1. Consider each of the following functions: f(x) = |x – 1| / (x – 1) g(x) = (1 – x) / (1 –  x) The domain for each function includes {x | x  1, x  0}, and each function is continuous on this set. Can either function be defined at x = 1 so that the function will be continuous on the set of nonnegative real numbers? If x 1, then f(x) =.–11 Since lim f(x) does not exist, f(x) cannot be defined at x = 1 so that the function will be continuous on the set of nonnegative real numbers. x1x1 If x  1, then g(x) =1 +  x. Since lim g(x) =, g(1) can be defined to be, and the function will be continuous on the set of nonnegative real numbers. x1x1 22

2. Consider the following function: z = f(x,y) = cos[1/(x 2 + y 2 )] The domain for this function is {(x, y) | (x, y)  (0, 0)}, and this function is continuous on this domain. Can this function be defined at (0, 0) so that the function will be continuous on R 2 ? Find the following level curves c = cos[1/(x 2 + y 2 )] : c = 2 c = 1 c = 1/  2 c = 0 c = –1/  2 c = –1 c = –2 The level curve is empty. x 2 + y 2 = 1/(2n  ), n = 1, 2, 3, … The level curve is empty. x 2 + y 2 = 1/[(2n + 1)  ], n = 0, 1, 2, 3, … x 2 + y 2 = 1/[(2n + k/4)  ], n = 0, 1, 2, 3, …, k = 1, 7 x 2 + y 2 = 1/[(2n + k/4)  ], n = 0, 1, 2, 3, …, k = 3, 5 x 2 + y 2 = 1/(n + 1/2) , n = 0, 1, 2, 3, …

3. Consider the following function: z = f(x,y) = cos[1/(x 2 + y 2 )] The domain for this function is {(x, y) | (x, y)  (0, 0)}, and this function is continuous on this domain. Can this function be defined at (0, 0) so that the function will be continuous on R 2 ? Find the level curve when y = 0 (that is, in the xz plane). z x xz = cos(1/x 2 )  2/   1/   2/(3  )  1/(2  )  2/(5  )  1/(3  ) 0 0 1 –1–1 0 –1

4. 1 Since lim cos ——— (x,y)  (0,0) x 2 + y 2 does not exist, then f(x,y) cannot be defined at (x,y) = (0,0) so that the function will be continuous on all of R 2. Consider the following function: z = g(x,y) = (x 2 + y 2 ) cos[1/(x 2 + y 2 )] The domain for this function is {(x, y) | (x, y)  (0, 0)}, and this function is continuous on this domain. Can this function be defined at (0, 0) so that the function will be continuous on R 2 ? It is difficult to graph the level curves c = (x 2 + y 2 ) cos[1/(x 2 + y 2 )].

5. z x xz = x 2 cos(1/x 2 )  2/   1/   2/(3  )  1/(2  )  2/(5  )  1(3  ) –1/  0 0 1/(2  ) 0 –1/(3  ) Consider the following function: z = g(x,y) = (x 2 + y 2 ) cos[1/(x 2 + y 2 )] The domain for this function is {(x, y) | (x, y)  (0, 0)}, and this function is continuous on this domain. Can this function be defined at (0, 0) so that the function will be continuous on R 2 ? Find the level curve when y = 0 (that is, in the xz plane).

6. 1 Since lim(x 2 + y 2 ) cos ——— (x,y)  (0,0) = 0, then g(0,0) can be defined to be 0, and the function will be continuous on all of R 2. Our statements thatlimf(x,y) does not exist, and (x,y)  (0,0) thatlimg(x,y)= 0 appear to be true from the graphical analysis, (x,y)  (0,0) but in general, a mathematically rigorous proof that a limit does, or does not, exist is required. Section 2.2 of the textbook consists of a mathematically rigorous discussion of limits in R n and continuity of functions from R n to R m. This material is covered in Math 432 (Real Analysis). In this course, we only briefly consider two special techniques involving functions from R 2 to R 1 : one for demonstrating that a limit is zero (0), and one for demonstrating that a limit does not exist. x 2 + y 2

