![]() Want to cite, share, or modify this book? This book uses theĬreative Commons Attribution-NonCommercial-ShareAlike License In Example 2.19, we look at simplifying a complex fraction. Example 2.17 illustrates the factor-and-cancel technique Example 2.18 shows multiplying by a conjugate. The next examples demonstrate the use of this Problem-Solving Strategy. If f ( x ) / g ( x ) f ( x ) / g ( x ) is a complex fraction, we begin by simplifying it.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.If f ( x ) f ( x ) and g ( x ) g ( x ) are polynomials, we should factor each function and cancel out any common factors.To do this, we may need to try one or more of the following steps: We then need to find a function that is equal to h ( x ) = f ( x ) / g ( x ) h ( x ) = f ( x ) / g ( x ) for all x ≠ a x ≠ a over some interval containing a.First, we need to make sure that our function has the appropriate form and cannot be evaluated immediately using the limit laws.Problem-Solving Strategy: Calculating a Limit When f ( x ) / g ( x ) f ( x ) / g ( x ) has the Indeterminate Form 0/0 lim x → 2 x + lim x → 2 1 ( lim x → 2 x ) 3 + lim x → 2 4 Apply the power law.lim x → 2 2 x 2 − 3 x + 1 x 3 + 4 = lim x → 2 ( 2 x 2 − 3 x + 1 ) lim x → 2 ( x 3 + 4 ) Apply the quotient law, making sure that. lim x → 2 x + lim x → 2 1 lim x → 2 x 3 + lim x → 2 4 Apply the sum law and constant multiple law. ![]() Lim x → 2 2 x 2 − 3 x + 1 x 3 + 4 = lim x → 2 ( 2 x 2 − 3 x + 1 ) lim x → 2 ( x 3 + 4 ) Apply the quotient law, making sure that. Again, we need to keep in mind that as we rewrite the limit in terms of other limits, each new limit must exist for the limit law to be applied. To find this limit, we need to apply the limit laws several times. We now practice applying these limit laws to evaluate a limit. Root law for limits: lim x → a f ( x ) n = lim x → a f ( x ) n = L n lim x → a f ( x ) n = lim x → a f ( x ) n = L n for all L if n is odd and for L ≥ 0 L ≥ 0 if n is even and f ( x ) ≥ 0 f ( x ) ≥ 0. Power law for limits: lim x → a ( f ( x ) ) n = ( lim x → a f ( x ) ) n = L n lim x → a ( f ( x ) ) n = ( lim x → a f ( x ) ) n = L n for every positive integer n. Quotient law for limits: lim x → a f ( x ) g ( x ) = lim x → a f ( x ) lim x → a g ( x ) = L M lim x → a f ( x ) g ( x ) = lim x → a f ( x ) lim x → a g ( x ) = L M for M ≠ 0 M ≠ 0 Product law for limits: lim x → a ( f ( x )
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