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Easy limit Help

  1. Sep 28, 2009 #1
    1. limit as x approaches -5 from the left
    of 3x/2x+10




    2. Relevant equations



    3. The attempt at a solution
    I know the solution is infinity but I don't know how to prove it. Should I use the formal definition or is there another way?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 28, 2009 #2

    lanedance

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    Homework Helper

    simply writing it as:
    [tex] \frac{3}{2} \frac{x}{(x+5)} [/tex]
    should be sufficient to show the numerator goes to -5*2, whilst denominator is goes to 2*0, so the answer isn't positive infinity...
     
  4. Sep 28, 2009 #3

    lanedance

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    Homework Helper

    otherwise fromally, you could show for any poistive integer N, you can choose d such that for
    -5-x < d then f(x) < -N

    though bit of overkill i think
     
  5. Sep 28, 2009 #4
    You can use the following theorems:

    For any real number a and positive integer m:

    [tex]\lim_{x \rightarrow a^+} \frac{1}{(x-a)^m} = \infty[/tex]

    [tex]\lim_{x \rightarrow a^-} \frac{1}{(x-a)^m} = \left\{ \begin{array}{rl}
    \infty, & \text{if }m \text{ is even} \\
    -\infty, & \text{if }m \text{ is odd}
    \end{array}[/tex]

    If [itex]\lim_{x \rightarrow a}f(x) = \infty[/itex] and [itex]\lim_{x \rightarrow a}g(x) = L > 0 \text{ or } \infty[/itex] then

    [tex]\lim_{x \rightarrow a}f(x) \cdot g(x) = \infty[/tex]

    (Note this last one works if approaching a from either the left or the right too.)

    These theorems are provable with delta-epsilon proofs.

    In you exercise let g(x) = 3x/2 and f(x) = 1/(x+5).

    I hope this helps.

    --Elucidus
     
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