Limits and Continuity


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1 Math 20C Multivariable Calculus Lecture Limits and Continuity Slide Review of Limit. Side limits and squeeze theorem. Continuous functions of 2,3 variables. Review: Limits Slide 2 Definition Given a function f(x, y) : D IR 2 IR and a point (x 0, y 0 ) IR 2, we write lim f(x, y) = L, (x,y) (x 0,y 0) if and only if for all (x, y) D close enough in distance to (x 0, y 0 ) the values of f(x, y) approaches L.
2 Math 20C Multivariable Calculus Lecture 2 A tool to show that a limit does not exist is the following: Theorem (Side limits) If f(x, y) L along a path C as (x, y) (x 0, y 0 ), and f(x, y) L 2 along a path C 2 as (x, y) (x 0, y 0 ), with L L 2, then lim f(x, y) (x,y) (x 0,y 0) does not exist. Slide 3 A tool to prove that a limit exists is the following: Theorem 2 (Squeeze) Assume f(x, y) g(x, y) h(x, y) for all (x, y) near (x 0, y 0 ); Assume Then lim f(x, y) = L = lim h(x, y), (x,y) (x 0,y 0) (x,y) (x 0,y 0) lim g(x, y) = L. (x,y) (x 0,y 0) Example: How to use the side limit theorem. Does the following limit exist? lim (x,y) (0,0) 3x 2 x 2 + 2y 2. () So, the function is f(x, y) = (3x 2 )/(x 2 + 2y 2 ). Let pick the curve C as the xaxis, that is, y = 0. Then, f(x, 0) = 3x2 x 2 = 3, then lim f(x, 0) = 3. (x,0) (0,0) Let us now pick up the curve C 2 as the yaxis, that is, x = 0. Then, then Therefore, the limit in () does not exist. f(0, y) = 0, lim f(x, 0) = 0. (x,0) (0,0) Notice that in the above example one could compute the limit for arbitrary lines, that is, C m given by y = mx, with m a constant. Then f(x, mx) = 3x 2 x 2 + 2m 2 x 2 = 3 + 2m 2,
3 Math 20C Multivariable Calculus Lecture 3 so one has that lim f(x, mx) = 3 (x,mx) (0,0) + 2m 2 is different for each value of m. Example: How to use the squeeze theorem. Does the following limit exist? lim (x,y) (0,0) x 2 y x 2 + y 2. (2) Let us first try the side limit theorem, to try to prove that the limit does not exist. Consider the curves C m given by y = mx, with m a constant. Then so one has that f(x, mx) = x2 mx x 2 + m 2 x 2 = mx + m 2, lim f(x, mx) = 0, (x,mx) (0,0) m IR. Therefore, one cannot conclude that the limit does not exist. However, this argument does not prove that the limit actually exists.this can be done with the squeeze theorem. First notice that x 2 x 2 + y 2, (x, y) IR2, (x, y) (0, 0). (proof: 0 y 2, then x 2 (x 2 + y 2 ).) Therefore, one has the inequality y x2 y x 2 + y 2 y, (x, y) IR2, (x, y) (0, 0). Then, one knows that lim y 0 y = 0, therefore the squeeze theorem says that lim (x,y) (0,0) x 2 y x 2 + y 2 = 0.
4 Math 20C Multivariable Calculus Lecture 2 4 Limits and Continuity Limit laws for functions of a single variable also holds for functions of two variables. Slide 4 If the limits lim x x0 f and lim x x0 g exist, then ( ) ( ± lim (f ± g) = x x 0 lim f x x 0 lim g x x 0 ( ) ( ) lim (fg) = lim f lim g. x x 0 x x 0 x x 0 ), Definition 2 (Continuity) A function f(x, y) is continuous at (x 0, y 0 ) if lim f(x, y) = f(x 0, y 0 ). (x,y) (x 0,y 0) Continuity Examples of continuous functions: Polynomial functions are continuous in IR 2, for example P 2 (x, y) = a 0 + b x + b 2 y + c x 2 + c 2 xy + c 3 y 2. Slide 5 Rational functions are continuous on their domain, for example, f(x, y) = P n(x, y) Q m (x, y), f(x, y) = x2 + 3y x 2 y 2 + y 4 x 2 y 2, x ±y. Composition of continuous functions are continuous, example f(x, y) = cos(x 2 + y 2 ).
