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# Calculus 1 week 3 Notes 191g

NMSU

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This 4 page Class Notes was uploaded by Brenna Graham on Sunday September 4, 2016. The Class Notes belongs to 191g at New Mexico State University taught by Dr. Andres Contreras in Fall 2016. Since its upload, it has received 17 views. For similar materials see Calculus I in Math at New Mexico State University.

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Date Created: 09/04/16

Evaluating Limits Algebraically EXAMPLE 1 multiplying by the conjugate lim( 1 − 22 ) x→1 x−1 x −1 Multiply by the conjugate: 1(x+1) 2 lim (x−1)(x+1) x −1 x→1 lim x21−2 x→1 x −1 x−1 lim x −1 x→1 lim 1 = 1 x→1 x+1 2 EXAMPLE 2 1 1 x→0( x ) =x0 Another ex 1 2 x→c( x ) x± ∞ limf(x) = f(c) if f is continuous x→c Assume f is continuous on h(c) and h is continuous at c If f(h(x)) is continuous at c 1 remember sin =/ arch sin x x2 3 EX: 2x−3 continuous except at 2 Rules: p(x) q(x) q = 0 Ex 2 arch sin + f(g(h(x)) x f = e g = x h = arch sin x Equations to remember lim 1−cosx= 0 x→0 x sinh lim h − 1 h→0 lim e −1= 1 h→0 h tan(8t) lt→0tan(7t) sin 8t cos 8t lim cos 7t t→0 lim sin 7ttcos 8t t→0 lim sin 8• limcos 7 t→0 sin 7tt→0 cos 8t lim sin 8t sin 8t 8t t→0 8t 7 lim ( 8t ) • lim ( sin 7t = lim sin 7t 1/ =88/7 t→0 t→0 t→0 8t 2 1 x→0x sin =x0 1 1 − 1 ≤ sin 2≤ x − x ≤ x sin ≤ x 2 x lim x−2 x→2 √ x− 4−x Remember: lim sinΘ = 1 lim 1−cosΘ = 0 Θ→0 Θ Θ→0 Θ Next lecture 2.7 Limits at infinity lim f(x) = L if f(x) → L as x increases x→+∞ lim f(x) = L if f(x) → L as x decreases x→−∞ lim f(x) means evaluate both sides separately x→+/− x→+∞ =+ ∞If x becomes larger without bound Y=L is the horizontal asymptote Ex f(x) = e x→+/−∞(x) =? lim e =+ ∞ x→∞ Rules n x→∞m x = + ∞ lim x−h = 0 x→∞ ∞ if n is even − ∞ if odd If p+q is a positive integer lim x p/q = {− 50 if odd undefined if q is even x→−∞ What can be said about asymtote behavior at certain points lim p(x)/q(x) x→∞ Polynomials Example 5x5−2x 9x6+19 9/2 −1/2 lim 2x 23x1/2 x→∞ x +x 4x−3 x→∞m √25x+4x 3 1. lim 33 +2 x→3 x −1 Solve by plugging in 3(27)+2 83 lim 27−1 = 26 x→3 2.lim √x−3 x→9 x−9 Solve by multiplying by congegate So √x−3 √x lim x−9 • x+9 = x + 13 x→9 Plug in 3(1)+2 = 5 √x−3 3. lim√ x−5−2 x→9 lim √x−3 • √x+5+2 x→9 √x−5−2 √x+5+2 √x−3 lim x−9 this was on top but other will be canceled −> x + 5 + 2 = 4 x→9 Bottom= 1 = 1/6 √x−3 tan 7x 4. lim sin 5x x→0 lim sin7x • 1 =1 x→0 cos 7x sin 5x 7x−10 5. lim 2 x→−∞ √16x +1 7x−10 4x+1 so − 7/4

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