Limits and Continuity: section 12.2

Transcription

1 Limits and Continuity: section 1. Definition. Let f(x,y) be a function with domain D, and let (a,b) be a point in the plane. We write f (x,y) = L if for each ε > 0 there exists some δ > 0 such that if (x,y) D and 0 < (x,y) (a,b) < δ, then f(x,y) L < ε. Note. To show f(x,y) = L we might equivalently show that f(x,y) L = 0 Algebraic Properties of the Limit Operator. [ ] (i) cf(x,y) = c (ii) (iii) [ f(x,y) ± g(x,y) f(x,y) [ ] f(x, y)g(x, y) = c being a constant ] = f(x,y) ± g(x,y) f(x,y) g(x,y) (iv) f(x,y) g(x,y) = f(x,y) g(x,y) Definition. If (a,b) is in the domain of f, then we say that f is continuous at (a,b) if the following holds: f (x,y) = f (a,b) Note. Sum, difference, and product of continuous functions is continuous. The quotient f(x,y) g(x,y) of two continuous functions at (a,b) is continuous at (a,b) provided that g(a,b) 0. Composite of continuous functions is continuous. Example. The function f(x,y) = sin(x y) x+y x + y > 0 : has the domain consisting of the points (x,y) satisfying { } D f = (x,y) x + y > 0 1

2 This function is continuous everywhere on its domain because it is the quotient of two continuous functions wit the denominator being nonzero on D. Therefore, one can write sin(x y) x 3 x+y = y 1 continuity sin(9 1) 3+1 = sin(8) Note. The functions f(x,y) = x y y and g(x,y) = x + y are examples of polynomials. The polynomial f is of degree 3, and g is of degree 1 (the largest degree present). In general, a polynomial is a combination of expressions x m y n where m 0 and n 0 are non-negative integers. Polynomials have the plane as their domain. Polynomials are continuous everywhere. Example. Find the it ( ) sin 1+x, or show that it does not exist. (x,y) (1,0) x +xy+1 Solution. ( ) sin 1+x (x,y) (1,0) x +xy+1 ( = sin = sin ( ) = sin(1) 1+x (x,y) (1,0) x +xy+1 ) the sine function is continuous so the it operator can go inside the sine function

3 Example. Find the it x+1 y+1, or show that it does not exist. (x,y) (3, 3) Solution. This is of the form 0 0. We rationalize it to find the answer: x+1 y+1 = x+1 y+1 (x,y) (3, 3) (x,y) (3, 3) x+1+ y+1 x+1+ y+1 (x+1) (y+1) (x,y) (3, 3) ()( x+1+ y+1) = (x,y) (3, 3) ()( x+1+ y+1) = = (x,y) (3, 3) 1 x+1+ y+1 drop x y = = 1 4 Example (section 1. exercise 35). Calculate the it Solution. This is of the form 0 0. By putting u = x + y, we have sin(x +y ) if it exists. x +y u = 0. so then sin(x + y ) sinu x + y = = 1 this it is proved in elementary Calculus u 0 u Example (section 1. exercise 6). Find all points of discontinuity of the function f(x,y) = Solution. We have f(x,y) = x + y xy(x + y). x+y x y+xy Note that the points (x,y) which satisfy either of xy = 0 or x + y = 0 are not in the domain of f so we cannot consider them as some points of discontinuity. So, the function is indeed continuous at any point of its domain. Note that the solution given in the Solution Manual is wrong. 3

4 [ ] Example (section 1. exercise 7). Evaluate the it (x,y) (a, a) cos(x + y) 1 sin (x + y) where 0 a π. Solution. [ ] (x,y) (a, a) cos(x + y) 1 sin (x + y) = cos(a + a) 1 sin (a + a) continuity of the functions sine and cosine = cos(a) 1 sin (a) = cos(a) cos (a) = cos(a) cos(a) = cos(a) cos(a) = 0 if 0 a π cos(a) + cos(a) = cos(a) if π a π Example (section 1. exercise 11). Calculate the it Solution. This is not of the form 0 0. x y (x,y) (, 1) x y = (x,y) (, 1) (x y)(x + y) x y Example (section 1. exercise 13). Evaluate the it x y if it exists. (x,y) (, 1) = (x + y) = 3 (x,y) (,1) x y if it exists. Solution. This it is of the indeterminate form 0 0 x y = ()(x+y) = (x + y) = 0 continuity of polynomials 4

