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Find the limits in Exercises \(1-12.\) $$\lim _{(x, y) \rightarrow(2,-3)}\left(\frac{1}{x}+\frac{1}{y}\right)^{2}$$

Short Answer

Expert verified
The limit is \( \frac{1}{36} \).

Step by step solution

01

Understanding the Problem

We need to find the limit of the function \( \left(\frac{1}{x}+\frac{1}{y}\right)^{2} \) as \( (x, y) \) approaches \( (2, -3) \). This means we are looking at how the value of the expression changes as \( x \) gets close to \( 2 \) and \( y \) gets close to \( -3 \).
02

Substitute Values

Substitute \( x = 2 \) and \( y = -3 \) into the expression \( \frac{1}{x} + \frac{1}{y} \). This gives us \( \frac{1}{2} + \frac{1}{-3} \).
03

Simplify the Expression

Compute the sum of the two fractions: \( \frac{1}{2} + \frac{1}{-3} = \frac{1}{2} - \frac{1}{3} \). To simplify this, find a common denominator, which is 6. This gives \( \frac{3}{6} - \frac{2}{6} = \frac{1}{6} \).
04

Square the Result

Now, square the result obtained from adding the two fractions. We have \( \left(\frac{1}{6}\right)^2 = \frac{1}{36} \).
05

Conclusion

The limit \( \lim _{(x, y) \rightarrow (2,-3)} \left(\frac{1}{x} + \frac{1}{y}\right)^{2} \) is \( \frac{1}{36} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Multivariable Calculus
Multivariable calculus deals with functions that depend on several variables. In this exercise, we look at a function with two variables, \(x\) and \(y\). This type of function is common in multivariable calculus as it allows us to examine how changes in multiple inputs affect an output. Such functions can represent real-world situations where outcomes depend on several factors.
As we compute the limits of these functions, we move closer to a point in the space defined by the variables. In our example, the focus is on the point \((2, -3)\). Understanding how the function behaves near this point can provide insights into the function's nature, such as continuity and differentiability, which are crucial for more advanced calculus concepts.
Key concepts in multivariable calculus include:
  • Partial derivatives: These express how a function changes with respect to one variable while keeping the others constant.
  • Gradient: A vector that represents both the direction and the rate of fastest increase.
  • Contour plots: Graphs that help visualize functions of two variables.
Focusing on how these concepts apply in varying conditions builds a solid foundation in multivariable calculus.
Limit Evaluation
Limit evaluation in calculus is the process of finding the value that a function approaches as the input approaches a certain point. It is one of the fundamental concepts of calculus, helping us understand the behavior of functions at specific points.
In this exercise, we need to find the limit of the function as \((x, y)\) goes to \((2, -3)\). This involves looking at how the function behaves as the inputs get closer and closer to these values.Key steps in evaluating limits include:
  • Substituting the variable values: Direct substitution can often find the limit, if the result is a defined value.
  • Simplifying expressions: Algebraic simplifications can help when dealing with complex fractions or expressions.
  • Using limit laws: These rules allow the computation of limits based on operations like addition, multiplication, or composition of functions.
By following a structured approach like this, evaluating limits becomes a systematic process, enhancing our understanding of the function's nature near the point in question.
Algebraic Simplification
Algebraic simplification is a method to make mathematical expressions easier to handle by reducing their complexity. In the given exercise, it's crucial to simplify the expression \(\frac{1}{x} + \frac{1}{y}\) before computing the limit.
Here, simplification involves finding a common denominator for the fractions \(\frac{1}{2}\) and \(\frac{1}{-3}\). The common denominator in this case is 6, allowing the expression to be rewritten as \(\frac{3}{6} - \frac{2}{6}\). This simplifies to \(\frac{1}{6}\).
The process of algebraic simplification includes:
  • Identifying factors: Recognizing common factors or variables that can be canceled or combined.
  • Combining like terms: Simplifying terms to reduce expression size.
  • Using algebraic identities: Applying known identities for easier computation.
Simplification makes calculations tractable and ensures they are free of unnecessary complexity, aiding precision in further mathematical operations like taking limits.

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Most popular questions from this chapter

In Exercises \(51-54,\) verify that \(w_{x y}=w_{y x}\) $$w=\ln (2 x+3 y)$$

The fifth-order partial derivative \(\partial^{5} f / \partial x^{2} \partial y^{3}\) is zero for each of the following functions. To show this as quickly as possible, which variable would you differentiate with respect to first: \(x\) or \(y ?\) Try to answer without writing anything down. $$ \begin{array}{l}{\text { a. } f(x, y)=y^{2} x^{4} e^{x}+2} \\ {\text { b. } f(x, y)=y^{2}+y\left(\sin x-x^{4}\right)} \\ {\text { c. } f(x, y)=x^{2}+5 x y+\sin x+7 e^{x}} \\ {\text { d. } f(x, y)=x e^{y^{2} / 2}}\end{array} $$

You will explore functions to identify their local extrema. Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level curves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. d. Calculate the function's second partial derivatives and find the discriminant \(f_{x x} f_{y y}-f_{x y}^{2}\) . e. Using the max-min tests, classify the critical points found in part (c). Are your findings consistent with your discussion in part (c)? $$f(x, y)=\left\\{\begin{array}{ll}{x^{5} \ln \left(x^{2}+y^{2}\right),} & {(x, y) \neq(0,0)} \\ {0,} & {(x, y)=(0,0)}\end{array}\right.$$ $$-2 \leq x \leq 2, \quad-2 \leq y \leq 2$$

Find all the second-order partial derivatives of the functions in Exercises \(41-50 .\) $$w=x^{2} \tan (x y)$$

Extreme temperatures on a sphere Suppose that the Celsius temperature at the point \((x, y, z)\) on the sphere \(x^{2}+y^{2}+z^{2}=1\) is \(T=400 x y z^{2} .\) Locate the highest and lowest temperatures on the sphere.

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