EVALUATING LIMITS ANALYTICALLY Last class, we learned that the limit of π(π₯) as π₯ approaches π does not depend on the value of π at π₯ = π. It may happen, however, that the limit is precisely π π , which means lim π π₯ = π(π) π₯βπ In such cases, the limit can be evaluated by direct substitution. THEOREM 2.1 Some Basic Limits Let π and π be real numbers and let π be a positive integer. Then, 1. lim π = π π₯βπ 2. lim π₯ = π π₯βπ 3. lim π₯ π = ππ π₯βπ Properties of Limits Handout Ex 3. Let: lim π π₯ = 4 π₯βπ Find: lim [π π₯ + β π₯ ] a) π₯βπ π π₯ b) lim π₯βπ β π₯ lim π π₯ = 0 lim β π₯ = β3 π₯βπ π₯βπ c) d) lim β π₯ 2 π₯βπ β π₯ lim π₯βπ π π₯ So, if itβs a polynomial or rational with nonzero denominator, just plug it inβ¦ In fact, always try to plug in π and see what happensβ¦ If you get a number back, then great! Thatβs the limit! If notβ¦ DIVIDING OUT (FACTORING) TECHNIQUE π₯2 + π₯ β 6 Ex 4. lim π₯ββ3 π₯+3 COMMON DENOMINATOR TECHNIQUE Ex 5. 1 1 β lim π₯ + 4 4 π₯β0 π₯ RATIONALIZING (CONJUGATE) TECHNIQUE Ex 6. π₯+1β1 lim π₯β0 π₯ THEOREM 2.9 Two Special Limits 1. sin π₯ lim =1 π₯β0 π₯ 2. 1 β cos π₯ lim =0 π₯β0 π₯ Ex 7. tan π₯ lim π₯β0 π₯ Ex 8. sin 4π₯ lim π₯β0 5π₯ Ex 9. 2βπ₯ lim+ π₯β2 |2 β π₯| So, in summary: 1. 2. 3. 4. 5. Try Direct Substitution! You might get lucky! If notβ¦ Factoring (Dividing Out) Common Denominator (if you see fractions in fractions) Rationalizing (if you see a radical) Remember special trig limits!
© Copyright 2026 Paperzz