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AP Calculus Chapter 1, Section 3
Evaluating Limits Analytically
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Some Basic Limits In some cases, the limit can be evaluated by direct substitution. lim π₯βπ π =π lim π₯βπ π₯ =π lim π₯βπ π₯ π = π π
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Evaluate the following limits
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Properties of Limits Let b and c be real numbers, let n be a positive integer, and let f and g be functions with the following limits. lim π₯βπ π π₯ =πΏ and lim π₯βπ π π₯ =πΎ Scalar multiple: lim π₯βπ ππ π₯ =ππΏ Sum or difference: lim π₯βπ π π₯ Β±π π₯ =πΏΒ±πΎ Product: lim π₯βπ π π₯ π π₯ =πΏπΎ Quotient: lim π₯βπ π(π₯) π(π₯) = πΏ πΎ β² provided πΎβ 0 Power: lim π₯βπ [π π₯ ] π = πΏ π
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lim π₯β2 ( 4π₯ 2 +3)
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Limits of Polynomials & Rational Functions
If p is a polynomial function and c is a real number, then lim π₯βπ π π₯ =π(π) If r is a rational function given by r π₯ =π(π₯)/π(π₯) and c is a real number such π(π)β 0, then lim π₯βπ π π₯ =π π = π(π) π(π)
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lim π₯β1 π₯ 2 +π₯+2 π₯+1
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The Limit of a Function Involving a Radical
Let n be a positive integer. The following limit is valid for all c if n is odd, and is valid for π>0 if n is even. lim π₯βπ π π₯ = π π
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The Limit of a Composite Function
If f and g are functions such that lim π₯βπ π π₯ =πΏ and lim π₯βπ π π₯ =π(πΏ) , then lim π₯βπ π π π₯ =π lim π₯βπ π π₯ =π(πΏ)
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Because lim π₯β0 π₯ 2 +4 = 0 2 +4=4 and lim π₯β4 π₯ =2 it follows that lim π₯β0 π₯ 2 +4 = 4 =2
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Limits of Trigonometric Functions
lim π₯βπ sin π₯ = sin π lim π₯βπ cos π₯ = cos π lim π₯βπ tan π₯ = tan π lim π₯βπ cot π₯ = cot π lim π₯βπ csc π₯ = csc π lim π₯βπ sec π₯ = sec π
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lim π₯β0 tan π₯ lim π₯βπ (π₯ cos π₯ ) lim π₯β0 sin 2 x
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Dividing Out Technique
lim π₯ββ3 π₯ 2 +π₯β6 π₯+3
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Rationalizing Technique
lim π₯β0 π₯+1 β1 π₯ Check your answer by using a table
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Then lim π₯βπ π(π₯) exists and is equal to L.
The Squeeze Theorem Basically says if you have two different function that have the same limit as π₯βπ, and you have a 3rd function that falls between the first two functions, the 3rd function will also have the same limit as π₯βπ. β(π₯)β€π π₯ β€π π₯ for all x in an open interval containing c, except possibly c itself, and if lim π₯βπ β π₯ =πΏ= lim π₯βπ π(π₯) Then lim π₯βπ π(π₯) exists and is equal to L.
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Special Trigonometric Functions
lim π₯β0 sin π₯ π₯ =1 lim π₯β0 1β cos π₯ π₯ =0
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Find the limit: lim π₯β0 tan π₯ π₯
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Find the limit: lim π₯β0 sin 4π₯ π₯
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Homework Pg. 67 β 69: #1 β 77 every other odd, 83, 87, 113
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