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Using L'Hopitals rule

  1. Jun 20, 2005 #1
    As a revision exercise we have been given.

    find

    [tex] \lim_{\substack{x\rightarrow infty} } (x+e^x) ^ (2/x) [/tex]

    ****Darn It it wont work, the above is (x+e^x) all raised to (2/x)


    our lecturer has given us a walkthrough the solution which is done by letting the function = y and taking logs of both sides, which gives a limit of infinity/infinity, so we can use L'hopitals rule to derive a limit for lny to infinity then invert it onto the index of e for the answer.

    This all makes sense to me, I just have one problem with the working.

    the function for lny comes out to [tex] \frac {2ln(x+e^x)}{x} [/tex]

    and the derivative of this is given as [tex]\frac {2(1+e^x)/(x+e^x)}{1}[/tex]

    The numerator makes sense to me to be the derivative of 2ln(x+e^x) but I don't understand how the denominator becomes so simple. It looks to me like the solution just differentiates the top and then differentiates the bottom?
    :uhh:
     
    Last edited: Jun 20, 2005
  2. jcsd
  3. Jun 20, 2005 #2
    You got the derivative for the lny function wrong, use the quotient rule.
     
  4. Jun 20, 2005 #3

    Yes, whozum that is the point of my post, if that derivative is wrong then the following solution given in my notes is wrong.

    So I am very confused.

    The derivative given by the quotient rule is quite messy and doesn't simplify the search for a limit like the derivative given in my notes has, I wondered if perhaps the said derivative using the quotient rule simplifies to the one given in my notes, but I struggle sometimes with log simplifications..

    Here is the derivative I got using the product rule which I though would be simpler than the quotient rule.

    ln [tex] y' = \frac {-2ln(x+e^x)} {x^2} + \frac {2(1+e^x)} {x} [/tex]
     
    Last edited: Jun 20, 2005
  5. Jun 20, 2005 #4
    Theres only one derivative to that function, and its obtained by the quotient rule.

    [tex] \frac {2ln(x+e^x)}{x} [/tex]

    [tex] \frac{d}{dx} \ above \ = \frac{(\frac{2x}{x+e^x})(1+e^x) - 2ln(x+e^x)}{x^2} [/tex]

    Show me what you have in your notes.
     
    Last edited: Jun 20, 2005
  6. Jun 20, 2005 #5
    Just two things whozum.

    1. I don't think you understood my question, I am trying to tell you that this:

    [tex]\frac {2(1+e^x)/(x+e^x)}{1}[/tex]

    is given to me by my lecturer as the solution. **It is not my solution. And I need to exhaust all the possibilites that it is right (including simplifictions that I hadn't thought of or a flaw in my understanding of the problem) before I write it off.


    Secondly I do think that the derivative can be rightly obtained using the product rule. Unless I am wrong that.

    [tex] \frac {2ln(x+e^x)}{x} = 2ln(x+e^x )* (1/x) [/tex]

    ?


    Hi again I just saw your edit asking for what is in my notes. Basically the notes tell me that after I derive,

    [tex]\frac {2(1+e^x)/(x+e^x)}{1}[/tex]

    the limit is still infinity/infinity so I apply L'Hopitals a second time to get

    [tex]2 \frac {(1+e^x)}{(1+e^x)}[/tex]

    Which gives a limit of lny to infinity of 2 and the limit of the original equation as e^2 but I hadn't really worked on that part yet.

    The thing that is coming to mind is wondering wether my lecturer is showing me to use the Algebra of limits theorem and take the derivatives separately and their respective limits separately also by relying on that rule.

    It seems to be possible but am not quite able to get my head around how it's done. Just to note, the actual question states: using any appropriate technique evaluate the limit, I take that to be a hint that you apply a few rules simultaneously to get the answer.
     
    Last edited: Jun 20, 2005
  7. Jun 20, 2005 #6

    Curious3141

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    Whozum, you are incorrect. The notes are right. The method proposed by the lecturer is to call the whole original expression y, and to define a new function z = lny, and to find the limit of z by LH. It's perfectly valid.

    [tex]z = \ln y = \frac {2\ln(x+e^x)}{x}[/tex]

    The quotient rule doesn't apply here, only the chain rule. Because they're only finding the derivative of the numerator, which is [itex]\frac{d}{dx}2\ln(x+e^x) = 2\frac{1 + e^x}{x + e^x}[/itex]. The derivative of the denominator is [itex]\frac{d}{dx}x = 1[/itex]

    We go through the steps :

    [tex]\lim_{x \rightarrow \infty}z = \lim_{x \rightarrow \infty}2\frac{1 + e^x}{x + e^x}[/tex]

    Applying LH rule 2 more times in succession like so :

    [tex]\lim_{x \rightarrow \infty}2\frac{1 + e^x}{x + e^x} = \lim_{x \rightarrow \infty}2\frac{e^x}{1 + e^x} = \lim_{x \rightarrow \infty}2\frac{e^x}{e^x} = 2[/tex]


    yields the limit of z as 2, which means the limit of the orig. expression is [itex]e^2[/itex].
     
