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Consider each of the following functions: f ( x ) = | x – 1| / ( x – 1)

Consider each of the following functions: f ( x ) = | x – 1| / ( x – 1) g ( x ) = (1 – x ) / (1 –  x ).

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Consider each of the following functions: f ( x ) = | x – 1| / ( x – 1)

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  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) = , and if x > 1, then f(x) = . –1 1 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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. 2 2 x1

  2. Consider the following function: z = f(x,y) = cos[1/(x2 + y2)] 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 R2? Find the following level curves c = cos[1/(x2 + y2)] : c = 2 c = 1 c = 1/2 c = 0 c = –1/2 c = –1 c = –2 The level curve is empty. x2 + y2 = 1/(2n) , n = 1, 2, 3, … x2 + y2 = 1/[(2n + k/4)] , n = 0, 1, 2, 3, … , k = 1, 7 x2 + y2 = 1/(n + 1/2) , n = 0, 1, 2, 3, … x2 + y2 = 1/[(2n + k/4)] , n = 0, 1, 2, 3, … , k = 3, 5 x2 + y2 = 1/[(2n + 1)] , n = 0, 1, 2, 3, … The level curve is empty.

  3. Consider the following function: z = f(x,y) = cos[1/(x2 + y2)] 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 R2? Find the level curve when y = 0 (that is, in the xz plane). z xz = cos(1/x2) 2/ 1/ 2/(3) 1/(2) 2/(5) 1/(3) 0 –1 0 x 1 0 –1

  4. 1 Since lim cos ——— does not exist, then x2 + y2 (x,y)(0,0) f(x,y) cannot be defined at (x,y) = (0,0) so that the function will be continuous on all of R2. Consider the following function: z = g(x,y) = (x2 + y2) cos[1/(x2 + y2)] 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 R2? It is difficult to graph the level curves c = (x2 + y2) cos[1/(x2 + y2)] .

  5. Consider the following function: z = g(x,y) = (x2 + y2) cos[1/(x2 + y2)] 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 R2? Find the level curve when y = 0 (that is, in the xz plane). z xz = x2 cos(1/x2) 2/ 1/ 2/(3) 1/(2) 2/(5) 1(3) 0 –1/ 0 x 1/(2) 0 –1/(3)

  6. 1 Since lim (x2 + y2) cos ——— = 0, then g(0,0) can be defined x2 + y2 (x,y)(0,0) to be 0, and the function will be continuous on all of R2. Our statements that lim f(x,y) does not exist, and (x,y)(0,0) that lim g(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 Rn and continuity of functions from Rn to Rm. This material is covered in Math 432 (Real Analysis). In this course, we only briefly consider two special techniques involving functions from R2 to R1: one for demonstrating that a limit is zero (0), and one for demonstrating that a limit does not exist.

  7. When evaluating lim f(x,y) all paths (an infinite number) must be considered. (x,y)→(x0,y0) 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) – (x0 , y0)|| = (x– x0)2 + (y– y0)2 approaches zero (0). Note: the sum and product rules for limits involving one variable also apply with limits involving two variables.

  8. Example Evaluate lim (2x2y – x3) / (x2 + y2) (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 x2 x2 + y2 which implies |x|   x2 + y2 = d ; Similarly, |y|  d. ||(x , y) – (x0 , y0)|| = ||(x,y) – (0,0)|| = (x – 0)2 + (y – 0)2 =  x2 + y2 = d lim |(2x2y – x3) / (x2 + y2)| = lim |(2x2y – x3)| / |(x2 + y2)| d0 (x,y)(0,0)  lim (2|x|2|y| + |x|3) / |x2 + y2|  lim (2d 3 +d 3) / d 2 d0 d0 = lim (3d) = 0 d0

  9. Example Evaluate lim x /  x2 + y2 (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). lim x /  x2 + y2 = lim t /  t2 + 02 = lim 1 = 1 (t,0)(0,0) 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). lim x /  x2 + y2 = lim 0 /  0 + t2 = lim 0 = 0 (0,t)(0,0) t0+ t0+ The limit does not exist, since a different limit can be obtained along different paths.

  10. Example Evaluate lim 2xy / (x2 + y2) (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). lim 2xy / (x2 + y2) = lim 2t(0) / (t2 + 02) = lim 0 = 0 (t,0)(0,0) 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). lim 2xy / (x2 + y2) = lim 2(0)t / (02 + t2) = lim 0 = 0 (0,t)(0,0) t0+ t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.

  11. We shall first consider this limit along the path where x = t > 0 and y = 0 with t approaching zero (0). lim 2xy / (x2 + y2) = lim 2t(0) / (t2 + 02) = lim 0 = 0 (t,0)(0,0) 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). lim 2xy / (x2 + y2) = lim 2(0)t / (02 + t2) = lim 0 = 0 (0,t)(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 / (x2 + y2) = lim 2t2 / (t2 + t2) = lim 1 = 1 (t,t)(0,0) 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.

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