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Math Help - Trouble with a L'hoptial question:

  1. #1
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    Trouble with a L'hoptial question:

    Thank you all again for your help. I've been studying with the L'hopital rule and for the most part haven't had any difficulty with it except for this one question. You guys have helped tremendously with this self-study course thus far.

    It may be more of an algebra issue for me than anything else, but here's the question:

    Use L'Hopital's rule to evaluate % MathType!MTEF!2!1!+-<br />
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% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaac  i<br />
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% amiEaiabgkHiTmaakaaabaGaamiEamaaCaaaleqabaGaaGOmaa  aaki<br />
% abgUcaRiaadIhaaSqabaaaaa!4335!<br />
$$\mathop {\lim }\limits_{x \to \infty } x - \sqrt {{x^2} + x} $$

    I am having difficulty getting this into a quotient where L'Hopital applies. For that reason I have two questions about this problem:

    1) Is % MathType!MTEF!2!1!+-<br />
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% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqt  ubsr<br />
% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9<br />
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% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaac  q<br />
% GHEisPcqGHsislcqGHEisPaeaacqGHEisPaaaaaa!3B43!<br />
$${{\infty  - \infty } \over \infty }$$ an indeterminate form? To me it would be undefined because of the \infty - \infty, which isn't quite the same as being in an indeterminate form for L'hopital as far as I know (which I'm more than willing to be wrong about).

    2) What would be the proper way to set this up for L'hopital?

    Thank you again.
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  2. #2
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    2) Multiply the whole expression inside the limit by:

    \frac{x+\sqrt{x^2 + x}}{x+\sqrt{x^2 + x}}

    Regards.
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  3. #3
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    Of course!!!! I don't know why I didn't think of that. That was a forehead slapping experience much like I expected it to be. Thank you so much!

    The limit then works out to infinity as I would expect.
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  4. #4
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    Am I correct about my first question, however? Undefined vs. an indeterminate form?
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  5. #5
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    I also want to verify my procedure...

    In the course of the problem I come across this:

    <br />
$${{4x + 1} \over {1 + (2x + 1)({1 \over {2\sqrt {{x^2} + x} }})}}$$

    I state that the $${{1 \over {2\sqrt {{x^2} + x} }}}$$ as x approaches infinity is equal to 0, and then use that 0 to cancel out the 2x+1, leaving me with just 4x+1 for an answer of infinity.

    Correct?
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  6. #6
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    I don't think your algebra is correct there. Moreover, I don't think the limit is infinite. After you've done what p0oint says, what do you get?
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  7. #7
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    You were right, I made a sign error when expanding and ended up with 2x^2 + x instead of just -x for my numerator.

    This problem has been quite frustrating for me.

    % MathType!MTEF!2!1!+-<br />
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% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqt  ubsr<br />
% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9<br />
% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x<br />
% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaac  i<br />
% GGSbGaaiyAaiaac2gaaSqaaiaadIhacqGHsgIRcqGHEisPaeqa  aOGa<br />
% amiEaiabgkHiTmaakaaabaGaamiEamaaCaaaleqabaGaaGOmaa  aaki<br />
% abgUcaRiaadIhaaSqabaaaaa!4335!<br />
$$\mathop {\lim }\limits_{x \to \infty } x - \sqrt {{x^2} + x} $$

    % MathType!MTEF!2!1!+-<br />
% feaagyart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuL  wBLn<br />
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqt  ubsr<br />
% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9<br />
% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x<br />
% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaac  q<br />
% GHEisPcqGHsislcqGHEisPaeaacqGHEisPaaaaaa!3B43!<br />
$${{\infty  - \infty } }$$ is undefined

    % MathType!MTEF!2!1!+-<br />
% feaagyart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuL  wBLn<br />
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqt  ubsr<br />
% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9<br />
% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x<br />
% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg  k<br />
% HiTmaakaaabaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUca  Riaa<br />
% dIhaaSqabaGccaGGOaWaaSaaaeaacaWG4bGaey4kaSYaaOaaae  aaca<br />
% WG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaamiEaaWcbeaa  aOqa<br />
% aiaadIhacqGHRaWkdaGcaaqaaiaadIhadaahaaWcbeqaaiaaik  daaa<br />
% GccqGHRaWkcaWG4baaleqaaaaakiaacMcaaaa!48E0!<br />
$$(x - \sqrt {{x^2} + x}) ({{x + \sqrt {{x^2} + x} } \over {x + \sqrt {{x^2} + x} }})$$

