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Chapter 2 | Limits
Figure 2.24 The graphs of f ( x ) and g ( x ) are identical for all x ≠1. Their limits at 1 are equal.
We see that
x 2 −1 x −1 = lim x →1
( x −1)( x +1) x −1
lim x →1
( x +1)
= lim
x →1
=2.
f ( x ) g ( x ) ,
The limit has the form lim x → a g ( x ) =0. (In this case, we say that f ( x )/ g ( x ) has the indeterminate form 0/0.) The following Problem-Solving Strategy provides a general outline for evaluating limits of this type. where lim x → a f ( x ) =0 and lim x → a Problem-Solving Strategy: Calculating a Limit When f ( x )/ g ( x ) has the Indeterminate Form 0/0 1. First, we need to make sure that our function has the appropriate form and cannot be evaluated immediately using the limit laws. 2. We then need to find a function that is equal to h ( x ) = f ( x )/ g ( x ) for all x ≠ a over some interval containing a . To do this, we may need to try one or more of the following steps: a. If f ( x ) and g ( x ) are polynomials, we should factor each function and cancel out any common factors. b. If the numerator or denominator contains a difference involving a square root, we should try multiplying the numerator and denominator by the conjugate of the expression involving the square root. c. If f ( x )/ g ( x ) is a complex fraction, we begin by simplifying it. 3. Last, we apply the limit laws.
The next examples demonstrate the use of this Problem-Solving Strategy. Example 2.17 illustrates the factor-and-cancel technique; Example 2.18 shows multiplying by a conjugate. In Example 2.19 , we look at simplifying a complex fraction.
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