    Last edited: Jun 20, 2005
  8. Jun 20, 2005 #7

    Curious3141

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    Your application of LH rule the second time is incorrect, please see my working.
     
  9. Jun 20, 2005 #8
    This is what I was tryin to wrap my head around but I couldnt quite get to it. I actually thought of it while i was going potty, but you've already got it.
     
  10. Jun 20, 2005 #9

    Thankyou for validating my notes for me Curious, could you please explain this step in more detail?
     
  11. Jun 20, 2005 #10

    I didn't get to that part yet, that is just what my notes gave as the next step.
     
  12. Jun 20, 2005 #11

    VietDao29

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    monet A, read your notes again thoroughly.
    1. L' Hospitals rule:
    [tex]\lim_{x \rightarrow \alpha} \frac{f(x)}{g(x)}[/tex]
    And you meet : [itex]\frac{0}{0}[/itex] Or [itex]\frac{\infty}{\infty}[/itex]
    g'(x) <> 0.
    And you'll have : [tex]\lim_{x \rightarrow \alpha} \frac{f(x)}{g(x)} = \lim_{x \rightarrow \alpha} \frac{f'(x)}{g'(x)}[/tex]
    2. To type something like [itex]x^{2x}[/itex]. Use {...}.
    3. What Curious is doing is to find the limit of z as x tends to infinity.
    Viet Dao,
     
  13. Jun 20, 2005 #12

    Curious3141

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    Actually, the easiest way (IMHO) to do this limit is with the binomial expansion and maybe a limited one time use of LH.

    [tex]y = (x + e^x)^{\frac{2}{x}} = e^2(1 + xe^{-x})^{\frac{2}{x}}[/tex]

    By LH rule, the expression [tex]xe^{-x} = \frac{x}{e^x}[/tex] tends to the limit 0 as x grows without bound.

    Since [itex]xe^{-x}[/tex] is "small" as we tend to the limit, we're justified in using a first order approximation in the binomial expansion of y, like so :

    [tex]y[/tex] approximates [tex] e^2 [1 + (\frac{2}{x})(xe^{-x})] = e^2(1 + 2e^{-x})[/tex], the limit of which is obviously [itex]e^2[/itex]

    EDIT : The Tex came out wrong because I couldn't remember the code for approx equal. Yeah, I'm dumb. :frown:
     
    Last edited: Jun 20, 2005
  14. Jun 20, 2005 #13

    Curious3141

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    Remember that [itex]\ln a^b = b\ln a[/itex] ? This is just the same thing, "bring the power down" and see whether you can get the same expresssion.
     
  15. Jun 20, 2005 #14
    Meaning that the two functions can be differentiated separately and that is entirely valid? I think I understand that now, my problem was probably that in my head I was seeing ...

    [tex] \frac{f'(x)}{g'(x)}[/tex]

    as

    [tex] (\frac{f(x)}{g(x)} )' [/tex]

    **I have written and rewritten this and I swear my latex does not have x^2x in it anywhere. My code says. I saw f'(x)/g'(x) in my head as (f(x)/g(x))' what is going on with this??? Ahh refreshed the page and the correction is there at last
    :approve:

    Thankyou


    I knew that. You're the third person to miss that I understood that part and thik that it was my problem.

    :rofl:
     
    Last edited: Jun 20, 2005
  16. Jun 20, 2005 #15

    Wrong step curious.

    I mean the step we all keep discussing but noone seems to understand that I dont get it.

    Why are we differentiating the numerator and the denominator separately? was my question, but I think I have finally understood the answer in my above post to Vietdao, can you check that and tell me if I am right.

    Oh, and we are not using binomial expansions in this course.
     
  17. Jun 20, 2005 #16

    Curious3141

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    That's basically LH rule. VietDao has already explained it, and you can just do a search on google for the formal statement of the rule and the conditions that the numerator and denominator functions must satisfy before the rule is valid.

    Basically, just differentiate the top, differentiate the bottom, and see whether the result makes any more sense. That's all there is to LH. :wink:

    I would've thought you'd have covered binomial expansions before LH. In any case, as the saying goes, there are many ways to skin a cat. The essence of good math is finding the best way. :biggrin:
     
  18. Jun 20, 2005 #17

    VietDao29

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    Hmm, you should refer your book for L'Hospital rule. I don't think you have understand it correctly. In L'Hospital rule, you want to fnd the differentiation of the numerator and denominator separately, while in quotient rule, you are differentiate the whole number.
    What I mean is:
    [tex]\frac{f'(x)}{g'(x)} \neq \left( \frac{f(x)}{g(x)} \right) '[/tex]
    Viet Dao
     
    Last edited: Jun 20, 2005
  19. Jun 20, 2005 #18

    Yes thankyou Vietdao, if you look back to my post I said that I didn't realise before that

    [tex] L'hopitals Rule \neq \left( \frac{f(x)}{g(x)} \right) '[/tex]

    but I do now remember that

    [tex] L'hopitals Rule = \frac{f'(x)}{g'(x)} [/tex]

    Just to reassure both you and curious that I do understand it all now and you have managed to help me with all that I needed.

    :smile:
     
    Last edited: Jun 20, 2005
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