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Find the limits in Exercises \(25-30.\) $$\lim _{P \rightarrow(\pi, 0,3)} z e^{-2 y} \cos 2 x$$

Short Answer

Expert verified
The limit is 3.

Step by step solution

01

Identify the Function and Limits

The function given is \( z e^{-2y} \cos 2x \) and we need to find the limit as \( P \) approaches \( (\pi, 0, 3) \). This means we need to substitute \( x = \pi \), \( y = 0 \), and \( z = 3 \) into the function.
02

Substitute Limit Values into the Function

Substitute \( x = \pi \), \( y = 0 \), and \( z = 3 \) into the function \( z e^{-2y} \cos 2x \):\[3 e^{-2(0)} \cos 2(\pi)\]
03

Simplify the Expression

Simplify the expression from Step 2. We begin by solving the exponent:\[e^{-2(0)} = e^0 = 1\]Then solve the cosine:\[\cos 2(\pi) = \cos(2\pi) = 1\]Thus, the expression becomes \(3 \times 1 \times 1\).
04

Calculate the Final Value

As a result of the simplification, we have:\[3 \times 1 \times 1 = 3\]This is the final value of the limit.

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Key Concepts

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

Multivariable Functions
Multivariable functions are mathematical expressions with multiple variables. In calculus, these functions require an understanding of partial derivatives and limits. The original exercise involved the multivariable function \( z e^{-2y} \cos 2x \). This expression involves three variables: \( x \), \( y \), and \( z \). Understanding how each variable interacts within the function is crucial for finding limits as they approach specific values.
  • Variables can represent different dimensions or directions in space.
  • Multivariable functions are used to model complex systems in engineering and science.
  • Approaching limits with multivariable functions often involves substitution and evaluating the function at specific points.
In our example, we are tasked with evaluating the limit of the function as the point \( P \) approaches \((\pi, 0, 3)\). This involves substituting each variable with its corresponding value to determine the output at that particular instance.
Substitution Method
The substitution method is a valuable technique in calculus, especially when working with limits and multivariable functions. It involves replacing variables in an equation with specific values to simplify the computation. For the problem at hand, the substitution method helps evaluate the limit of the function \( z e^{-2y} \cos 2x \) by substituting \( x = \pi \), \( y = 0 \), and \( z = 3 \) into the function.
  • This step is crucial for simplifying complex expressions and directly finding the limits.
  • Substitution allows us to see the impact each variable has on the outcome.
With the given function, by replacing the variables with their respective values from the limit point, the expression simplifies to \( 3 e^{-2(0)} \cos 2(\pi) \). This transformation is essential for calculating the final result without unnecessary complications.
Trigonometric Functions
Trigonometric functions, like \( \cos \theta \), are fundamental in calculus, especially when dealing with periodic or wave-like phenomena. In the exercise, we used the cosine function during the substitution method. Knowing the properties and values of trigonometric functions at key points is crucial.
  • The cosine function is periodic with a period of \( 2\pi \).
  • \( \cos 2\pi = 1 \) is a special value often encountered in problems involving limits and oscillatory behavior.
  • Understanding trigonometric identities helps simplify expressions.
In the solution provided, the calculation \( \cos 2(\pi) \) was simplified to \( 1 \), which contributed significantly to the overall simplification process of the limit evaluation. Mastery of these functions is essential for tackling advanced calculus problems that involve periodic functions.

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Most popular questions from this chapter

Two dependent variables Find \(\partial x / \partial u\) and \(\partial y / \partial u\) if the equations \(u=x^{2}-y^{2}\) and \(v=x^{2}-y\) define \(x\) and \(y\) as functions of the independent variables \(u\) and \(v,\) and the partial derivatives exist. (See the hint in Exercise \(69 . )\) Then let \(s=x^{2}+y^{2}\) and find \(\partial s / \partial u .\)

Use a CAS to perform the following steps implementing the method of Lagrange multipliers for finding constrained extrema: a. Form the function \(h=f-\lambda_{1} g_{1}-\lambda_{2} g_{2},\) where \(f\) is the function to optimize subject to the constraints \(g_{1}=0\) and \(g_{2}=0\) . b. Determine all the first partial derivatives of \(h\) , including the partials with respect to \(\lambda_{1}\) and \(\lambda_{2},\) and set them equal to \(0 .\) c. Solve the system of equations found in part (b) for all the unknowns, including \(\lambda_{1}\) and \(\lambda_{2}\) . d. Evaluate \(f\) at each of the solution points found in part (c) and select the extreme value subject to the constraints asked for in the exercise. Minimize \(f(x, y, z)=x y+y z\) subject to the constraints \(x^{2}+y^{2}-\) \(2=0\) and \(x^{2}+z^{2}-2=0.\)

If \(z=x+f(u),\) where \(u=x y,\) show that \begin{equation}x \frac{\partial z}{\partial x}-y \frac{\partial z}{\partial y}=x.\end{equation}

Use a CAS to perform the following steps implementing the method of Lagrange multipliers for finding constrained extrema: a. Form the function \(h=f-\lambda_{1} g_{1}-\lambda_{2} g_{2},\) where \(f\) is the function to optimize subject to the constraints \(g_{1}=0\) and \(g_{2}=0\) . b. Determine all the first partial derivatives of \(h\) , including the partials with respect to \(\lambda_{1}\) and \(\lambda_{2},\) and set them equal to \(0 .\) c. Solve the system of equations found in part (b) for all the unknowns, including \(\lambda_{1}\) and \(\lambda_{2}\) . d. Evaluate \(f\) at each of the solution points found in part (c) and select the extreme value subject to the constraints asked for in the exercise. Minimize \(f(x, y, z)=x^{2}+y^{2}+z^{2}\) subject to the constraints \(x^{2}-x y+y^{2}-z^{2}-1=0\) and \(x^{2}+y^{2}-1=0.\)

In Exercises \(53-60,\) sketch a typical level surface for the function. $$ f(x, y, z)=x^{2}+y^{2}+z^{2} $$

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