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Chapter 12: Problem 26

Evaluate the following limits. $$\lim _{(u, v) \rightarrow(8,8)} \frac{u^{1 / 3}-v^{1 / 3}}{u^{2 / 3}-v^{2 / 3}}$$

Short Answer

Expert verified
Answer: The limit of the function as (u, v) approaches (8, 8) is \(\frac{1}{4}\).

Step by step solution

01

Factor the function

First, let's rewrite the given function as: $$f(u, v) = \frac{u^{1 / 3}-v^{1 / 3}}{u^{2 / 3}-v^{2 / 3}}$$ In order to evaluate the limit, we need to find a common factor in both the numerator and the denominator. The common factor in this case is the difference of cubes. Remember the identity for the difference of two cubes: $$a^3-b^3=(a-b)(a^2+ab+b^2)$$ So let's rewrite the function with this factorization: $$f(u, v) = \frac{(u - v)\left[(u^{1 / 3})^2+u^{1 / 3}v^{1 / 3}+(v^{1 / 3})^2\right]}{(u - v)\left[(u^{2 / 3})^2+u^{2 / 3}v^{2 / 3}+(v^{2 / 3})^2\right]}$$
02

Simplify the function

Next, we can simplify the function by canceling out the common factor (u - v), which results in: $$f(u, v) = \frac{(u^{1 / 3})^2+u^{1 / 3}v^{1 / 3}+(v^{1 / 3})^2}{(u^{2 / 3})^2+u^{2 / 3}v^{2 / 3}+(v^{2 / 3})^2}$$
03

Evaluate the limit

Now, we can directly apply the limit as (u, v) approaches (8, 8): $$\lim _{(u, v) \rightarrow(8,8)} f(u, v) = \frac{(8^{1 / 3})^2+8^{1 / 3}\cdot8^{1 / 3}+(8^{1 / 3})^2}{(8^{2 / 3})^2+8^{2 / 3}\cdot8^{2 / 3}+(8^{2 / 3})^2}$$
04

Simplify and find the result

Finally, we can now simplify the expression and find the result: $$\lim _{(u, v) \rightarrow(8,8)} f(u, v) = \frac{(2)^2+2\cdot2+(2)^2}{(4)^2+4\cdot4+(4)^2} = \frac{12}{48} = \frac{1}{4}$$ So the limit of the given function as (u, v) approaches (8, 8) is \(\frac{1}{4}\).

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Most popular questions from this chapter

In its many guises, the least squares approximation arises in numerous areas of mathematics and statistics. Suppose you collect data for two variables (for example, height and shoe size) in the form of pairs \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right)\) The data may be plotted as a scatterplot in the \(x y\) -plane, as shown in the figure. The technique known as linear regression asks the question: What is the equation of the line that "best fits" the data? The least squares criterion for best fit requires that the sum of the squares of the vertical distances between the line and the data points is a minimum. Generalize the procedure in Exercise 70 by assuming that \(n\) data points \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right)\) are given. Write the function \(E(m, b)\) (summation notation allows for a more compact calculation). Show that the coefficients of the best-fit line are $$ \begin{aligned} m &=\frac{\left(\sum x_{k}\right)\left(\sum y_{k}\right)-n \sum x_{k} y_{k}}{\left(\sum x_{k}\right)^{2}-n \sum x_{k}^{2}} \text { and } \\ b &=\frac{1}{n}\left(\sum y_{k}-m \Sigma x_{k}\right) \end{aligned}, $$ where all sums run from \(k=1\) to \(k=n\).

Show that the plane \(a x+b y+c z=d\) and the line \(\mathbf{r}(t)=\mathbf{r}_{0}+\mathbf{v} t,\) not in the plane, have no points of intersection if and only if \(\mathbf{v} \cdot\langle a, b, c\rangle=0 .\) Give a geometric explanation of this result.

Find an equation for a family of planes that are orthogonal to the planes \(2 x+3 y=4\) and \(-x-y+2 z=8\)

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 12.5 and 12.6 are consistent with the results in parts \((a)-(d)\). $$f(x, y)=\sqrt{|x y|}$$

Use the gradient rules of Exercise 81 to find the gradient of the following functions. $$f(x, y, z)=(x+y+z) e^{x y z}$$
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