# 3.1 b - MATH 1081  Wednesday, January 26

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Unformatted text preview: MATH 1081  Wednesday, January 26    Chapter 3 – Section 1(continued)    LIMITS    Homework #2 (due 1/31):  Section 3.1 #8, 34, 38, 40, 42, 46,   48, 50    Clicker Check­in:  Choose any letter to check in now.  Last time, we examined the notion of a limit be consulting the  graph of the function at the specified point.    Today, we will examine how to evaluate a limit with just an  algebraic expression representing the function.       Of course, the result still expresses the same idea as when  looking at the graph.  We just compute the result in a different  manner.    x 2 − 25 Consider the function  f ( x ) = .  At  x = 5 , this function  x−5 does not exist.    However, if we plug in values of  x  near  5 , we see the following  results.    x  4.95   5  5.05   5.1  5.2   4.8   4.9   f ( x )  9.95   DNE  10.05   10.1  10.2   9.8   9.9       According to this table of values, what are the values of  f ( x )   tending towards for  x  closer and closer to  5 ?  x 2 − 25 In the case of  f ( x ) = , when  x = 5  we have the problem  x−5 of division by  0 .      However, for any  x ≠ 5 , we have the common (non‐zero) factor  of ( x − 5 )  that can be cancelled out to yield ( x + 5 ) .    So, although  f ( 5 )  doesn’t exist,    x 2 − 25 ( x − 5 ) ( x + 5 ) = lim x + 5 = 10 .  lim = lim ( ) x→ 5 x − 5 x→ 5 x→ 5 x−5   There are actually rules for computing limits that essentially  boil down to the idea that MANY functions are nice and the  value of the limit at  x = a  is actually equal to the value of the  function at  x = a .    So, if we can compute  f ( a )  by plugging in  a  to the function and  get a real number, then that will be the value of the limit of  f ( x )  as  x → a .    However, if evaluating   f ( a )  results in an illegal operation, like  division by 0, we cannot be so hasty to draw a conclusion as  x 2 − 25 the example with  f ( x ) =  showed.  x−5 This is from page  155 in your text.   Example:  Compute the limit or determine that it does not   exist.    1. lim 3x 2 − 5 x + 1  x→ 2 2. 3. 4. 5. x → −3 lim 2 x   x2 − 2x + 1   lim 2 x →1 x −1 ex ( x + 4 )   lim x → − 4 x 2 + 8 x + 16 lim h→ 0 x+h − h   h What issue(s) do we run into when computing  lim x ?   x→ 0 Or  lim x ?  x → −8       Do we have the same issue(s) when computing  lim 3 x ?  x→ 0 Or  lim 3 x ?  x → −8   As for computing limits as  x  goes to  +∞  or  −∞ , we have to  consider what happens to the expression defining the function  when  x  is VERY large in magnitude.    For polynomial and rational functions, we can use the idea that  the term(s) with the largest exponent will be so much larger  than all other terms when  x  gets big, that none of the other  terms even matter.    The book describes a method of dividing by the highest power  of  x .  We can use a more intuitive method just by inspecting  and isolating the terms with the largest exponents, since when  the value of  x  gets VERY large, the term with the largest  exponent will dominate all the otherterms.    However, if you prefer a specific set of steps to follow,  reference the “Finding Limits At Infinity” box on page 163.    Example:  Compute the limit or determine that it does not   exist.    1. lim 4 x 5 − 9 x 3 + 12 x + 1000   x→ ∞ 2. 3. 4. 5. x → −∞ lim − 2 x 6 + 25 x 5 + 7 x 4 − 90 x 3 + 1  3x 2 − 5 x + 7   lim x → ∞ 9 x 3 + 17 x 5 x 5 − 4 x 2 + 9 x − 10   lim 4 3 2 x → −∞ 3x + 8 x − 2 x + x + 12 4 x 4 − 5 x 3 + 2 x 2 − 3x + 1   lim 4 x→ ∞ 2 x + 200 Clicker Check­out:  Choose any letter to check out now.    In recitation tomorrow is Quiz 1, covering Sections 1.1 & 2.1.    Next week – Sections 3.2 & 3.3.  ...
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## This note was uploaded on 05/01/2011 for the course MATH 1081 taught by Professor Johanson during the Spring '08 term at Colorado.

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