Chapter 2 Review. Find the limit by substituting algebraically: 1.lim x→ 4 2. lim x→ 0 (x+5)(x-4) x-4 (x+5) 4+5 = 9 4-0 1+2 4 3.

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Presentation transcript:

Chapter 2 Review

Find the limit by substituting algebraically: 1.lim x→ 4 2. lim x→ 0 (x+5)(x-4) x-4 (x+5) 4+5 =

Find the discontinuous points 3.4. (x+6)(x-3) x-3 (x+6) none X = -9

Find the limit: 5. lim x→ 0 6. lim x→ 0 4xsin3x x sin5x x + sin3x 3x 4x sin5x 5x = 5 sin6x cos6x ● 1 7x sin6x 6x 1__ cos6x 6x 7x 6767

Find the limit: 7. lim x→∞ 8. lim x→ ∞ 6x +sin x x x 3x _ 8 x x lim Θ→ 0 5cosΘ 5·1 = 5

9. Find horizontal asymptote: 10. Find the vertical asymptote: Horizontal: lim x→ ∞ x 2 _ 25 x 2 x 2 2x 2 _ 9x _ 5 x 2 x 2 x Vertical: (x+5)(x-5) (2x+1)(x-5) y = x = -1 2