Chapter 2 Section 6

Chapter 2 Section 6 - -= =-→ → → → θ Example 8...

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1 Chapter 2 Section 6 Limits of Trig Functions and the Squeeze Theorem
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2 Objectives 1. Find limits involving trig functions. 2. Find limits using the Squeeze Theorem
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3 Notation review: sin 2 (x) = [sin(x)] 2 sin x 2 = sin(x 2 ) [sin(x)] -1 = 1/sin(x) = csc(x) sin -1 (x) = arcsin(x)
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4 x x Consider x ) sin( lim 0 x -1 -0.5 -0.1 -0.01 0 0.01 0.1 0.5 1 sin(x)/x 0.84 0.96 0.998 0.99998 dne 0.99998 0.998 0.96 0.84 x -1 -0.5 -0.1 -0.01 0 0.01 0.1 0.5 1 [1-cos(x)]/x -0.46 -0.24 -0.05 -0.005 dne 0.005 0.05 0.24 0.46 Note: “x” must be in radians.
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5 -4 -3 -2 -1 1 2 3 4 5 -4 -3 -2 -1 1 2 3 4 x y y = sin(x)/x 1 ) sin( lim 0 = x x x 0 1 0 = - x x x ) cos( lim -4 -3 -2 -1 1 2 3 4 5 -3 -2 -1 1 2
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6 3 2 6 4 lim 6 1 4 1 lim 6 6 ) 6 sin( 4 4 ) 4 sin( lim ) 6 sin( ) 4 sin( lim : 0 0 0 0 = = = = = x x x x x x x x x x x x x x Example
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7 0 1 0 lim sin cos 1 lim 1 1 sin cos 1 lim sin cos 1 lim : 0 0 0 0 = = - = ×
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Unformatted text preview: -= =-→ → → → θ Example 8 Example: Given 3x < f(x) < x 3 + 2 for 0< x< 2, find lim x 1 f(x) We must use the Squeeze Theorem. 9 Squeeze Theorem If L (x) < f(x) < U (x) when x is near “c” except possibly at “c” and lim x c L (x) = lim x c U (x) = N then lim x c f(x) = N. 10 Example: Given 3x < f(x) < x 3 + 2 for 0< x< 2, find lim x 1 f(x) lim x 1 3x = 3 lim x 1 x 3 + 2 = 3 therefore, lim x 1 f(x) = 3 by the Squeeze Theorem 11 ) sin( lim : x x e x Example π + →...
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This note was uploaded on 09/28/2009 for the course MATH 1550 taught by Professor Wei during the Spring '08 term at LSU.

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Chapter 2 Section 6 - -= =-→ → → → θ Example 8...

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