Lesson 4: Calculating Limits
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  • 1. Section 2.3 Computation of Limits Math 1a October 1, 2007
  • 2. Limit Laws Suppose that c is a constant and the limits lim f (x) and lim g (x) x→a x→a exist. Then 1. lim [f (x) + g (x)] = lim f (x) + lim g (x) x→a x→a x→a 2. lim [f (x) − g (x)] = lim f (x) − lim g (x) x→a x→a x→a 3. lim [cf (x)] = c lim f (x) x→a x→a 4. lim [f (x)g (x)] = lim f (x) · lim g (x) x→a x→a x→a
  • 3. Limit Laws, continued lim f (x) f (x) = x→a 5. lim , if lim g (x) = 0. x→a g (x) lim g (x) x→a x→a
  • 4. Limit Laws, continued lim f (x) f (x) = x→a 5. lim , if lim g (x) = 0. x→a g (x) lim g (x) x→a x→a n n 6. lim [f (x)] = lim f (x) x→a x→a
  • 5. Limit Laws, continued lim f (x) f (x) = x→a 5. lim , if lim g (x) = 0. x→a g (x) lim g (x) x→a x→a n n 6. lim [f (x)] = lim f (x) (follows from 3 repeatedly) x→a x→a
  • 6. Limit Laws, continued lim f (x) f (x) = x→a 5. lim , if lim g (x) = 0. x→a g (x) lim g (x) x→a x→a n n 6. lim [f (x)] = lim f (x) (follows from 3 repeatedly) x→a x→a 7. lim c = c x→a 8. lim x = a x→a
  • 7. Limit Laws, continued lim f (x) f (x) = x→a 5. lim , if lim g (x) = 0. x→a g (x) lim g (x) x→a x→a n n 6. lim [f (x)] = lim f (x) (follows from 3 repeatedly) x→a x→a 7. lim c = c x→a 8. lim x = a x→a 9. lim x n = an x→a √ √ 10. lim n x = n a x→a
  • 8. Limit Laws, continued lim f (x) f (x) = x→a 5. lim , if lim g (x) = 0. x→a g (x) lim g (x) x→a x→a n n 6. lim [f (x)] = lim f (x) (follows from 3 repeatedly) x→a x→a 7. lim c = c x→a 8. lim x = a x→a 9. lim x n = an (follows from 6 and 8) x→a √ √ 10. lim n x = n a x→a
  • 9. Limit Laws, continued lim f (x) f (x) = x→a 5. lim , if lim g (x) = 0. x→a g (x) lim g (x) x→a x→a n n 6. lim [f (x)] = lim f (x) (follows from 3 repeatedly) x→a x→a 7. lim c = c x→a 8. lim x = a x→a 9. lim x n = an (follows from 6 and 8) x→a √ √ 10. lim n x = n a x→a n 11. lim f (x) = lim f (x) (If n is even, we must additionally n x→a x→a assume that lim f (x) > 0) x→a
  • 10. Direct Substitution Property Theorem (The Direct Substitution Property) If f is a polynomial or a rational function and a is in the domain of f , then lim f (x) = f (a) x→a
  • 11. Limits do not see the point! (in a good way) Theorem If f (x) = g (x) when x = a, and lim g (x) = L, then lim f (x) = L. x→a x→a
  • 12. Limits do not see the point! (in a good way) Theorem If f (x) = g (x) when x = a, and lim g (x) = L, then lim f (x) = L. x→a x→a Example x 2 + 2x + 1 Find lim , if it exists. x +1 x→−1
  • 13. Limits do not see the point! (in a good way) Theorem If f (x) = g (x) when x = a, and lim g (x) = L, then lim f (x) = L. x→a x→a Example x 2 + 2x + 1 Find lim , if it exists. x +1 x→−1 Solution x 2 + 2x + 1 = x + 1 whenever x = −1, and since Since x +1 x 2 + 2x + 1 lim x + 1 = 0, we have lim = 0. x +1 x→−1 x→−1
  • 14. Finding limits by algebraic manipulations Example √ x −2 Find lim . x −4 x→4
  • 15. Finding limits by algebraic manipulations Example √ x −2 Find lim . x −4 x→4 Solution √ √ √ 2 Write the denominator as x − 4 = x − 4 = ( x − 2)( x + 2).
  • 16. Finding limits by algebraic manipulations Example √ x −2 Find lim . x −4 x→4 Solution √2 √ √ Write the denominator as x − 4 = x − 4 = ( x − 2)( x + 2). So √ √ x −2 x −2 = lim √ √ lim x→2 x − 4 x→2 ( x − 2)( x + 2) 1 1 = lim √ = 4 x +2 x→2
  • 17. Finding limits by algebraic manipulations Example √ x −2 Find lim . x −4 x→4 Solution √2 √ √ Write the denominator as x − 4 = x − 4 = ( x − 2)( x + 2). So √ √ x −2 x −2 = lim √ √ lim x→2 x − 4 x→2 ( x − 2)( x + 2) 1 1 = lim √ = 4 x +2 x→2 Example√ √ x− 3a 3 Try lim . x −a x→a
  • 18. Two More Important Limit Theorems Theorem If f (x) ≤ g (x) when x is near a (except possibly at a), then lim f (x) ≤ lim g (x) x→a x→a (as usual, provided these limits exist). Theorem (The Squeeze/Sandwich/Pinching Theorem) If f (x) ≤ g (x) ≤ h(x) when x is near a (as usual, except possibly at a), and lim f (x) = lim h(x) = L, x→a x→a then lim g (x) = L. x→a
  • 19. We can use the Squeeze Theorem to make complicated limits simple.
  • 20. We can use the Squeeze Theorem to make complicated limits simple. Example 1 Show that lim x 2 sin = 0. x x→0
  • 21. We can use the Squeeze Theorem to make complicated limits simple. Example 1 Show that lim x 2 sin = 0. x x→0 Solution We have for all x, 1 −x 2 ≤ x 2 sin ≤ x2 x The left and right sides go to zero as x → 0.