Calculus-Sec-13-2-Solution

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13.2

13.2 Limits and Continuity of Multivariable Functions

by SGLee - HSKim, VLang, KHKim

1-7. Find the limit, if it exists, or show that the limit does not exist.

Let and along the line . So the limit of does not exist. For example, for , and for , .

Let , so that when . Therefore the limit,

$\therefore\lim_{(x,y)\rightarrow\(0,0)} \frac{y^4}{x^4+3y^4}=\lim_{x\rightarrow0}\frac{m^4x^4}{x^4+3m^4x^4}=\frac{m^4}{1+3m^4}$ .

The limit does not depend on $x$ or $y$ ; thus, there is no limit.

2. .

http://matrix.skku.ac.kr/cal-lab/cal-12-1-1.html

As and , .

The limit exists, and it is .

Let $x=rcos\theta$ and $y=rsin\theta$ , so that $(x,y)\rightarrow (0,0)$ when $r\rightarrow 0$ . Therefore the limit,

$\therefore \lim_{(x,y)\rightarrow(0,0)}\frac{x^3}{2x^2+6y^4}=\lim_{r\rightarrow0}\frac{r^3cos^3\theta}{2r^2cos^2\theta+6r^4sin^4\theta}$ $=\lim_{r\rightarrow0}\frac{r^2}{r^2}\left ( \frac{rcos^3\theta}{2cos^2\theta+6r^2sin^2\theta} \right )\rightarrow 0$

The limit does exist, and it is 0.

http://matrix.skku.ac.kr/cal-lab/cal-12-1-Exs3.html

<Cal-Book Solution>

Answer: The limit does not exist.

Let , so that when . Therefore the limit,

$\therefore\lim_{(x,y)\rightarrow(0,0)}\frac{x^2-y^2}{x^2+y^2}=\lim_{x\rightarrow0}\frac{x^2-m^2x^2}{x^2+m^2x^2}=\lim_{x\rightarrow0}\frac{x^2}{x^2}\frac{1-m^2}{1+m^2}=\frac{1-m^2}{1+m^2}$