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Chapter 12: Problem 79

Evaluate the following limits. $$\lim _{(x, y) \rightarrow(0,2)}(2 x y)^{x y}$$

Short Answer

Expert verified
The limit of the given function as (x, y) approaches (0, 2) is equal to 1.

Step by step solution

01

Rewrite the limit using logarithms

We have the limit: $$\lim _{(x, y) \rightarrow(0,2)}(2 x y)^{x y}$$ We can rewrite this using the natural logarithm (ln): $$\ln \left( \lim_{(x, y) \rightarrow (0, 2)} (2xy)^{xy} \right) = \lim_{(x, y) \rightarrow (0, 2)} xy\ln(2xy)$$ So now we need to find the limit: $$\lim_{(x, y) \rightarrow(0, 2)}xy\ln(2xy)$$
02

Apply L'Hopital's Rule

Since the function is in an indeterminate form, we can apply L'Hopital's Rule by differentiating the function with respect to both x and y and then finding the limit. First, let's rewrite the limit as a fraction: $$\lim_{(x, y) \rightarrow(0, 2)}\frac{\ln(2xy)}{\frac{1}{xy}}$$ Now, we differentiate the numerator and the denominator with respect to both x and y: Numerator: $$\frac{\partial}{\partial x}\left( \ln(2xy) \right) = \frac{2y}{2xy} = \frac{1}{x}$$ $$\frac{\partial}{\partial y}\left( \ln(2xy) \right) = \frac{2x}{2xy} = \frac{1}{y}$$ Denominator: $$\frac{\partial}{\partial x}\left( \frac{1}{xy} \right) = -\frac{1}{x^2y}$$ $$\frac{\partial}{\partial y}\left( \frac{1}{xy} \right) = -\frac{1}{xy^2}$$ Our function becomes: $$\lim_{(x, y) \rightarrow(0, 2)}\frac{\frac{1}{x}+\frac{1}{y}}{-\frac{1}{x^2y}-\frac{1}{xy^2}}$$
03

Evaluate the limit

Now, substitute the values of the limit (0, 2) into the function: $$\lim_{(x, y) \rightarrow(0, 2)}\frac{\frac{1}{0}+\frac{1}{2}}{-\frac{1}{0^2\cdot2}-\frac{1}{0\cdot2^2}}$$ The function is not defined when x = 0, but we can analyze what happens when x approaches 0: $$\lim_{x \rightarrow 0}\frac{\frac{1}{x}+\frac{1}{2}}{-\frac{1}{x^2\cdot2}-\frac{1}{x\cdot2^2}}$$ Applying L'Hopital's Rule again, we have: Numerator: $$\frac{\partial}{\partial x}\left(\frac{1}{x}+\frac{1}{2}\right) = -\frac{1}{x^2}$$ Denominator: $$\frac{\partial}{\partial x}\left(-\frac{1}{x^2\cdot2}-\frac{1}{x\cdot2^2}\right) = \frac{3}{x^3\cdot2^2}$$ So our limit becomes: $$\lim_{x \rightarrow 0}\frac{-\frac{1}{x^2}}{\frac{3}{x^3\cdot2^2}}$$ Now, simplifying and evaluating the limit: $$\lim_{x \rightarrow 0}\frac{-4x^3}{3x^2} = \lim_{x \rightarrow 0}\frac{-4x}{3} = \frac{-4 \cdot 0}{3} = 0$$ Therefore, the limit is: $$\ln\left(\lim_{(x, y) \rightarrow (0, 2)} (2xy)^{xy}\right) = 0$$ To find the original limit, we take the exponent of both sides: $$\lim_{(x, y) \rightarrow (0, 2)} (2xy)^{xy} = e^0 = 1$$