    % MathType!MTEF!2!1!+-<br />
% feaagyart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuL  wBLn<br />
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqt  ubsr<br />
% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9<br />
% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x<br />
% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaac  a<br />
% WG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaamiEamaakaaa  baGa<br />
% amiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaadIhaaSqaba  Gccq<br />
% GHRaWkcaWG4bWaaOaaaeaacaWG4bWaaWbaaSqabeaacaaIYaaa  aOGa<br />
% ey4kaSIaamiEaaWcbeaakiabgkHiTiaadIhadaahaaWcbeqaai  aaik<br />
% daaaGccqGHsislcaWG4baabaGaamiEaiabgUcaRmaakaaabaGa  amiE<br />
% amaaCaaaleqabaGaaGOmaaaakiabgUcaRiaadIhaaSqabaaaaa  aa!4E34!<br />
$${{{x^2} - x\sqrt {{x^2} + x}  + x\sqrt {{x^2} + x}  - {x^2} - x} \over {x + \sqrt {{x^2} + x} }}$$

    % MathType!MTEF!2!1!+-<br />
% feaagyart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuL  wBLn<br />
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqt  ubsr<br />
% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9<br />
% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x<br />
% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0Yaa  S<br />
% aaaeaacaWG4baabaGaamiEaiabgUcaRmaakaaabaGaamiEamaa  Caaa<br />
% leqabaGaaGOmaaaakiabgUcaRiaadIhaaSqabaaaaaaa!3DB6!<br />
$$ - {x \over {x + \sqrt {{x^2} + x} }}$$

    % MathType!MTEF!2!1!+-<br />
% feaagyart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuL  wBLn<br />
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqt  ubsr<br />
% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9<br />
% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x<br />
% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0Yaa  S<br />
% aaaeaacqGHEisPaeaacqGHEisPaaaaaa!39D2!<br />
$$ - {\infty  \over \infty }$$ is indeterminate

    % MathType!MTEF!2!1!+-<br />
% feaagyart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuL  wBLn<br />
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqt  ubsr<br />
% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9<br />
% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x<br />
% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaac  i<br />
% GGSbGaaiyAaiaac2gaaSqaaiaadIhacqGHsgIRcqGHEisPaeqa  aOGa<br />
% eyOeI0YaaSaaaeaacaWG4baabaGaamiEaiabgUcaRmaakaaaba  Gaam<br />
% iEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaadIhaaSqabaaa  aOWa<br /> <br /> <br />
$$\mathop {\lim }\limits_{x \to \infty }  - {x \over {x + \sqrt {{x^2} + x} }}\mathop  = \limits^{L'H}  - {1 \over {1 + {1 \over 2}{{({x^2} + x)}^{ - {1 \over 2}}}(2x + 1)}} =  - {1 \over {1 + (2x + 1)({1 \over {2\sqrt {{x^2} + x} }})}}$$

    So using the method I had before, that would yield a limit of -1 (not infinity), but I'm not sure if it's legal to declare that % MathType!MTEF!2!1!+-<br />
% feaagyart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuL  wBLn<br />
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqt  ubsr<br />
% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9<br />
% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x<br />
% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikamaal  a<br />
% aabaGaaGymaaqaaiaaikdadaGcaaqaaiaadIhadaahaaWcbeqa  aiaa<br />
% ikdaaaGccqGHRaWkcaWG4baaleqaaaaakiaacMcaaaa!3CC7!<br />
$${({1 \over {2\sqrt {{x^2} + x} }})}$$ =0 as x approaches infinity.

    When I check my answer on my TI-89, it's giving me an answer of -1/2 . I'm not following it.

    What other horrific errors have I made?

