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Derivatives & Limits

  1. May 2, 2010 #1
    1. The problem statement, all variables and given/known data

    lim (e^(7x)-1)/x^2

    3. The attempt at a solution

    I typed in "does not exist" and it was wrong.
    Last edited: May 2, 2010
  2. jcsd
  3. May 2, 2010 #2


    Staff: Mentor

    You are missing dx. The rest is correct.
    I get 1/6 for the limit as well.
    I get the same. For this problem, the right-hand limit is infinity and the left-hand limit is -infinity, so the two-sided limit does not exist.
    I don't either. All I can suggest is to make sure the problems you posted here are the same ones that are in your book or wherever you got them.
  4. May 2, 2010 #3
    Last edited: May 2, 2010
  5. May 2, 2010 #4
    2. I suspected as much! It's the limit as x approaches NEGATIVE three.
    3. Use L'Hopital's rule...once
    Last edited: May 2, 2010
  6. May 2, 2010 #5
    Concerning #3,

    Since [tex]\lim_{x\rightarrow 0} \frac{e^(7x)-1}{x^2} = \frac{0}{0}[/tex] (called an indeterminate), you have to use L'Hopital's rule, which states that [tex]\lim_{x\rightarrow c} \frac{f(x)}{g(x)} = \lim_{x\rightarrow c} \frac{f'(x)}{g'(x)}[/tex]

    So, find the derivative of f(x) and g(x) and use apply the rule above.
  7. May 2, 2010 #6
    wow was that a stupid mistake and for number 3 people were telling me that I couldn't use l'hopital's rule that's why I didn't.

    so here's what I got
    1. for number one do I just have to put dx at the end??? like this: 7(x^4-7)^6 * 4x^3 dx or like this 28x^3(x^4-7)^6 dx (it's my last chance to put the correct answer and I don't want to put the wrong thing in)
    2. I factored it out and got lim 1/(x+3)
    then I get 1/0..so does that mean positive infinity or something
    3. this is what I did before: lim 7e^7x/2x = 7/0, so is that infinity?
    Before I was thinking I should apply the rule twice because I get 7/0 and If I did apply it twice I get: lim 49e^(7x)/2= 49/2
    Last edited: May 2, 2010
  8. May 2, 2010 #7


    Staff: Mentor

    For #1 I would go with 28x^3(x^4 - 7)^6 dx, but the other expression is equal to this, so either should be marked as correct.

    For #2, the limit is taken as x --> -3 from the right, so the limit is +infinity.

    For #3, after applying L'Hopital's rule once you get 7e^(7x)/(2x) (which is what you show). Is the left side limit (x --> 0-) the same as the right side limit (x --> 0+)?
  9. May 2, 2010 #8
    no, x --> 0- goes to negative infinity and x --> 0+ goes to positive infinity, so the limit does not exist, but I put "does not exist" and it was wrong....
    Last edited: May 2, 2010
  10. May 2, 2010 #9


    Staff: Mentor

    I don't see how that answer could be marked wrong. Can you talk to your instructor?

    Are you supposed to write something like DNE?
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