Midterm SOS 237 package

# D efinition 42 let a the sum b the product c the

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Unformatted text preview: ng simpler functions. To do this, we ﬁrst deﬁne some operations on functions. D EFINITION 4.2. Let (a) the sum (b) the product (c) the quotient and by by by if . We deﬁne 6 WATERLOO SOS E XAM -AID: MATH237 For , we also deﬁne the composition by for all such that . As with the limit theorems, our continuity theorems from single variable functions are preserved when we move up into higher dimensions. In particular, we have the following: T HEOREM 4.3. Suppose (if ), and are continuous at , and are continuous at . is continuous at . Then , , The following functions are “known” (i.e. can be easily proven, left as exercise) to be continuous on their domains: (a) the constant functions (b) the coordinate functions (c) the logarithm function (d) the exponential function (e) the trigonometric functions (f) the inverse trigonometric functions (g) the absolute value function E XAMPLE 9. On what set is continuous? and are continuous, . Then by composing with S OLUTION. Applying our continuity theorems to the coordinate functions, and hence their quotient is continuous except at , where cosine which is continuous, is continuous on . We can use these in conjunction with our work on limits to decide if functions are continuous, and on what set. E XAMPLE 10. Deﬁne by if if On what set is continuous? is continuous at in example 8. By work done in example 9, . Hence, is continuous on all of . 7 S OLUTION. Recall we showed is also continuous except at WATERLOO SOS E XAM -AID: MATH237 E XERCISE 6. Deﬁne continuous? by for . Can be deﬁned at so that is 5 Limits Revisited (3.3) Using our knowledge of continuity and its deﬁnition, certain limits become easier by simply evaluating the function at the point. E XAMPLE 11. Evaluate where . S OLUTION. By continuity theorems (see example 9), uous at . By the deﬁnition of continuity, is continuous except at , hence it is contin- R EMARK 5.1. Notice we may have implicitly assumed continuity and utilized the above method of calculating limits in the previous section. For example, we use facts like (recall example 5). This is true, and we might be tempted to argue that since is continuous, our continuity theorem ensures that is continuous and thus by the deﬁnition of continuous, we have . However, we need to prove is continuous ﬁrst, which is equivalent to proving ! In fact, we never used continuity results in the previous section. We implicitly used the formal deﬁnition of limits in the sense. Indeed, for any , take . Then if , we have so . To use the squeeze theorem, you should bound the function by a function which is a basic continuous function. 6 Partial Derivatives (4.1-4.2) . The partial derivatives of at are deﬁned by D EFINITION 6.1. Let provided the limit exists. For example, if , then the partials at are: 8 WATERLOO SOS E XAM -AID: MATH237 provided the limits exist. Notice the deﬁnition of a partial derivative is the same as usual differentiation in one variable, where we ﬁx all other variables. In two variables, geometrically is the slope of the tangent of the curve formed by the cross section holding the other variable ﬁxed at the point, and physically, it is a measure of the change in a function in one coordinate direction. Usually, we will try to calculate the partial derivatives by using the standard differentiation rules in one variable. E XAMPLE 12. Let point. be deﬁned by . Calculate and at an arbitrary S OLUTION. By differentiation rules for single variable, Sometimes, we cannot apply the rules of differentiation, and must instead use the deﬁnition of the partial derivatives. E XAMPLE...
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## This note was uploaded on 11/30/2010 for the course MATH 235/237 taught by Professor Wilkie during the Spring '10 term at Waterloo.

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