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Chapter 12: Q 41. (page 931)

In Exercises 41鈥�46, use polar coordinates to analyze the given limits.
$\lim_{(x, y) 鈫� (0, 0)} \frac{x^{2}}{x^{2} + y^{2}}$

Short Answer

Expert verified

The $\lim_{(x, y) 鈫� (0, 0)} \frac{x^{2}}{x^{2} + y^{2}}$ does not exist.

Step by step solution

01

Step 1. Given information.

We have given expression: $\lim_{(x, y) 鈫� (0, 0)} \frac{x^{2}}{x^{2} + y^{2}}$

02

Step 2. Use polar coordinates to analyze the given limits.

The relation between the rectangular coordinates $(x, y)$ and the polar coordinates $(r, 胃)$ is

$x = r \cos 胃 y = r \sin 胃$

On substituting values of $x$ and $y$ we get.

role="math" localid="1653208081348" $\lim_{(x, y) 鈫� (0, 0)} \frac{x^{2}}{x^{2} + y^{2}} = \lim_{r 鈫� 0} \frac{{(r \cos 胃)}^{2}}{{(r \cos 胃)}^{2} + {(r \sin 胃)}^{2}} = \lim_{r 鈫� 0} \frac{r^{2} \cos^{2} 胃}{r^{2} \cos^{2} 胃 + r^{2} \sin^{2} 胃} = \lim_{r 鈫� 0} \frac{r^{2} \cos^{2} 胃}{r^{2} (\cos^{2} 胃 + \sin^{2} 胃)} = \lim_{r 鈫� 0} \cos^{2} 胃 = \cos^{2} 胃$

Since the value of $\lim_{(x, y) 鈫� (0, 0)} \frac{x^{2}}{x^{2} + y^{2}}$ depends on $胃$ . Therefore the value of data-custom-editor="chemistry" $\lim_{(x, y) 鈫� (0, 0)} \frac{x^{2}}{x^{2} + y^{2}}$ is different for different value of $胃$ .

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Most popular questions from this chapter

In Exercises, find the maximum and minimum of the function f subject to the given constraint. In each case explain why the maximum and minimum must both exist.

$f (x, y, z) = x + y + z when x^{2} + 4 y^{2} + 16 z^{2} = 64$

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$Extrema : Find the local maxima, local minima, and saddle points of the given functions f (x, y) = x^{3} 鈭� 12 xy 鈭� y^{3} .$

$A chain rule for functions of two variables : Let f (x, y) = x^{2} y^{3}, x = s \cos t, and y = s \sin t . Find \frac{鈭� f}{鈭� s} and \frac{鈭� f}{鈭� t} using the preceding results .$

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