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Solving problem l'Hospital Rule

  1. Nov 19, 2009 #1
    1. The problem statement, all variables and given/known data
    solve


    2. Relevant equations
    f(x)/g(x) = f'(x)/g'(x)


    3. The attempt at a solution
    1. [tex] $\displaystyle \Large \lim _{x\to \infty }(1+3x)^{1/x}$ [/tex] = 0
    because 1+3x is infinity..
    infinity^(1/infinity)
    small#/big# = 0
    so infinity^0 = 1

    2. [tex] $\displaystyle \Large \lim _{x\to 1}\frac{\ln x}{x^2+x-2}$ [/tex] = 1/3
    (1/x) / (2x+1).....

    3. [tex] $\displaystyle \Large \lim _{x\to \infty }\frac{x^2}{e^x}$ [/tex] = infinity
    cuz if plugging in infinity, i get infinity2/infinity = infinity

    im i right so far?
     
  2. jcsd
  3. Nov 19, 2009 #2

    Mark44

    Staff: Mentor

    1 and 3 are wrong, but 2 is right. The following are all indeterminate forms, which means that they aren't numbers, and the limits that are in these forms can come out to be anything.
    [tex][\infty^0],~\left[\frac{0}{0}\right],~\left[\frac{\infty}{\infty}\right],~[\infty - \infty][/tex]

    Use L'Hopital's Rule on #3, and you should see that the limit isn't infinite. #1 takes a fair amount of time to explain, so see if your textbook has an example that's similar. It involves taking logs before you use L'Hopital's Rule.
     
  4. Nov 19, 2009 #3
    for #3.. if i plug in e^infinity it says error..
    so it should be undefined right?

    for #1
    the rule states:
    "If the expression
    [tex]
    \frac{\lim_{x \rightarrow a} f(x)}{\lim_{x \rightarrow a} g(x)}
    [/tex]
    has the form 0/0 or infinity/infinity, then l'Hôpital's rule states that
    f(x)/g(x) = f'(x)/g'(x)
    provided that that second limit exists."

    what about
    [tex]
    [\infty^0],~[\infty - \infty]
    [/tex]

    i know that for
    [tex]
    [\infty - \infty]
    [/tex]
    it should be 1/x - 1/sinx or something like that
     
  5. Nov 19, 2009 #4

    Mark44

    Staff: Mentor

    No, not for #3, and besides e^infinity is not a number. I don't know what you're plugging that into, but #3 has a limit. Use L'Hopital's Rule. If necessary, use it again.
    L'Hopital's Rule doesn't apply to these indeterminate forms. For the first one, the usual strategy is to let y = your expression, and then take the log of both sides. That will typically give you a product that you can rewrite as a quotient that is one of the two indeterminate forms [0/0] or [infinity/infinity], and then use L'Hopital's Rule. For the second one, you can sometimes rearrange things to get a quotient on which to use L'Hopital's Rule.
    Sort of.
    [tex]\lim_{x \rightarrow 0}\left[ 1/x - 1/sin(x)\right][/tex] is an indeterminate form of this type.
     
  6. Nov 19, 2009 #5
    oh i get it now.
    so #1 would be lny = 1/3 ln(1+3x)
    and after solving i get 'infinity'

    for #3 it is 2/e^infinity ==> 0
    so #3 would be 0

    im i correct?

    and what if i have a question like this:
    [tex] $\displaystyle \Large \lim _{x\to \infty}(\sqrt{x^2+x}-x)$ [/tex]
    i get [tex] [\infty - \infty] [/tex]
    now how do i use [tex]
    \lim_{x \rightarrow 0}\left[ 1/x - 1/sin(x)\right]
    [/tex]?

    would it be, (sinx-x)/xsinx and now apply the rule f'(x)/g'(x) ?
     
  7. Nov 19, 2009 #6

    Mark44

    Staff: Mentor

    #1 is still wrong, but #3 is right, so now you have two out of the three of them. I'll talk about #1 at the end of this post.

    For this problem
    [tex] \lim _{x\to \infty}(\sqrt{x^2+x}-x)[/tex]
    multiply by 1 in the form of [sqrt(x^2 + x) + x]/[sqrt(x^2 + x) + x]. That will give you a quotient that might be in a form so that you can use L'Hopital's Rule.


    For #1, let y = (1 + 3x)1/x
    Take the log of both sides : ln y = ln[(1 + 3x)1/x]
    Your result of 1/3 ln(1 + 3x) is incorrect. What should you get?
    Notice that there is no limit process going on. After a while, we'll bring limits back in.
     
  8. Nov 19, 2009 #7
    oh that was my mistake.. for #1 i will get 1/x ln(1 + 3x) and since 1/x is 0 that makes the whole thing 0.. so the final answer is 0.

    for the question
    [tex] \lim _{x\to \infty}(\sqrt{x^2+x}-x)[/tex]

    i get x/(sqrt{x^2+x}+x)
    from there if i use the rule, i get 1/(0.5(x2+x)-0.5*(2x+1)+1)
    would that equal 0? cuz 1/infinity?
     
