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MTH132-Quiz1-Solutions

# MTH132-Quiz1-Solutions - -= = when we have 4 1 2 2 1 2 2 1...

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MTH132 Section 5 & 18, Quiz 1 Sept 3, 2008 Instructor: Dr. W. Wu Name (Print Clearly): Student Number: Instructions: Answer the following questions in the space provided. There is more than adequate space provided to answer each question. The total time allowed for this quiz is 15 minutes. 1 [6 marks] . For the function f(x) graphed below, find the following limits or answer “DNE” if they do not exist. (Ex 2.1 # 3 page 81) = - ) ( lim 1 x f x -1 = ) ( lim 0 x f x 0 = ) ( lim 1 x f x DNE 2 [6 marks] . Find the limits. 2008 4 ) 3 ( lim + - x x (Ex 2.2 #12 page 89) 1 ) 1 ( ) 3 4 ( 2008 2008 = - = + - = (use DSP) x x x x - - 2 4 lim 2 4 (Ex 2.2 #30 page 89) 16 ) 4 2 ( 4 ) 2 ( lim 4 ) 2 )( 4 ( lim ) 2 )( 2 ( ) 2 )( 4 ( lim 4 4 4 = + = + = - + - = + - + - x x x x x x x x x x x x x x (use “eliminating of zero denominator” and “DSP”)

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3 [4 marks] . If x x f / 1 ) ( = , please evaluate the limit of the form h x f h x f h ) ( ) ( lim 0 - + , when . 2 - = x (Ex 2.2 #46 page 90) x h x x h x h h x h x h h x x h x h x h x f h x f ) ( 1 ) ( ) ( ) (
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Unformatted text preview: +-=-+ =-+ when, we have 4 / 1 ) 2 ( 2 1 ) 2 ( 2 1 lim ) ( ) ( lim-= +---= +---=-+ → → h h x f h x f h h 4 [4 marks] . Please use the precise definition of limit to prove 2 1 3 lim 1 =-→ x x . (Supplemental Ex 2.3 #7) 1) For all ε , solve <-| ) ( | L x f . So 3 / 1 3 / 1 3 3 3 3 3 | 2 1 3 | + < <-⇒ + < <-⇒ <-<-⇒ <--x x x x 2) We know here 1 = a , so 3 / 1 3 / 1 3 / 1 =-+ =-+ a and 3 / ) 3 / 1 ( 1 ) 3 / 1 ( =--=--a . Let 3 / } 3 / , 3 / min{ δ = = . 3) Then for all x, <--⇒ <-< | 2 ) 1 3 ( | | 1 | x x . Therefore by the precise definition of limit We prove that 2 1 3 lim 1 =-→ x x ....
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MTH132-Quiz1-Solutions - -= = when we have 4 1 2 2 1 2 2 1...

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