5 Math 20C Multivariable Calculus Lecture 2 5 Partial derivatives Slide 6 Definition of Partial derivatives. Higher derivatives. Examples of differential equations. Partial derivatives Definition 3 (Partial derivative) Consider a function f : D IR 2 R IR. Slide 7 The partial derivative of f(x, y) with respect to x at (a, b) D is denoted as f x (a, b) and is given by f x (a, b) = lim [f(a + h, b) f(a, b)]. h 0 h The partial derivative of f(x, y) with respect to y at (a, b) D is denoted as f y (a, b) and is given by f y (a, b) = lim [f(a, b + h) f(a, b)]. h 0 h
6 Math 20C Multivariable Calculus Lecture 2 6 So, to compute the partial derivative of f(x, y) with respect to x at (a, b), one can do the following: First, evaluate the function at y = b, that is compute f(x, b); second, compute the usual derivative of single variable functions; evaluate the result at x = a, and the result is f x (a, b). Example: Find the partial derivative of f(x, y) = x 2 + y 2 /4 with respect to x at (, 3).. f(x, 3) = x 2 + 9/4; 2. f x (x, 3) = 2x; 3. f x (, 3) = 2. To compute the partial derivative of f(x, y) with respect to y at (a, b), one follows the same idea: First, evaluate the function at x = a, that is compute f(a, y); second, compute the usual derivative of single variable functions; evaluate the result at y = b,, and the result is f y (a, b). Example: Find the partial derivative of f(x, y) = x 2 + y 2 /4 with respect to y at (, 3).. f(, y) = + y 2 /4; 2. f y (, y) = y/2; 3. f y (, 3) = 3/2. Discuss the geometrical meaning of the partial derivative using the graph of the function. Partial derivatives The partial derivatives of f computed at any point (x, y) D define the functions f x (x, y) and f y (x, y), the partial derivatives of f. More precisely: Slide 8 Definition 4 Consider a function f : D IR 2 R IR. The functions partial derivatives of f(x, y) are denoted by f x (x, y) and f y (x, y), and are given by the expressions f x (x, y) = lim [f(x + h, y) f(x, y)], h 0 h f y (x, y) = lim [f(x, y + h) f(x, y)]. h 0 h
7 Math 20C Multivariable Calculus Lecture 2 7 Examples: f(x, y) = ax 2 + by 2 + xy. f x (x, y) = 2ax y, = 2ax + y. f y (x, y) = 0 + 2by + x, = 2by + x. f(x, y) = x 2 ln(y), f x (x, y) = 2x ln(y), f y (x, y) = x2 y. f(x, y) = x 2 + y2 4, f x (x, y) = 2x, f y (x, y) = y 2. f(x, y) = 2x y x + 2y, f x (x, y) = = = f y (x, y) = = 2(x + 2y) (2x y) (x + 2y) 2, 2x + 4y 2x + y (x + 2y) 2, 5y (x + 2y) 2. (x + 2y) (2x y)2 (x + 2y) 2, 5x (x + 2y) 2. f(x, y) = x 3 e 2y + 3y, f x (x, y) = 3x 2 e 2y, f y (x, y) = 2x 3 e 2y + 3, f yy (x, y) = 4x 3 e 2y, f yyy (x, y) = 8x 3 e 2y, f xy = 6x 2 e 2y, f yx = 6x 2 e 2y.
8 Math 20C Multivariable Calculus Lecture 2 8 Higher derivatives Slide 9 Higher derivatives of a function f(x, y) are partial derivatives of its partial derivatives. For example, the second partial derivatives of f(x, y) are the following: f xx (x, y) = lim h 0 h [f x(x + h, y) f x (x, y)], f yy (x, y) = lim h 0 h [f y(x, y + h) f y (x, y)], f xy (x, y) = lim h 0 h [f x(x + h, y) f x (x, y)], f yx (x, y) = lim h 0 h [f y(x, y + h) f y (x, y)].
9 Math 20C Multivariable Calculus Lecture 2 9 Higher derivatives Slide 0 Theorem 3 (Partial derivatives commute) Consider a function f(x, y) in a domain D. Assume that f xy and f yx exists and are continuous in D. Then, f xy = f yx. Examples of differential equations Differential equations are equations where the unknown is a function, and where derivatives of the function enter into the equation. Examples: Laplace equation: Find φ(x, y, z) : D IR 3 IR solution of Slide φ xx + φ yy + φ zz = 0. Heat equation: Find a function T (t, x, y, z) : D IR 4 IR solution of T t = T xx + T yy + T zz. Wave equation: Find a function f(t, x, y, z) : D IR 4 IR solution of f tt = f xx + f yy + f zz.
10 Math 20C Multivariable Calculus Lecture 2 0 Exercises: Verify that the function T (t, x) = e t sin(x) satisfies the onespace dimensional heat equation T t = T xx. Verify that the function f(t, x) = (t x) 3 satisfies the onespace dimensional wave equation T tt = T xx. Verify that the function below satisfies Laplace Equation, φ(x, y, z) = x2 + y 2 + z 2.
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