Topic 9: Calculus Option Limits of Sequences and Functions lim [π(π₯) + π(π₯)] = lim π(π₯) + lim π(π₯) $→& Algebraic limit laws: $→& lim [π(π₯) − π(π₯)] = lim π(π₯) − lim π(π₯) $→& $→& $→& lim [π(π₯) × π(π₯)] = lim π(π₯) × lim π(π₯) $→& $→& lim 2 $→& $→& $→& lim π(π₯) π(π₯) 3 = $→& π(π₯) lim π(π₯) $→& lim [ππ(π₯)] = π lim π(π₯) $→& $→& L’Hopital’s rule states if the following occurs: π(π) π π₯π’π¦ = π→π π(π) π π(π₯) ±∞ lim = $→& π(π₯) ±∞ or Differentiate top and bottom fraction separately π(π₯) πΏ′π» π′(π₯) lim $→& π(π₯) = $→& π′(π₯) lim The Squeeze Theorem states if: Use when involving trigonometric functions π(π) ≤ π(π) ≤ π(π) and $→& 1 =∞ 0> 1 =0 ∞ ∞ = πππππππππ ∞ 0 = πππππππππ 0 A closed interval [π, π] includes end points An open interval (π, π) excludes end points A function is continuous at π₯ = π when: lim π(π₯) = lim π(π₯) = limR π(π₯) lim π(π₯) = πΏ Then: 1 = πππππππππ 0 Interval Notation: lim π(π₯) = lim β(π₯) = πΏ $→& 1 = −∞ 07 $→&Q $→& $→& $→& if lim |πS | = 0, then lim πS = 0 The Absolute Value Theorem states: S→T S→T π(π₯) must be continuous at π₯V A function is differentiable at a point ππ if: π(π₯) must not have a “sharp point” at π₯V the tangent to π(π₯) at π₯V must be vertical The derivative of a function π(π) at the point ππ can be found if the limit exists by: If not, then the function is not differentiable at that point π(ππ + π) − π(ππ ) π→π π π′(ππ ) = π₯π’π¦ or π(π₯) − π(π₯V ) X→V π₯ − π₯V π′(π₯V ) = lim π(π₯) is continuous on [π, π] Rolle’s Theorem requires three conditions: a and b are found from the given interval π(π₯) is differentiable on (π, π) π(π₯) is continuous on [π, π] Mean Value Theorem requires three conditions: π(π₯) is differentiable on ]π, π[ a and b are found from the given interval π(π)= π(π) If three conditions are met then there must be a point π such that π[ (π) = π π ∈]π, π[ If these conditions are met, then there exists a point π such that: π [ (π) = π(π) − π(π) π−π Topic 9: Calculus Option Improper Integrals The Fundamental Theorem of Calculus states if π(π) is continuous on [π, π] where: π(π) = π ∫π π(π) π π π≤π≤π then The p-series states for: T ] πΉ′(π₯) = π(π₯) _ Converges if π > 1 1 ππ₯ π₯^ Diverges if π ≤ 1 And for any function where π(π) such that π′(π) = π(π) g h g g ] π(π₯) ππ₯ = ] π(π₯) ππ₯ + ] π(π₯) ππ₯ π ] π(π) π π = π(π) − π(π) π & Where πΉ(π₯) is the antiderivative of π(π₯) & h & ] = −] & g T T Integrals in the form ∫π π(π) π π are known as improper integrals Convergent when ∫& π(π₯) ππ₯ = 0 ππ πππππ‘π π£πππ’π Improper integrals are either convergent or divergent: T Divergent when ∫& π(π₯) ππ₯ = ±∞ When integrating improper integrals, use π₯π’π¦ and replace ∞ with π π→T T o ] π(π₯) ππ₯ = lim ] π(π₯) ππ₯ = o→T & & If 0 ≤ π(π₯) ≤ π(π₯) for all π₯ ≥ π then: The Comparison Test for improper Integrals states: T T ∫& π(π₯) ππ₯ is convergent if ∫& π(π₯) ππ₯ is convergent T T ∫& π(π₯) ππ₯ is divergent if ∫& π(π₯) ππ₯ is divergent or a decreasing function π(π₯) for all π₯ > π, then there is an upper and lower sum such that: T T T q π(π) < ] π(π₯) ππ₯ < q π(π) & st&>_ st& The Rienmann Sum states: For an increasing function π(π₯) for all π₯ > π, then there is an upper and lower sum such that: T T T q π(π) < ] π(π₯) ππ₯ < q π(π) st& & st&>_ Topic 9: Calculus Option Series Part 1 T A series consists of ππ + ππ + ππ … It is denoted by: T q πS = π q πs = πS St_ st_ For a total sum For a partial sum If lim πS ≠ 0 or does not exist then ∑ πS diverges S→T The Divergence Test states: If lim πS = 0 then ∑ πS may converge or may diverge S→T Geometric Infinite Series: Converges if |π| < 1 Diverges if |π| ≥ 1 T q ππ S7_ Sum of infinite geometric series: St_ Key series types: Telescoping Series: Partial Fractions P-Series: = Find πS equation π 1−π Take lim πS Simplify πS _ S→T Converges if π > 1 S} _ (Harmonic Series: S) Diverges if π ≤ 1 If πS ≤ πS for all of π and ∑ πS converges, then ∑ πS also converges The Comparison Test states given for two series of positive terms: If πS ≥ πS for all of π and ∑ πS diverges, then ∑ πS also diverges 1. Check if positive (0 ≤ πS) 2. Find πS (πS generally is πS take away a useless term) Steps: 3. Determine whether πS is smaller or larger than πS 3. Find if πS is convergent or divergent If ∑ &~ g~ exists and πS and πS are positive to use limit comparison test: Then both series either converges of diverges The Limit Comparison Test States: 1. Find πS (πS generally is πS take away a useless term) Steps: & 2. Find if πS and πS are positive and if ∑ g~ exists ~ T The Integral Test states for π(π) = ππ , if then q πS St_ Positive Decreasing Continuous 1. Check criteria: If it’s positive decreasing and continuous T ] St_ π(π₯)ππ₯ Will both converge or diverge 2. Then integrate πS Topic 9: Calculus Option Series Part 2 The Alternating Series Test states for: T if lim |πS | = 0 and |πS>_ | < |πS |, then the series will converge S→T q(−π)π ππ πtπ If: ∑ ππ converges, then π₯π’π¦ πΊπ = πΊ π→T However, if π does not approach ∞ then the Error of a Sum can be found: The Absolute Converge states: For ∑ πS, if ∑|πS | = |π_ | + |π• | + β― is convergent, then ∑ πS is absolutely convergent Use if series is not always positive, or not alternating, or when dealing with trig 1. Write ∑ πS as ∑|πS | For alternative series only: Note: |πΉπ | = |πΊ − πΊπ | ≤ ππ>π Absolute convergence, is stronger than convergence |πΈππππ| ≤ |ππ>π | If a series is absolute convergent, it must be convergent ππ>π ≤ πππππ πππππππππ Conditional convergence is when series converges, but is not absolutely convergent &~RŒ 1. lim ‹ S→T The ratio test stages if: Don’t need to use divergent test for ratio test &~ ‹ < 1, ∑ πS is absolutely convergent &~RŒ 2. lim ‹ S→T &~ &~RŒ 3. lim ‹ S→T &~ ‹ > 1, ∑ πS is divergent ‹ = 1, ∑ πS is inconclusive
