Chapter 2. Limits and Continuity

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Ashlie Montgomery
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1 Chapter 2 Limits and Continuity 4 11

2 2.1 Limits: One variable to two variables From one variable calculus you will be aware of the idea of a it of a function at a point. Essentially, if f : IR IR then we write f (x) = l x a if f (x) is close to l whenever x is close to a. (You might like to review chapter 5 of your first year calculus notes.) For example, looking at the table x f (x) = x2 4 x it does seem clear that x 2 f (x) = 4 even though the function cannot be calculated at the it point 2. We have a similar idea for functions of two variables. Z A function f : IR 2 IR has the it l at (a,b) if f (x,y) is close to l whenever (a,b) is close to (a,b). For this we write f (x,y) = l. (x,y) (a,b) 2.1 Example Let f (x,y) = x2 +x y 2 +y x+y. What is f (x,y)? (x,y) f (x,y) (1, 0) 2 (0.1, 0.2) 0.9 (0.2, 0.1) 1.1 (0.05, 0.01) (0.001, 0.001) 1 ( , 0) It seems clear that the it of our two variable function is 1 which we write as f (x,y) = 1. Notice that we had to be careful in our approach to the origin and avoid points of the form (x, x) in the left hand column as such points are not in the domain of the function. 12

3 However, we have to be very careful about its in IR 2. Consider the following: 2.2 Example We look for a it of the function f (x,y) = x y at the origin. (x,y) f (x,y) = x/y (0.01, 0.2) 0.05 (0.0001, 0.02) ( , 0.01) The right hand column approaches 0 as the left hand column approaches (0,0). One might conclude from this that the it is 0. But what about the following table for the same function: (x,y) f (x,y) = x/y (0.01, 0.01) 1 (0.0001, ) 1 ( , ) 1 Again the left hand column approaches the origin while the right hand column approaches 1. It is not hard to see why we have got two different its. In the second case, each point in the left hand column has the form (x,y) = (a,a). It is not surprising then that f (x,y) = x y = a = 1(for x 0) a while in the first case each point (x,y) is of the form (a 2,2a) and so f (x,y) = x y = a2 2a = a/2 which of course gets closer to 0 when (x,y) = (a 2,2a) gets close to (0,0). We say that the it of f along the approach (a 2,2a) (0,0) is 0, and write Similarly we write f (x,y) = 0. (x,y)=(a 2,2a) (x,y)=(a,a) f (x,y) = x=y f (x,y) = 1. Now we cannot allow a function to have two different its at a point. So if a function has different its along different approaches then we say the function does not have a it at the point. 1 The moral of the story is: 1 You can debate among yourselves if it is entirely logical to say a function has no it because it has more than one it! 13

4 Z A function f : IR 2 IR can have different its along different approach paths. If this happens, we say that the it of the function does not exist. This means we can easily recognise if a function does not have a it: if we find different its along two approach paths then the function does not have a it. 2.3 Note (a) For the it of f at a to exist, it is not necessary for f (a) to be defined and, if it is defined, it is not necessary that f (a) = l. (b) In the 1-variable case, where we can only approach a along the real line, left and right hand its must exist and be equal for the it to exist. In the several variable case, we can approach (a 1,a 2 ) along many different curves. For the it to exist, the it along every curve must exist and they must all be equal. This makes the analysis of its much more tricky. 2.4 Example Consider x 2. There are many ways of approaching the + y2 it point (0, 0). If for example we approach along the x-axis, i.e. we consider points of the form (x,y) = (a,0) approaching (0,0) then we have (x,y)=(a,0) (0,0) Similarly if we approach along the y-axis: (x,y)=(0,a) (0,0) x 2 + y 2 = x 0 x 2 + y 2 = a 0 0 a 2 = 0. 0 a 2 = 0. However if we approach along the diagonal line x = y, i.e. (x, y) = (a, a), then (x,y)=(a,a) (0,0) x 2 + y 2 = a 0 a 2 2a 2 = 1 2. Since we get different its along different curves, the it does not exist. 2.5 Example Consider (x 1,x 2 ) (0,0) x 2. There are many ways of approaching the + y2 it point (0, 0). If for example we approach along the x-axis, i.e. we consider points of the form (x 1,0) approaching (0,0) then we have (x 1,0) (0,0) x 2 + y 2 = x 0 0 x 2 1 = 0. Similarly if we approach along the y-axis: (0,x 2 ) (0,0) x 2 + y 2 = 0 x 2 0 x 2 =

5 However if we approach along the diagonal x = y then (x 1,x 1 ) (0,0) x 2 + y 2 = x 2 x 0 2x 2 = 1 2. Since we get different its along different curves, the it does not exist. 2x 2 y 2.6 Example x 4. Along x = 0 and along y = kx, (that is, any straight + y2 line path to the origin) the it is zero. However if we approach along the quadratic curve y = x 2 the it is 1. Therefore, the it does not exist. 15

