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I need some limit help

  1. Sep 17, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the limit:
    lim (e2x-1)/tanx
    x[tex]\rightarrow[/tex]0
     
  2. jcsd
  3. Sep 17, 2009 #2

    Mark44

    Staff: Mentor

    This is the indeterminate form [0/0]. Do you know L'Hopital's Rule?
     
  4. Sep 17, 2009 #3
    No, we have not learned this rule yet. So, since it is indeterminate form, the limit would be 0, or does not exist?
     
  5. Sep 17, 2009 #4

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It means you need to be careful. Knowing it's of the form 0/0 only tells you that it could be any value, and you need another method of evaluating it than the standard split the limit up trick.

    Basically what you have is

    [tex] \lim \frac{e^{2x}-1}{sin(x)}*cos(x)[/tex]

    the cos(x) part stays near 1, so is fine. You need a trick to manipulate [tex] \frac{e^{2x}-1}{sin(x)}[/tex] to make it more amenable to calculation, either through Taylor series around 0 or some other method
     
  6. Sep 17, 2009 #5

    Mark44

    Staff: Mentor

    If L'Hopital's Rule hasn't been presented yet, it's a safe bet that you don't know about Taylor's series or Maclaurin series yet. Other than these techniques, I don't have any advice. Is this a problem in the book, or is it one your instructor gave for extra credit? The best advice I can offer is to contact him/her for a hint.
     
  7. Sep 18, 2009 #6

    VietDao29

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    Homework Helper

    Well, in limit problems, when dealing with ln, and exponential function, there are 2 well-known limits you should memorize (I know it should be somewhere in your book, and proving it is not very hard, you can give it a try if you want):

    1. [tex]\lim_{x \rightarrow 0} \frac{e ^ x - 1}{x} = 1[/tex] (Limit 1)
    2. [tex]\lim_{x \rightarrow 0} \frac{\ln(1 + x)}{x} = 1[/tex] (Limit 2)

    When working with sin(x), there's one well-known (the most well-known, I may say) limit:

    [tex]\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1[/tex]. (Limit 3)

    Ok, so now, back to your problem. First, we can try to expand every terms out:

    [tex]\lim_{x \rightarrow 0} \frac{e ^ {2x} - 1}{\tan(x)} &= \lim_{x \rightarrow 0} \left( {\color{red}\frac{e ^ {2x} - 1}{\sin(x)}} \times {\color{blue}\cos(x)} \right)[/tex]

    Pay attention to the red term, it's (0/0), hence one of the Indeterminate Forms, right? The blue term will tend to cos(0) = 1, which does not cause much troubles. So, here the problematic term is the red one.

    Now, think of a way to make use of (Limit 1), and (Limit 3) to eliminate the Indeterminate Form 0/0. It should be easy if you know the trick. Well, let's see how far can you go. :)
     
  8. Sep 18, 2009 #7
    Note that [tex] e^{2x} - 1 = \int_0^{2x} e^u du = 2xe^c [/tex] for some c in between 0 and 2x, by the MVT of integral calculus. Thus [tex] \frac{e^{2x} - 1}{sinx} cosx = \frac{2xe^c}{sinx} cosx = \frac{x}{sinx} 2e^c cosx [/tex].

    Remember that [tex] \frac{x}{sinx} [/tex] goes to 1 for x approaching 0, so now let x go to 0 in the entire new expression. Also remember that c must also go to 0 as x goes to 0.
     
  9. Sep 18, 2009 #8

    Mark44

    Staff: Mentor

    This advice might not be helpful to the OP if he is studying differential calculus and hasn't gotten to integral calculus yet.
     
  10. Sep 18, 2009 #9
    e^2x - 1 = (e^x)^2 - 1 = (e^x - 1)(e^x + 1)

    => (e^2x - 1)/tanx = (e^x-1)/sinx * (e^x+1)cosx.
    What can you do now to be able to use the known limits? For example multiply by 1=a/a=b/b=c/c=...
     
  11. Sep 18, 2009 #10

    lurflurf

    User Avatar
    Homework Helper

    for x!=0
    [exp(2x)-1]/tan(x)=[(exp(x)-1)/x][exp(x)+1][cos(x)]/[sin(x)/x]
    you should know
    exp'(0)=lim [(exp(x)-1)/x]=1
    sin'(0)=lim [sin(x)/x]=1
    exp(0)=1
    cos(0)=1
     
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