Chapter 2 | Limits
163
Example 2.16 Evaluating a Limit of a Rational Function
2 x 2 −3 x +1 5 x +4 .
Evaluate the lim x →3
Solution Since 3 is in the domain of the rational function f ( x ) = 2 x
2 −3 x +1 5 x +4
, we can calculate the limit by substituting
3 for x into the function. Thus,
2 x 2 −3 x +1 5 x +4
lim x →3
= 10 19 .
⎛ ⎝ 3 x 3 −2 x +7 ⎞ ⎠ .
2.12
Evaluate lim
x →−2
Additional Limit Evaluation Techniques As we have seen, we may evaluate easily the limits of polynomials and limits of some (but not all) rational functions by direct substitution. However, as we saw in the introductory section on limits, it is certainly possible for lim x → a f ( x ) to exist when f ( a ) is undefined. The following observation allows us to evaluate many limits of this type: If for all x ≠ a , f ( x ) = g ( x ) over some open interval containing a , then lim x → a f ( x ) = lim x → a g ( x ). To understand this idea better, consider the limit lim x →1 x 2 −1 x −1 . The function f ( x ) = x 2 −1 x −1 = ( x −1)( x +1) x −1 and the function g ( x ) = x +1 are identical for all values of x ≠1. The graphs of these two functions are shown in Figure 2.24 .
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