Section 2 3 Computing limits This is an outline of some ways to evaluate lim f x x a or to determine it does not exist Case 1 f x is given by an algebraic rule and f a is defined then f a is the limit In other words if substituting a for x gives a meaningful answer then that is the limit Examples x 1 lim x 0 x 2 1 lim x 2 x 2 2 3 x 10 x 3 lim x 3 Examples of non meaningful expressions are 0 x 0 0 1 0 0 0 and there are others that we will discuss if they come up nonzero Substitution gives 0 In this case the limit does not exist as a number the magnitude of the function goes to infinity 0 If Substitution gives 0 This is indeterminate the limit could be any number or not exist Do Algebra p x Case 2 Case 2 f x is a rational function nonzero substitution gives q x and q a is 0 but p a is not 0 Then so the limit does not exist 0 Case 3 f x is a rational function p x q x 0 and substitution gives 0 This means both p a and q a have x a as a factor Factor the numerator and denominator and cancel common factors Substitute again and re evaluate it Example x x 2 2 9 x 6 x lim x 3 x 2 2 0 9 Substitution of 3 for x gives x 6 x 3 x 3 x 3 x 2 x 3 x 2 if x 3 taking the limit Now substitution gives Example x x 2 2 2x 4x 4 x lim x 2 x 2 2 4x 4 x 2 does not exist 2 5 This is OK since we do not let x equal 3 in lim f x x 3 6 5 0 2x x x 2 6 0 x x 2 Substitution of 2 for x gives if x 2 0 Now substituting x 2 gives 2 0 so the limit Case 4 Conjugating is needed to simplify Recall that x a x a x x and a a x c 2 d 2 a x x a c d c d a x a x so also This is a called conjugating Example Find lim x 1 x 1 x 1 lim x 1 lim x 1 x x 1 x 1 0 Substitution gives x 1 1 x 1 1 lim x 1 x 1 0 so we conjugate and cancel 1 2 Case 5 The above do not fit Look at the graph or try to use the squeeze theorem for some possible strategies Find lim sin x 0 1 x if it exists Look at the graph and you will see that the function oscillates between 1 and 1 and takes on all values in between no matter how close we get to x 0 The limit does not exist Find 1 lim x sin x 0 x 1 1 sin 1 x so 1 x sin x x x and x both approach 0 so the limit is 0 The function is squeezed between two functions which have the same limit
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