Chapter 2 | Limits
169
Figure 2.26 This graph shows a function f ( x ). a. Since f ( x ) =4 x −3 for all x in (−∞, 2), replace f ( x ) in the limit with 4 x −3 and apply the limit laws: lim x →2 − f ( x ) = lim x →2 − (4 x −3) =5. b. Since f ( x ) = ( x −3) 2 for all x in (2, +∞), replace f ( x ) in the limit with ( x −3) 2 and apply the limit laws: lim x →2 + f ( x ) = lim x →2 + ( x −3) 2 =1. c. Since lim x →2 − f ( x ) =5 and lim x →2 + f ( x ) =1, we conclude that lim x →2 f ( x ) does not exist.
⎧ ⎩ ⎨ − x −2 if x <−1 2 if x =−1 x 3 if x >−1
2.17
Graph f ( x ) =
f ( x ).
lim x →−1 −
and evaluate
We now turn our attention to evaluating a limit of the form lim x → a f ( x ) g ( x ) ,
f ( x ) = K , where K ≠0 and
where lim x → a
lim x → a g ( x ) =0. That is, f ( x )/ g ( x ) has the form K /0, K ≠0 at a . Example 2.23 Evaluating a Limit of the Form K /0, K ≠0 Using the Limit Laws
x −3 x 2 −2 x
Evaluate lim x →2 −
.
Solution Step1. After substituting in x =2, we see that this limit has the form −1/0. That is, as x approaches 2 from the
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