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Chapter 12: Q 43. (page 931)

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

Short Answer

Expert verified

The value of $\lim_{(x, y) 鈫� (0, 0)} \frac{x^{2} y^{2}}{x^{2} + y^{2}}$ is $0$ .

Step by step solution

01

Step 1. Given information.

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

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.

$\lim_{(x, y) 鈫� (0, 0)} \frac{x^{2} y^{2}}{x^{2} + y^{2}} = \lim_{r 鈫� 胃} \frac{r^{2} \cos^{2} 胃 (r^{2} \sin^{2} 胃)}{r^{2} \cos^{2} 胃 + r^{2} \sin^{2} 胃} = \lim_{r 鈫� 胃} \frac{r^{4} \cos^{2} 胃 \sin^{2} 胃}{r^{2} (\cos^{2} 胃 + \sin^{2} 胃)} = \lim_{r 鈫� 胃} r^{2} \cos^{2} 胃 \sin^{2} 胃 = 0$

Since the value of $\lim_{(x, y) 鈫� (0, 0)} \frac{x^{2} y^{2}}{x^{2} + y^{2}}$ is $0$ .

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