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Stuck on limits

  1. Aug 11, 2009 #1
    Hello everyone

    I have a question thus:-

    Prove

    lim x -> 0 x^3/e^(x^3) -1 = 1

    I have tried to tackle it as the previous questions but this has e in it whereas the others didn't.
    If I put x = 0 we have 0/0 which is undefined.

    I just don't know how to start.....?

    Many Thanks

    James
     
  2. jcsd
  3. Aug 11, 2009 #2

    daniel_i_l

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    Gold Member

    Do you know L'Hopital's rule?
     
  4. Aug 11, 2009 #3
    Err unfortunately not. This is new to me and doesn't make much sense at the mo. I need to be able to work it out without a calculator....?

    Not sure what to do at all
     
  5. Aug 11, 2009 #4

    uart

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    Science Advisor

    Hi James, remember to use brackets where they're needed.

    The lim x -> 0, x^3/e^(x^3) -1 is not equal 1, as written it's actually equal to -1.

    However lim x -> 0, x^3/(e^(x^3) -1) = 1.

    It would be best if you know either L'hopitals rule or Taylor series expansion to do this one.
     
    Last edited: Aug 11, 2009
  6. Aug 11, 2009 #5
    Thanks for the help!

    I'm not sure if I may have 'stumbled' across something, we are given a basic limit

    lim x -> 0 (e^x -1)/x =1

    Now this looks remarkably close to my question. I'm thinking if I let u = x^3 then I have

    lim x -> 0 u/(e^u -1) which is the reciprocal of the given basic limit. I'm also sure that I have read the following property somewhere

    lim x -> c f(x) = L

    Then

    lim x -> c 1/f(x) = 1/L

    So taking my theory further as the given limit is 1 and my function with the substitution of u = x^3 is the reciprocal then 1/1 =1 which 'fits', but how do I write the limit containing my substitution.....?

    Or is the above just total b*ll*cks....!

    Many Thanks

    James
     
  7. Aug 11, 2009 #6
    I haven't followed you 100%, but I can tell you here that you need to be careful. Letting u = x^3, it LOOKS like the x disappears from the top of the equation, but u is in fact dependent on x. It's a function of x. If your goal is to use the reciprocal rule, I'm pretty sure c has to be constant with respect to x (yet u is not).
     
  8. Aug 11, 2009 #7
    Shame - thought I was onto something...

    Just haven't a clue
     
  9. Aug 11, 2009 #8
    You do not need L'Hospital's Rule or Taylor series is you've already been shown that [itex]\lim_{x\rightarrow 0}\frac{e^x-1}{x}=1[/itex].

    You are correct that if [itex]\text{If }\lim_{x\rightarrow a}f(x) = L \neq 0,\text{ then }\lim_{x\rightarrow a}\frac{1}{f(x)}=\frac{1}{L}[/itex].

    What you also need (and alluded to in your post) is the Change of Variables Theorem (or one of them at least):

    Change of Variables

    If [itex]\lim_{x\rightarrow a}g(x) = b \text{ and }\lim_{u\rightarrow b}f(u) = c[/itex]
    then [itex]\lim_{x\rightarrow a}(f\circ g)(x)=c[/itex] provided either

    (1) f is continuous at b, OR
    (2) there exists an open interval containing a such that for all [itex]x \neq a[/itex] in the interval, [itex]g(x) \neq b[/itex].

    This last theorem is rarely ever shown to calculus students (and should be) but is used with reckless abandon in examples and exercises.

    What you have is [itex]f(u) = \frac{e^u-1}{u} \text{ and }g(x) = x^3[/itex]. Unfortunately f is not continuous at 0, but x3 is invertible throughout the real line so we've satisfied part (2) of the theorem and hence

    [tex]\lim_{x\rightarrow 0}\frac{e^{x^3}-1}{x^3}=\lim_{u\rightarrow 0}\frac{e^u-1}{u}=1[/tex] (Note [itex]u\rightarrow 0 \text{ as } x\rightarrow 0[/itex]).

    The result you seek can be derived from there through the reciprocal.

    I hope this is helpful.

    --Elucidus
     
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