5 Note. we learned in elementary Calculus that if a it exists then the left-hand it and right-hand it exist and are equal. A similar result holds for multivariable functions: If a it f(x,y), then the it on any path ending at (a,b) must exist and these its all must be equal. Example (section 1. exercise 9). Evaluate the it Solution. We take the following two paths: x 6 y 3x 6 + y = y x 0 y = 1 on the path x=0 x 6 y 3x 6 + y = x 6 x 0 3x 6 = 1 3 on the path y=0 Since we get different values on different paths, the it Note. The next three examples use the concept of equivalence. Example. Evaluate the it x 6 y 3x 6 +y if it exists. x 6 y 3x 6 does not exist. + y x + sin y x + y or show that it does not exist. Solution. As we know form elementary Calculus, we have y 0 siny y = 1, therefore the expressions siny and y are equivalent as y 0, in the sense that as long as multiplication or division is concerned, x + sin y we can use y for siny. So now, instead of calculating the it x + y we may calculate x + y this it: x + y y=x x + y x + y = x 0 x + x x + x = x x 0 3x = 3 y=0 x + y x + y = x 0 x + 0 x + 0 = x 0 x = 1 x Conclusion: the it does not exists as we get different values on different paths. 5

6 Example. Evaluate the it xycosy 3x or show that it does not exist. + y Solution. We know that y 0 cosy = 1, therefore we may substitute 1 for cosy in evaluating the it (cosy and 1 are equivalent), and we may study the following it instead: xy 3x + y y=x xy 3x + y = x 0 x x 3x + x = x 0 4x = 1 4 y=0 xy 3x + y = x 0 0 3x + 0 = 0 Conclusion: the it does not exists as we get different values on different paths. 6

7 Here are two examples where changing to polar coordinates will help: Example. Show that Solution. x y x + y = 0 Method 1 (by changing to polar coordinates). We have: x = rcosθ y = rsinθ x + y = r and note that r is the distance of the point (x,y) from the origin, therefore when (x,y) 0 we have r 0 and vice versa. So: x y x + y = (r cosθ) (r sinθ) r 0 r Method. Since 0 x x + y, we have 0 bounded. Then: 0 x y x + y 0 = x y x + y = x y Then by the Sandwich Theorem, we have x y x + y = 0 Example. Evaluate the it Solution. ( x x + y = = r cosθ sinθ r 0 }{{} = 0 bounded x x 1, in other words, the term x + y x + y is ) x + y }{{ } is bounded y }{{} tends to zero = 0 x y x + y 0 = 0. Equivalently we have: x sin y x or show that it does not exist. + y 7

8 We substitute y for siny as y 0, and therefore we study the following it instead x y x + y Method 1 (by changing to polar coordinates). x y x + y = r 0 (r cos θ)(r sin θ) (r cos θ) + (r sin θ) = r 0 r ( = r cos ) θ r 0 cos θ + sin sin θ = see the next line θ cos θ cos θ+sin θ cos θ sin θ cos θ + sin θ But note that the quotient is bounded as the numerator is less than the denominator. On the ( ) other hand, the term sin θ is bounded, hence the product sin θ is bounded. So, we continue: = r r 0 }{{} tends to zero cos θ cos θ + sin θ sin θ }{{} is bounded Method. Since 0 x x + y, we have 0 bounded. Then: 0 x y x + y 0 = x y x + y = x y Then by the Sandwich Theorem, we have x y x + y = 0 ( x x + y = cos θ cos θ+sin θ x x + y 1, in other words, the term x x + y is ) x + y }{{ } is bounded y }{{} tends to zero = 0 x y x + y 0 = 0. Equivalently we have: 8