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Chapter 13: Problem 23

Evaluate the following limits. $$\lim _{(x, y) \rightarrow(2,2)} \frac{y^{2}-4}{x y-2 x}$$

Short Answer

Expert verified
Answer: The limit is 2.

Step by step solution

01

Identify the limit point and rewrite the limit expression

Given the limit expression, we can see that the limit point is \((2, 2)\). Let's rewrite the expression as: $$\lim _{(x, y) \rightarrow(2,2)} \frac{y^{2}-4}{x y-2 x}$$
02

Plug in the limit point into the function

Plug in the coordinates \((2, 2)\) into the function to identify the form: $$\frac{2^2-4}{(2)(2) - 2(2)} = \frac{4-4}{4-4} = \frac{0}{0}$$ Since we get an indeterminate expression of the form \(\frac{0}{0}\), we need to simplify the expression further.
03

Simplify the numerator and denomitor before attempting to solve for the limit

Factor the numerator and the denominator: $$\frac{y^2 - 4}{xy - 2x} = \frac{(y + 2)(y - 2)}{x(y - 2)}$$
04

Cancel out common factors

Divide the \((y-2)\) terms in the numerator and denominator to further simplify the expression: $$\frac{(y + 2)(y - 2)}{x(y - 2)} = \frac{y + 2}{x}$$
05

Plug in the limit point again after simplification

Now, plug in the point \((2, 2)\) into the simplified expression: $$\lim _{(x, y) \rightarrow(2,2)} \frac{y + 2}{x} = \frac{2 + 2}{2} = \frac{4}{2}$$
06

Evaluate the limit

Now that the expression has been simplified, the limit becomes a simple calculation: $$\frac{4}{2} = 2$$ So, the limit of the given function as \((x, y) \rightarrow (2, 2)\) is \(2\).

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Most popular questions from this chapter

In the advanced subject of complex variables, a function typically has the form \(f(x, y)=u(x, y)+i v(x, y),\) where \(u\) and \(v\) are real-valued functions and \(i=\sqrt{-1}\) is the imaginary unit. A function \(f=u+i v\) is said to be analytic (analogous to differentiable) if it satisfies the Cauchy-Riemann equations: \(u_{x}=v_{y}\) and \(u_{y}=-v_{x}\) a. Show that \(f(x, y)=\left(x^{2}-y^{2}\right)+i(2 x y)\) is analytic. b. Show that \(f(x, y)=x\left(x^{2}-3 y^{2}\right)+i y\left(3 x^{2}-y^{2}\right)\) is analytic. c. Show that if \(f=u+i v\) is analytic, then \(u_{x x}+u_{y y}=0\) and \(v_{x x}+v_{y y}=0\)

Use the method of your choice to ate the following limits. $$\lim _{(x, y) \rightarrow(1,1)} \frac{x^{2}+x y-2 y^{2}}{2 x^{2}-x y-y^{2}}$$

The domain of $$f(x, y)=e^{-1 /\left(x^{2}+y^{2}\right)}$$ excludes \((0,0) .\) How should \(f\) be defined at (0,0) to make it continuous there?

Find the points (if they exist) at which the following planes and curves intersect. $$\begin{aligned}&2 x+3 y-12 z=0 ; \quad \mathbf{r}(t)=\langle 4 \cos t, 4 \sin t, \cos t\rangle\\\&\text { for } 0 \leq t \leq 2 \pi\end{aligned}$$

Consider the following functions \(f.\) a. Is \(f\) continuous at (0,0)\(?\) b. Is \(f\) differentiable at (0,0)\(?\) c. If possible, evaluate \(f_{x}(0,0)\) and \(f_{y}(0,0)\) d. Determine whether \(f_{x}\) and \(f_{y}\) are continuous at \((0,0).\) e. Explain why Theorems 5 and 6 are consistent with the results in parts \((a)-(d).\) $$f(x, y)=\sqrt{|x y|}$$
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