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.

1.

<Solution from Cal-Book>

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

<Detailed Solution>

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 $\inline \small y$; thus, there is no limit.

2. .

<Solution from Cal-Book>

As   and  .

The limit exists, and it is .

<Detailed Solution>

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

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

The limit does exist, and it is 0.

3.

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

<Cal-Book Solution>

Answer: The limit does not exist.

<Detailed Solution>

Let , so that  when .  Therefore the limit,

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

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

4.  .

<Cal-Book Solution>

After plotting the function, we know it converges. So the order of  and  does not matter.

<Detailed Solution>

5.

Let  and  along the line .

The limit does not exist.

6.

After plotting the function, we know it converges.

Let  and  along the line .

7.

So .

The limit exists, and it is .

8. What value of  for   will make the function continuous at .

=>

Since  , then  by Squeeze Theorem.

Hence, if we define  will be continuous at .

9-10. Let each of the following functions have the value 0 at the origin.

9.

The limit exists and it is .

So  is continuous at the origin.

10.

The limit does not exist.

11-12. Find  and the set on which  is continuous.

11.

Since  is continuous for all  and  is continuous for all

is continuous on a set  by using Theorem 5 in Section 13.2.

12.

Since  is continuous for all  and  is continuous for all

is continuous on a set  by using Theorem 5 in Section 13.2.