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					                  Warm Up
• Find the limit as x approaches 1 for the
  following functions
• f(x) = 3
• f(x) = x
• f(x) = 3x +5
• f(x) = x2 +4x – 2
• f(x) = (x3 – 1)/(x – 1)
Alternatives to Direct
     Substitution
   September 10, 2010
                  Objective
• SWBAT use an alternative to direct
  substitution to evaluate the limit of a function
              Vocabulary
Rationalize   Multiplying by a conjugate to
              eliminate a radical from an
              expression

Conjugate     Two identical binomials that have
              opposite signs
         Alternative Strategies
• When evaluating using direct substitution
  produced 0 in the denominator, direct
  substitution will not work to find a limit.
• If you have a polynomial in the numerator,
  factor the polynomial.
• If the numerator and denominator now share
  a factor, cancel it out and try direct
  substitution again.
         Alternative Strategies
• If you have a radical (square root) in the
  numerator, multiply the numerator and
  denominator by the conjugate of the
  numerator.
• This should eliminate the radical and change
  the denominator so you can use direct
  substitution.
                  Example
         x3  1
1. lim
    x 1 x  1
                      Example
         x2  x  6
2. lim
   x 3    x 3
                     Example
          x 1  1
3. lim
   x 0      x
                              Practice
            x  3x
               2
                                             x 3 1
    1. lim                             lim
                                    1. x 1
       x 0   x                              x 1


               x                              2 x  2
   2. lim 2
        x 1 x  x                2. lim
                                       x 0      x


               2x 2  x  3                    x 1  2
    3.   lim
                                  3. lim
         x 1     x 1                x 3     x 3



                             
                          Practice
             x3  8                          x5 5
        lim                          lim
        x 2 x  2
                                     x 0      x

             2 x                            x53
        lim 2                         lim
        x 2 x  4                    x 4    x4
                         
            x2  x  6
      lim
      x 3   x2  9
                          
           x 2  5x  4
      lim 2
      x 4 x  2x  8


				
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posted:3/29/2013
language:English
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