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Key Concepts

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

Indeterminate Forms
In calculus, limits can sometimes yield expressions called indeterminate forms. These are expressions where substitution of limit values into the formula results in forms like \(\frac{0}{0}\), \(\infty - \infty\), \(0 \times \infty\), among others. These forms do not directly give us information about the behavior of the function as it approaches a certain point. An important thing to note is that simple substitution won't work, so we need special techniques to deal with them. For instance, the function \((2xy)^{xy}\) as \((x, y)\rightarrow(0,2)\) initially appears as \(0^0\), which is another indeterminate form. To evaluate the limit, we often employ algebraic manipulation, L'Hopital's rule, or series expansion as strategies to resolve these forms.
L'Hopital's Rule
L'Hopital's Rule is a tool in calculus used to find limits of indeterminate forms such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). Jean-Baptiste L'Hopital, a French mathematician, proposed that if you have a limit in one of these forms, you can differentiate the numerator and the denominator separately, then take the limit again. However, it is crucial to ensure that the original limit satisfies the conditions for applying this rule, which involves ensuring the functions are differentiable near the limit point. In our problem, after transforming the expression with \(\ln\), we ended up with a fraction {{\textstyle \frac{\ln(2xy)}{\frac{1}{xy}}}}. This is transformed into another fraction expression which is resolved using L'Hopital's once more. Each step involved differentiating correctly to justify using L'Hopital's Rule effectively.
Multivariable Limits
When dealing with limits in functions of more than one variable, such as in this exercise, understanding the approach in each variable dimension is vital. Multivariable limits can be complex because you often need to consider different paths by which the variables approach the given limit point. For our particular problem, \((x, y) \rightarrow (0, 2)\), the strategy used differentials for both variables. Paths, such as direct substitution and approaching one variable first to then check continuity or use symmetry arguments, are other methods that might come into play. It's more challenging than single-variable limits because the function's value might depend on the path taken to the point. Therefore, careful handling and manipulation of partial derivatives are crucial.
Logarithmic Functions
Logarithmic functions, specifically the natural logarithm \(\ln\), play a pivotal role in many calculus problems to simplify complex exponential expressions into manageable forms. In the given exercise, we transformed \((2xy)^{xy}\) using the property that \((a)^b = e^{b\ln(a)}\). By taking the natural logarithm, the exponent \(xy\) becomes a multiplier, specifically \(xy\ln(2xy)\). This transformation allowed us to convert the original problem into a form that was suitable for L'Hopital's Rule. The logarithmic function is particularly helpful in resolving indeterminate forms because it can transform products into sums, which simplifies differentiation and limit calculation. Mastery of logarithmic properties is indispensable for succeeding in higher-level calculus.

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Most popular questions from this chapter

A classical equation of mathematics is Laplace's equation, which arises in both theory and applications. It governs ideal fluid flow, electrostatic potentials, and the steadystate distribution of heat in a conducting medium. In two dimensions, Laplace's equation is $$\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0.$$ Show that the following functions are harmonic; that is, they satisfy Laplace's equation. $$u(x, y)=x\left(x^{2}-3 y^{2}\right)$$

A classical equation of mathematics is Laplace's equation, which arises in both theory and applications. It governs ideal fluid flow, electrostatic potentials, and the steadystate distribution of heat in a conducting medium. In two dimensions, Laplace's equation is $$\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0.$$ Show that the following functions are harmonic; that is, they satisfy Laplace's equation. $$u(x, y)=\tan ^{-1}\left(\frac{y}{x-1}\right)-\tan ^{-1}\left(\frac{y}{x+1}\right)$$

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Find the points (if they exist) at which the following planes and curves intersect. $$y=2 x+1 ; \quad \mathbf{r}(t)=\langle 10 \cos t, 2 \sin t, 1\rangle, \text { for } 0 \leq t \leq 2 \pi$$

Given three distinct noncollinear points \(A, B,\) and \(C\) in the plane, find the point \(P\) in the plane such that the sum of the distances \(|A P|+|B P|+|C P|\) is a minimum. Here is how to proceed with three points, assuming that the triangle formed by the three points has no angle greater than \(2 \pi / 3\left(120^{\circ}\right)\). a. Assume the coordinates of the three given points are \(A\left(x_{1}, y_{1}\right)\) \(\underline{B}\left(x_{2}, y_{2}\right),\) and \(C\left(x_{3}, y_{3}\right) .\) Let \(d_{1}(x, y)\) be the distance between \(A\left(x_{1}, y_{1}\right)\) and a variable point \(P(x, y) .\) Compute the gradient of \(d_{1}\) and show that it is a unit vector pointing along the line between the two points. b. Define \(d_{2}\) and \(d_{3}\) in a similar way and show that \(\nabla d_{2}\) and \(\nabla d_{3}\) are also unit vectors in the direction of the line between the two points. c. The goal is to minimize \(f(x, y)=d_{1}+d_{2}+d_{3} .\) Show that the condition \(f_{x}=f_{y}=0\) implies that \(\nabla d_{1}+\nabla d_{2}+\nabla d_{3}=0\) d. Explain why part (c) implies that the optimal point \(P\) has the property that the three line segments \(A P, B P,\) and \(C P\) all intersect symmetrically in angles of \(2 \pi / 3\) e. What is the optimal solution if one of the angles in the triangle is greater than \(2 \pi / 3\) (just draw a picture)? f. Estimate the Steiner point for the three points (0,0),(0,1) and (2,0).
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