7. When evaluating limf(x,y) all paths (an infinite number) must be considered. (x,y)→(x 0,y 0 ) To show the limit does not exist, one only needs to show that the limit is not the same along two different paths. To show that L is the limit, one needs to show that |f(x,y) – L| approaches zero (0) as d = ||(x, y) – (x 0, y 0 )|| =  (x – x 0 ) 2 + (y – y 0 ) 2 approaches zero (0). Note: the sum and product rules for limits involving one variable also apply with limits involving two variables.

8. Example Evaluatelim (2x 2 y – x 3 ) / (x 2 + y 2 ) (x,y)  (0,0) We shall show that this limit is zero (0). Note that for all values of x and y, we have that x 2  x 2 + y 2 which implies |x|   x 2 + y 2 = d ; Similarly, |y|  d. ||(x, y) – (x 0, y 0 )|| = ||(x,y) – (0,0)|| =  (x – 0) 2 + (y – 0) 2 =  x 2 + y 2 = d lim|(2x 2 y – x 3 ) / (x 2 + y 2 )| (x,y)  (0,0) = lim |(2x 2 y – x 3 )| / |(x 2 + y 2 )| d0d0  lim (2|x| 2 |y| + |x| 3 ) / |x 2 + y 2 | d0d0  lim(2d 3 + d 3 ) / d 2 d0d0 = lim(3d) d0d0 = 0

9. Example Evaluatelim x /  x 2 + y 2 (x,y)  (0,0) We shall first consider this limit along the path where x = t > 0 and y = 0 with t approaching zero (0). limx /  x 2 + y 2 = (t,0)  (0,0) lim t /  t 2 + 0 2 = t0+t0+ lim 1 = 1 t0+t0+ We shall now consider the limit along the path where x = 0 and y = t > 0 with t approaching zero (0). limx /  x 2 + y 2 = (0,t)  (0,0) lim 0 /  0 + t 2 = t0+t0+ lim 0 = 0 t0+t0+ The limit does not exist, since a different limit can be obtained along different paths.

10. Example Evaluatelim 2xy / (x 2 + y 2 ) (x,y)  (0,0) We shall first consider this limit along the path where x = t > 0 and y = 0 with t approaching zero (0). We shall now consider the limit along the path where x = 0 and y = t > 0 with t approaching zero (0). Since both paths have the same limit, we must consider other paths in order to show that a different limit can be obtained along different paths. lim 2xy / (x 2 + y 2 ) = (t,0)  (0,0) lim 2t(0) / (t 2 + 0 2 ) = t0+t0+ lim 0 = 0 t0+t0+ lim 2xy / (x 2 + y 2 ) = (0,t)  (0,0) lim 2(0)t / (0 2 + t 2 ) = t0+t0+ lim 0 = 0 t0+t0+

11. We shall first consider this limit along the path where x = t > 0 and y = 0 with t approaching zero (0). We shall now consider the limit along the path where x = 0 and y = t > 0 with t approaching zero (0). lim 2xy / (x 2 + y 2 ) = (t,0)  (0,0) lim 2t(0) / (t 2 + 0 2 ) = t0+t0+ lim 0 = 0 t0+t0+ lim 2xy / (x 2 + y 2 ) = (0,t)  (0,0) lim 2(0)t / (0 2 + t 2 ) = t0+t0+ lim 0 = 0 t0+t0+ We shall now consider the limit along the path where x = y = t > 0 with t approaching zero (0). lim 2xy / (x 2 + y 2 ) = (t,t)  (0,0) lim 2t 2 / (t 2 + t 2 ) = t0+t0+ lim 1 = 1 t0+t0+ The limit does not exist, since a different limit can be obtained along different paths.

12. The discussion of continuity in Section 2.2 of the text is a straightforward extension of a discussion of continuity in one-variable calculus.