Pre-Calc Homework Solutions 43

Pre-Calc Homework Solutions 43 - Section 2.1 10. lim y2 43...

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10. lim y →2 } y 2 1 y 1 5 y 2 1 6 } 5 } 2 2 1 2 5 1 (2 2 ) 1 6 } 5 } 2 4 0 } 5 5 Graphical support: [ 2 3, 3] by [ 2 5, 10] 11. lim y 2 3 } y 2 1 y 2 4 2 y 3 1 3 } 55 } 0 6 } 5 0 Graphical support: [ 2 5, 5] by [ 2 5, 5] 12. lim x 1/2 int x 5 int } 1 2 } 5 0 Note that substitution cannot always be used to find limits of the int function. Its use here can be justified by the Sandwich Theorem, using g ( x ) 5 h ( x ) 5 0 on the interval (0, 1). Graphical support: [ 2 4.7, 4.7] by [ 2 3.1, 3.1] 13. lim x 2 2 ( x 2 6) 2/3 5 ( 2 2 2 6) 2/3 5 ˇ 3 ( 2 w 8 w ) 2 w 5 ˇ 3 6 w 4 w 5 4 Graphical support: [ 2 10, 10] by [ 2 10, 10] 14. lim x 2 ˇ x w 1 w 3 w 5 ˇ 2 w 1 w 3 w 5 ˇ 5 w Graphical support: [ 2 4.7, 4.7] by [ 2 3.1, 3.1] 15. lim x 0 ( e x cos x ) 5 e 0 cos 0 5 1 ? 1 5 1 Graphical support: [ 2 4.7, 4.7] by [ 2 3.1, 3.1] 16. lim x p /2 ln (sin x ) 5 ln 1 sin } p 2 } 2 5 ln 1 5 0 Graphical support: [ 2 p , p ] by [ 2 3, 1] 17. You cannot use substitution because the expression ˇ x w 2 w 2 w is not defined at x 52 2. Since the expression is not defined at points near x 2, the limit does not exist.
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