  9. Nov 19, 2009 #8

    Mark44

    Staff: Mentor

    In fact, the limit in your last problem is 1/2,not 0, so sorry to burst your bubble.
    You don't need L'Hopital's Rule for this problem.
    [tex] \lim _{x\to \infty}(\sqrt{x^2+x}-x)~=~ \lim _{x\to \infty}(\sqrt{x^2+x}-x)\frac{\sqrt{x^2+x}+x}{\sqrt{x^2+x}+x}~=~\lim _{x\to \infty}\frac{x}{\sqrt{x^2+x}+x}[/tex]
    Factor x^2 out of the two terms in the radical, and bring it out of the radical as x.
    [tex]\lim _{x\to \infty}\frac{x}{x\sqrt{1+1/x}+x}~=~\lim _{x\to \infty}\frac{x}{x(\sqrt{1+1/x}+1)}~=~ 1/2[/tex]

    In the next-to-last step you can cancel the x in the numerator with the x in the denominator, leaving you with 1 in the numerator and and expression that approaches 2 in the denominator.

    For #1, you're closer, but no cigar, and also the limit is not zero.
    If y = ln(1 + 3x)1/x
    ln y = (1/x)*ln(1 + 3x)*3 -------chain rule!!!
    = (3*ln(1 + 3x))/x

    This is true for all reasonable x (i.e., x > -1/3), so it's also true in the limit as x gets large without bound.

    So lim lny = lim (3*ln(1 + 3x))/x with the limit taken as x gets large without bound, in which case both numerator and denominator get large without bound, so this is the indeterminate form [infinty/infinity].

    See if you can finish this off. Here's what you need to do:
    1. Evaluate the limit on the right, keeping in mind that this is the indeterminate form [infinity/infinity]. Call this value A.
    2. If lim lny = A, then ln(lim y) = A as well. As long as the function is continuous (the ln function is), you can switch the order of the function and the limit.
    3. Since ln (lim y) = A, what is lim y? Keep in mind that what we wanted all along was lim y = lim (1 + 3x)1/x.
     
  10. Nov 19, 2009 #9
    how did you get
    "ln y = (1/x)*ln(1 + 3x)*3 -------chain rule!!!"?

    i got ln y = (1/x)*ln(1 + 3x) = ln(1 + 3x) / x
    ==> f'(x)/g'(x) ==>(3/(1+3x))/x = 3x/(1+3x)
    ==> plug in 'infinity'
    3infinity / 1+3infinity = 3/3 = 1

    or if it would be infinity/infinity which means i have to apply the rule again,
    3/3 = 1 anyways..

    in yours
    ln y = (1/x)*ln(1 + 3x)*3 where is the last 3 coming from?
     
    Last edited: Nov 19, 2009
  11. Nov 20, 2009 #10

    Mark44

    Staff: Mentor

    Sorry about that - was thinking about the next step and got ahead of myself.
    That should be:
    If y = ln(1 + 3x)1/x
    ln y = (1/x)*ln(1 + 3x) = [ln(1 + 3x)]/x

    So lim lny = lim (ln(1 + 3x))/x with the limit taken as x gets large without bound, but no you never, never "plug in" infinity. That's what limits are for.

    This is how you write it so that you aren't plugging in infinity:
    [tex]\lim_{x \to \infty} ln y~=~\lim_{x \to \infty} \frac{ln(1 + 3x)}{x}~~~~\left[\frac{\infty}{\infty} \right][/tex]
    [tex]\Rightarrow ln(\lim_{x \to \infty} y)~=~\lim_{x \to \infty} \frac{\frac{3}{1 + 3x}}{1}~=~0\right][/tex]

    In the first line above, the [itex][\infty/\infty][/itex] notation is just a reminder to myself that I have an indeterminate form of the type I can use L'Hopital's Rule on. Don't clutter up your work with the f'(x)/g'(x) stuff. If you feel you need to include it, put it off to the side somewhere so that it's not in the main flow of your logic.

    In the second line I have switched the order of the limit operation and log function, and have applied L'Hopital's Rule -- that's where the chain rule came in.

    The second equation above says that the log of the limit I want is zero, so what is the value of the limit? IOW, ln[lim (1 + 3x)1/x)] = 0, so what's the value of this limit?
     
  12. Nov 20, 2009 #11
    ln[lim (1 + 3x)1/x)] = 0
    ==> lim (1 + 3x)1/x)= e^0
    ==> lim (1 + 3x)1/x)= 1

    thanks i got it!
    i really appreciate ur help!
     
  13. Nov 20, 2009 #12

    Mark44

    Staff: Mentor

    Sure, you're welcome. And I appreciate your tenacity. You made a lot of mistakes along the way in this thread, but you hung in there and kept working at it and trying to do better, which is to be admired.
     
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