    And was my methodology correct even if my algebra sucked?
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  8. #8
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    Everything looks great up to and including your application of l'Hospital's Rule. However, when you let x\to\infty, your method is a little lacking. It is correct to say that \lim_{x\to\infty}\frac{1}{2\sqrt{x^{2}+x}}=0. However, that fact does not help you here, because the fact is, you actually have to take the limit, in the denominator, of \frac{2x+1}{2\sqrt{x^{2}+x}} as x\to\infty, which is like \infty/\infty. Therefore, the nice quotient rule of limits does not apply because the individual limits do not exist. How should you evaluate this new limit, do you think?
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  9. #9
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    Quote Originally Posted by p0oint View Post
    2) Multiply the whole expression inside the limit by:

    \frac{x+\sqrt{x^2 + x}}{x+\sqrt{x^2 + x}}

    Regards.
    Once you do this, it actually makes a lot more sense to just evaluate the limit directly using the sandwich theorem. To be honest, I tried doing it with L'Hospital's Rule and just got back something I wanted to sandwich anyways.
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  10. #10
    MHF Contributor chisigma's Avatar
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    L'Hopital rule is directly applicable only for the 'indeterminate forms' \frac{0}{0} or \frac{\infty}{\infty}. For an 'indeterminate form' of the type \infty - \infty You have to use the identity...

    \displaystile f(x) - g(x) = \frac{\frac{1}{g(x)} - \frac{1}{f(x)}}{\frac{1}{ f(x)\ g(x)}} (1)

    In your case is f(x) = x and g(x)= \sqrt {x^{2} + x} and \lim_{x \rightarrow \infty} f(x) = \lim_{x \rightarrow \infty} g(x) = \infty so that is...

     f(x) - g(x) = \frac{\frac{1}{\sqrt {x^{2} + x}} - \frac{1}{x}}{\frac{1}{ x\ \sqrt{x^{2} + x}}} (2)

    Now You can apply l'Hopital rule to the expression (2)... not a very pratical way to find the limit that is - \frac{1}{2} ...

    Kind regards

    \chi \sigma
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  11. #11
    MHF Contributor Also sprach Zarathustra's Avatar
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    l'Hopital in French is hospital!
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  12. #12
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    Thank you for the responses... I didn't think I'd be so lucky as to be able to cancel that one out Ackbeet, however I thought I'd try it and see if I was wrong since it's felt like I've tried everything else. I tried to continue to apply L'Hopital from where I got in your question but it was a mess as I had to rearrange the function to take care of the standalone 1 in the numerator. After an application or two, it started to get uglier and uglier!

    Chisigma, that identity is a new one for me (not surprisingly). All I've been doing thus far is putting everything into a quotient and then chugging away with L'Hopital until the problem yields me a limit. With that method, however, I've been running what seems to be endless applications of L'Hopital with ridiculous fractions that I'm undoubtedly destroying with spotty algebra. I'm sure that will help quite a bit. I've brought it back to my professor since I have hunted through my textbook, notes relating to the assignment, and everything in hopes to find some sort of reference to that identity that I missed, but I haven't. Is that a common one?

    It seems that the consensus is that L'Hopital sucks for this function. Perhaps my professor made a mistake on this question.
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  13. #13
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    Malaclypse: don't give up! I wouldn't use l'Hospital's rule twice on your original problem, only once. Try this on for size:

    \lim_{x\to\infty}(x-\sqrt{x^{2}+x})
    =\lim_{x\to\infty}\left[(x-\sqrt{x^{2}+x})\frac{x+\sqrt{x^{2}+x}}{x+\sqrt{x^{  2}+x}}\right]
    =\lim_{x\to\infty}\left(\frac{x^{2}-x^{2}-x}{x+\sqrt{x^{2}+x}}\right)
    =-\lim_{x\to\infty}\left(\frac{x}{x+\sqrt{x^{2}+x}}\  right)
    =-\lim_{x\to\infty}\left(\frac{1}{1+\sqrt{1+1/x}}\right).

    That last step I got from dividing numerator and denominator by x. Now, I think, you can use your limit theorems to finish. Do you see your way forward?

    [EDIT]: I guess I'm not using l'Hospital's rule at all here. Do you have to use l'Hospital's rule on this problem, or can you solve as I've outlined here?
    Last edited by Ackbeet; July 8th 2010 at 01:48 AM. Reason: Using/Not Using l'Hospital's Rule?
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  14. #14
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    The question states that you must use L'Hopital, unfortunately.
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  15. #15
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    Ok. Can you combine l'Hospital's Rule with other methods?
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