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Finding the Limit

  1. Dec 5, 2008 #1
    Hi, I've got two questions here that I'm stuck on.
    The first:

    2. Relevant equations
    Find the Limit as x->0 of [(e^4x)-1-4x]/x^2


    3. The attempt at a solution
    So far:
    I got that the Limit as x->0 is indeterminate form 0/0 so I tried L'Hopital's:

    to find the Limit as x->0 of [(e^4x)-4]/2x

    but that gave me a Limit of -3/0 . So I think that means that there's no limit. But the book insists that the Limit is 8. and I have no idea how it got there.


    the second question I'm confused about is

    2. Relevant equations
    Find the Limit as x->2 of (x+2)/[(x-2)^4] , and explain why L'Hopital's Rule does not apply.


    3. The attempt at a solution
    so
    The Lim x->2 = 4/0 so it's not 0/0 indeterminate so L'Hopital's isn't applicable.

    But after that I'm a bit stuck on how to get the Limit. I know that because x can't equal 0 in the denominator that x cannot equal 2. So there's a vertical asymptote there.
    So...I know I can look at it from each side
    The Limit as x->0- and as x->0+ .
    So-
    The Limit as x->0+=infinity
    The Limit as x->0-=infinity.
    is that right? and is their an easier way to find that then just plugging in numbers near x=2?

    -Thanks, any help would be much appreciated!
     
  2. jcsd
  3. Dec 5, 2008 #2

    nicksauce

    User Avatar
    Science Advisor
    Homework Helper

    The derivative of e^(4x) is 4e^(4x)... correct this and it should work out.
     
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