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Unformatted text preview: XERCISE 3. Sketch the crosssections for 3 . WATERLOO SOS E XAM AID: MATH237 2 Limit Theorems (2.12.2)
. . That is, for Notation: We will use underline to denote a vector, for example Notation: We will use to denote Euclidean distance in , we have In particular, for we have for Fortunately, our limit theorems from single variable functions are preserved when we move up into higher dimensions. In particular, we have the following: T HEOREM 2.1. Let (a) (b) . If and . . both exist, then (c) if . C OROLLARY 2.2. If exists, then the limit is unique. It can be difﬁcult to use the formal deﬁnition of a limit to ﬁnd one or prove it exists. The following theorem can be used quite frequently to help us in that regard. T HEOREM 2.3 [S QUEEZE T HEOREM ]. Let for all in some neighborhood of and . If there exists a function , then . such that 3 Proving a Limit Does (Not) Exist (2.32.4) You may notice that the Squeeze Theorem relies on the fact that we begin with a candidate for the limit, . If this is not given, we can test the limiting behaviour of as we approach along straight lines of different slopes. If the limiting value is always , this is a possible candidate, and we can try to apply Squeeze Theorem. Note that it is possible for a function to have a limit along every straight line, yet still not have a limit in the formal sense, so it is important to prove the limit exists. Remember when testing for a limit, use curves that do indeed pass through the point. If we wanted the limit , it would not help to use lines like . Another trick for rational functions is to examine the degrees of the numerator and denominator. For limits going to , with potential limit , if the degree (maximum sum of the exponents of each factor in a term) is greater in the numerator, the limit will probably exist. If it is less in the numerator, the limit will probably not exist. If they are the same, nothing can be said in general. 4 WATERLOO SOS E XAM AID: MATH237 E XAMPLE 5. Determine whether exists, and if so ﬁnd its value. S OLUTION. Taking lines we get We then guess the limit is . Since , . Then Since we get by the Squeeze Theorem. E XAMPLE 6. Determine whether exists, and if so ﬁnd its value. S OLUTION. Taking lines we get This covers every nonvertical line. Finally for , This seems to indicate the limit should be 0. However, notice the degree (sum of the degrees of and ) on the numerator is while it is on the denominator. This suggests the denominator goes to zero faster, so that the limit does not exist. Indeed, along the curve , we have for for so the limit does not exist. E XERCISE 4. Determine whether (a) (b) (c) . . , for . exists for the following functions, and if so ﬁnd its value. 5 WATERLOO SOS E XAM AID: MATH237 4 Continuity Theorems (3.13.2) The deﬁnition of continuity for multivariable functions is very similar to that of singlevariable functions. D EFINITION 4.1. A function at every point in a set is continuous at , we say that if and only if . If is continuous is coninuous on . Note that by stating the equality, we are implicitly requiring the limit to exit, the value to be deﬁned at , and for the equality itself to hold. E XAMPLE 7. Let S OLUTION. Notice be deﬁned by is not even deﬁned at . Determine whether is continuous at . , so it will not be continuous. E XAMPLE 8. Let be deﬁned by if if Determine whether is continuous at . S OLUTION. By example 5, , so is continuous at . E XERCISE 5. Let Determine whether be deﬁned by is continuous at . if if Some functions can be shown to be continuous by relying on information about how it is formed usi...
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 Spring '10
 WILKIE
 Scalar, Sets

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