluate the following limits. $$\lim _{(u, v) \rightarrow(1,-1)} \frac{10 u v-2 v^{2}}{u^{2}+v^{2}}$$

Short Answer

Expert verified
Answer: The limit of the function as (u, v) approaches (1, -1) is -6.

Step by step solution

01

Check if the limit exists

To determine if the limit exists, we will first check if the function is continuous at the point (1, -1). A function is continuous if the following three conditions are satisfied: 1. The function is defined at the point (u=v=0). 2. The limit of the function exists at the point (u=v=0). 3. The value of the function at the point (u=v=0) is the same as the limit of the function as (u, v) approaches (1, -1).
02

Substitute (u, v) = (1, -1) into the function

Let's substitute (u, v)=(1,-1) into the given function to see if it is defined at the point (1, -1): $$\frac{10(1)(-1)-2(-1)^{2}}{1^{2}+(-1)^{2}} = \frac{-10-2}{1+1} = \frac{-12}{2} = -6$$ The function is defined at the point (1, -1), and its value is -6.
03

Determine if the limit exists

We will now try to find the limit of the function as (u, v) approaches (1, -1). We can rewrite the function as: $$\lim _{(u, v) \rightarrow(1,-1)} \frac{10 u v-2 v^{2}}{u^{2}+v^{2}} = \lim_{(u, v) \rightarrow (1, -1)} \frac{2v(5u-v)}{u^{2}+v^{2}}$$ For the limit to exist, the expression should have the same limit when approaching the point (1,-1) from different paths.
04

Examine the limit along different paths

Let's consider two different paths: Path 1: Let u = 1 + t, and v = -1 + t Then, as t approaches 0, (u, v) approaches (1, -1). So, the limit along path 1 is: $$\lim_{t \rightarrow 0} \frac{2(-1+t)(5(1+t)-(-1+t))}{(1+t)^2+(-1+t)^2} = -6$$ Path 2: Let u = 1 + t, and v = -1 - t Then, as t approaches 0, (u, v) approaches (1, -1). So, the limit along path 2 is: $$\lim_{t \rightarrow 0} \frac{2(-1-t)(5(1+t)-(-1-t))}{(1+t)^2+(-1-t)^2} = -6$$ Since the limit is the same along both paths, we can safely conclude that the limit exists and is equal to -6. Therefore, the limit of the given function as (u, v) approaches (1, -1) is -6. $$\lim _{(u, v) \rightarrow(1,-1)} \frac{10 u v-2 v^{2}}{u^{2}+v^{2}} = -6$$

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