lecture8 - Lecture 8: Evaluating Limits Chapter 2, Section...

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Lecture 8: Evaluating Limits Chapter 2, Section 8 Limit Laws Suppose lim x ! a f ( x ) and lim x ! a g ( x ) exist and c is a constant. We have the following limit laws: 1. lim x ! a [ f ( x ) § g ( x ) ] = 2. lim x ! a cf ( x ) = 3. lim x ! a f ( x ) g ( x ) = 4. If lim x ! a g ( x ) 6 = 0, lim x ! a f ( x ) g ( x ) = We can extend Law 3 to get: 5. If n is a positive integer, lim x ! a [ f ( x ) ] n =
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Note the following: 6. lim x ! a c = 7. lim x ! a x = We then have the following cases of the limit laws, where n is a positive integer: 8. lim x ! a x n = 9. lim x ! a n p x = 10. lim x ! a n p f ( x ) = NOTE: For laws 9 and 10, if n is even, assume that a > 0.
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ex. Evaluate lim x 1 (2 x 2 + 3 x ¡ 2) : Note that if f ( x ) = 2 x 2 + 3 x ¡ 2, lim x 1 f ( x ) = lim x 1 (2 x 2 + 3 x ¡ 2) is the same as the value of the function at x = ¡ 1, f ( ¡ 1). That is, we can flnd the limit by direct substitution. Direct Substitution Property If f ( x ) is a polynomial and a is in the domain of f then lim x ! a f ( x ) = f ( a ). This property can be verifled from the limit laws.
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What about a rational function R ( x ) = p ( x ) q ( x ) where p and q are polynomials? If q ( a ) 6
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lecture8 - Lecture 8: Evaluating Limits Chapter 2, Section...

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