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Math Help - Limits without use of L'Hospital and series expansion

  1. #1
    Senior Member pankaj's Avatar
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    Limits without use of L'Hospital and series expansion

    Without using series expansion of e^x and binomial expansion of \sqrt{1+x^2} evaluate the limit :
    \lim_{x\to 0}\frac{e^x-x-\sqrt{1+x^2}}{x^3}
    Also,you cannot use the L'Hospital rule.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by pankaj View Post
    Nobody wants to solve this limit!
    Its not that we don't want to solve it...its just that we can't think of a clever way to do it! We're discussing that same limit here.

    --Chris
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  3. #3
    Senior Member pankaj's Avatar
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    Thanks Chris.I am myself on it.
    Last edited by pankaj; September 27th 2008 at 08:07 AM.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by pankaj View Post
    Without using series expansion of e^x and binomial expansion of \sqrt{1+x^2} evaluate the limit :
    \lim_{x\to 0}\frac{e^x-x-\sqrt{1+x^2}}{x^3}
    Also,you cannot use the L'Hospital rule.
    The rules do not prevent you from using a series expansion of:

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

    Or avoiding series the fact that f(0)=0,\ f'(0)=0,\ f''(0)=0 and the definition of (3rd) derivative to give:

    f(x)=\frac{x^3}{6}+o(x^3)

    (here o(x^3) denotes a function such that the limit as x goes to 0 of this function divided by x^3 goes to zero) which will allow the evaluation of the limit.

    I suspect though that you might consider all this to be cheating as this is tantamount to either L'Hopital's rule and/or the use of series)

    RonL
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  5. #5
    Senior Member pankaj's Avatar
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    Please check my solution

    The limit does not change if x is replaced by -x

    i.e. L=\lim_{x\to 0}\frac{e^x-x-\sqrt{1+x^2}}{x^3}

    L=\lim_{x\to 0}\frac{-e^{-x}-x+\sqrt{1+x^2}}{x^3}

     <br />
L=\lim_{x\to 0}\frac{e^x-e^{-x}-2x}{2x^3}<br />

    Putting x = 3t

    L = \lim_{t\to 0}\frac {e^{3t} - e^{ - 3t} - 6 t }{54 t^3}

    54L = \lim_{t\to 0}\frac {(e^{t} - 1)^3 + (3e^{2t} - 3e^{t} + 1) - (e^{ - t} - 1)^3 - 3e^{ - 2t} + 3e^{ - t} - 1 - 6 t }{ t^3}

     <br />
54L = \lim_{t\to 0}(\frac {e^t - 1}{t})^3 + (\frac {e^{ - t} - 1}{ - t})^3 + \frac {3e^{2t} - 3e^{ - 2t} - 3e^{t} + 3e^{ - t} - 6t}{t^3}<br />

     <br />
54L = \lim_{t\to 0}(\frac {e^t - 1}{t})^3 + (\frac {e^{ - t} - 1}{ - t})^3 + 48\frac {e^{2t} - e^{ - 2t} - 4t}{2(2t)^3} - 6\frac {e^t - e^{ - t} - 2t}{2(t)^3}<br />

    54L = 1 + 1 + 48L - 6L = 2 + 42L

     <br />
12L = 2<br />

     <br />
L = \frac {1}{6}<br />
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by pankaj View Post
    Please check my solution

    The limit does not change if x is replaced by -x

    i.e. L=\lim_{x\to 0}\frac{e^x-x-\sqrt{1+x^2}}{x^3}

    L=\lim_{x\to 0}\frac{-e^{-x}-x+\sqrt{1+x^2}}{x^3}

     <br />
L=\lim_{x\to 0}\frac{e^x-e^{-x}-2x}{2x^3}<br />

    Putting x = 3t

    L = \lim_{t\to 0}\frac {e^{3t} - e^{ - 3t} - 6 t }{54 t^3}

    54L = \lim_{t\to 0}\frac {(e^{t} - 1)^3 + (3e^{2t} - 3e^{t} + 1) - (e^{ - t} - 1)^3 - 3e^{ - 2t} + 3e^{ - t} - 1 - 6 t }{ t^3}

     <br />
54L = \lim_{t\to 0}(\frac {e^t - 1}{t})^3 + (\frac {e^{ - t} - 1}{ - t})^3 + \frac {3e^{2t} - 3e^{ - 2t} - 3e^{t} + 3e^{ - t} - 6t}{t^3}<br />

     <br />
54L = \lim_{t\to 0}(\frac {e^t - 1}{t})^3 + (\frac {e^{ - t} - 1}{ - t})^3 + 48\frac {e^{2t} - e^{ - 2t} - 4t}{2(2t)^3} - 6\frac {e^t - e^{ - t} - 2t}{2(t)^3}<br />

    54L = 1 + 1 + 48L - 6L = 2 + 42L

     <br />
12L = 2<br />

     <br />
L = \frac {1}{6}<br />
    You seem to be using:

     <br />
\lim_{u \to 0} \frac{e^u-1}{u}=1<br />

    How do you justify this without the use of L'Hopital or a series expansion?

    CB
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  7. #7
    Senior Member pankaj's Avatar
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    I thought it was a standard formula or a definition
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  8. #8
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    Quote Originally Posted by CaptainBlack View Post
    You seem to be using:

     <br />
\lim_{u \to 0} \frac{e^u-1}{u}=1<br />

    How do you justify this without the use of L'Hopital or a series expansion?

    CB
    Can't we use definition of the derivative?

    \lim_{u \to 0} \frac{e^u - e^0}{u-0} = (e^u)'~\text{at}~u = 0
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by Chop Suey View Post
    Can't we use definition of the derivative?

    \lim_{u \to 0} \frac{e^u - e^0}{u-0} = (e^u)'~\text{at}~u = 0
    Actually I think not, that is a 1 term series expansion, if yoy allow this you have to allow the method using the definition of the thrird derivative on the original problem.

    CB
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  10. #10
    Math Engineering Student
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    Well, there's a way to justify the value of that limit. Try to find it.
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  11. #11
    Grand Panjandrum
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    Quote Originally Posted by Krizalid View Post
    Well, there's a way to justify the value of that limit. Try to find it.
    But we know what the limit is!

    RonL
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  12. #12
    Math Engineering Student
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    Haha, yeah, sorry, I misexpressed myself. I actually mean, a substitution method that's all.
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