6 2.2 Algebra of Limits This is similar to the 1-variable theory. Suppose then f (x 1,x 2 ) = l and g(x 1,x 2 ) = m (x 1,x 2 ) (a 1,a 2 ) (x 1,x 2 ) (a 1,a 2 ) (i) (ii) f (x 1,x 2 ) ± g(x 1,x 2 ) = l ± m, (x 1,x 2 ) (a 1,a 2 ) f (x 1,x 2 )g(x 1,x 2 ) = lm, (x 1,x 2 ) (a 1,a 2 ) (iii) If m 0 then f (x 1,x 2 ) (x 1,x 2 ) (a 1,a 2 ) g(x 1,x 2 ) = l m, (iv) If g : IR IR satisfies t l g(t) = w then (x,y) (a,b) g( f (x,y)) = w. To summarise: Z The it of a sum/product/quotient/composition is the sum/product/quotient/composition of the its (as long as we don t do something silly like divide by zero!) Note that if f 1 is a one variable function and x a f 1 (x) = l then we can consider f 1 as a two variable function via f (x,y) := f 1 (x). A useful fact is that In other words, f (x,y) = f 1(x) = f 1 (x) = l. (x,y) (a,b) (x,y) (a,b) x a Z If the function in the it depends on only one variable then we can treat the it as a one variable it: f (x) = f (x) (x,y) (a,b) x a f (y) = f (y) (x,y) (a,b) y b (= f (x)). x b Combined with the algebra of its, this allows us to compute many its. 16

7 2.7 Example 1. Find the two-variable it (x,y) (2,3) x 2 + y. (x,y) (2,3) x2 + y = (x,y) (2,3) x2 + y (x,y) (2,3) = x 2 x 2 + y 3 y = = 7. 6 sin(x 2 + y 7) 2. (x,y) (2,3) x 2 + y 7 This can be written as g( f (x,y)) (x,y) (2,3) where f (x,y) = x 2 +y 7 and g(t) = sin(t) t. Since we know from the first part that sin(x) (x,y) (2,3) f (x,y) = 0 and we know the one variable it x 0 x = 1, we can conclude from property (iv) of the algebra of its that this it is More difficult its This method of breaking a it into one variable its and using the algebra of its does not allow us to answer all its however because it breaks down in the 0 0 case. In this case one can hope to make progress by factoring. 2.8 Example Find x 2 +x y 2 +y x+y. (See Example 2.1.) The algebra of its gives 0 0 so we try to factorise. Hence x2 +x y 2 +y x+y x 2 + x y 2 + y = (x + y)(x y) + x + y = (x + y)(x y + 1). = (x+y)(x y+1) x+y which is 1 by the algebra of its. = x y+1 (whenever x+y 0) and the it becomes x y + 1 However, it has to be said that finding factors of expressions with two variables is not easy. 2 b 2.9 Exercise 1. Can you finish the following factorisations? (a) x 3 + 3x 2 y = x(...), (b) x y 2 1 = (x + y + 1)(...) (c) x x + y + y 2 = (x + y)(...) 2 By common consent among mathematicians, even factoring numbers is hard! 17

8 (d) x 2 4y 2 = (x + 2y)(...) (e) 3x 2 5 2y 2 + x 2y = (x 2y)(...) (f) x 3 y 3 = (x y)(...) 2. Using the algebra of its find the following its. State clearly which of the algebra of it rules you are using. (a) (x,y) (2,3) (x + y) 2. (b) (x,y) (0,1) x y Using the algebra of its and the factorisations found above, can you find the following its. (a) (x,y) (0,2) x 3 +3x 2 y 2x (b) (x,y) ( 1,0) x 2 +2+y 2 1 x+y+1, (c) (x,y) (2,1) 3x 2 5 2y 2 +x 2y x 2 4y 2. 18

9 2.3 Continuity 2.10 Definition Let f : D IR n IR and a D. We say that f IS CONTINUOUS AT a if f (x) = f (a). x a Particular implications of continuity at a are: (a) x a f (x) exists, (b) f is defined at a, (c) The values that exist in (a) and (b) are equal. We say f is continuous on D if it is continuous at every point of D. Sums and products (and quotients if one avoids division by zero) of continuous functions are continuous, and any continuous 1-variable function is continuous when considered as a function of two variables. Thus, for example, is continuous on its domain. f (x,y) = x3 y + sin(y 2 ) x

3 Algebraic Methods. we can differentiate both sides implicitly to obtain a differential equation involving x and y:

3 Algebraic Methods. we can differentiate both sides implicitly to obtain a differential equation involving x and y: 3 Algebraic Methods b The first appearance of the equation E Mc 2 in Einstein s handwritten notes. So far, the only general class of differential equations that we know how to solve are directly integrable