Calculus Help Stuck Part 2?

b.Determine the lim x-->4 (x^2+x-20/8-2x)

What Im stuck on is this

f(4)= 4^2+4-20 / 8-2(4)
f(4)= 0/0

x^2+x-20/8-2x = (x-4)(x+5)/-2(x-4)

What do I do next I'm so confused ik i would eliminate the x-4 from numerator and denominator but what would i do with the -2 that belongs to (x-4) in denominator.

I feel lost??

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asked by Marc
  1. Step1 for me in any limit question is actually sub in the approach value
    If you get a real number, that's your answer,
    all done!

    so for x = 4, we get 0/0, ok, we got work to do

    step: if you get 0/0 for simple rational expressions like the above, IT WILL FACTOR. Sometimes you have to do some fancy stuff first, but eventually it will factor.
    For ours, it is simply

    Step3: Factor, simplify and repeat Step1

    lim x-->4 (x^2+x-20/8-2x) , x--->4
    = lim (x+5)(x-4)/(2(4-x)) , notice x-4 and 4-x are opposites, so (x-4)/(4-x) = -1

    = lim -1(x+5)/2 , x ---> 4
    = -1(4+5)/2
    = -4.5

    Here is a cool trick for limits.
    Pick a number very close to the approach value, e.g. x = 4.001
    stick that in your calculator's memory
    and evaluate the original expression
    I got -4.5005
    My answer is correct

    I use this often before I start my algebra to predict what answer I should get.

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    posted by Reiny
  2. Ok thanks that helped a lot!

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    posted by Marc
  3. so basically to get the negative 1 i just divide -4/4 correct?

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    posted by Marc
  4. careful:
    what happened to your x's in
    (x-4)/(4-x) ?

    What really happened is this

    (x-4)/(4-x)
    = -1(4-x)/(4-x)
    = -1[ (4-x)/(4-x) ]
    = -1[ 1 ]
    = -1

    Any number divided by its opposite = -1

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    posted by Reiny

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