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Math Help - limit

  1. #1
    Member Nacho's Avatar
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    limit

    prove without l´hopital that:

    <br />
\mathop {\lim }\limits_{x \to \infty } x\left( {1 - x\ln \left( {\frac{{x + 1}}<br />
{x}} \right)} \right) = \frac{1}<br />
{2}<br />

    thanks
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  2. #2
    Lord of certain Rings
    Isomorphism's Avatar
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    Quote Originally Posted by Nacho View Post
    prove without l´hopital that:

    <br />
\mathop {\lim }\limits_{x \to \infty } x\left( {1 - x\ln \left( {\frac{{x + 1}}<br />
{x}} \right)} \right) = \frac{1}<br />
{2}<br />

    thanks
    You can use the power series of \ln (1 +t) = t - \frac{t^2}{2} + \frac{t^3}{3} - \frac{t^4}{4} + ..., with some justification. And then rearrange and simplify to take limits
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  3. #3
    MHF Contributor
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    Quote Originally Posted by Nacho View Post
    prove without l´hopital that:

    \mathop {\lim }\limits_{x \to \infty } x\left( {1 - x\ln \left( {\frac{{x + 1}}
    {x}} \right)} \right) = \frac{1}
    {2}
    " alt="
    \mathop {\lim }\limits_{x \to \infty } x\left( {1 - x\ln \left( {\frac{{x + 1}}
    {x}} \right)} \right) = \frac{1}
    {2}
    " />


    thanks

    Can I use Calculus but not L'Hopital's rule?
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  • #4
    Member Nacho's Avatar
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    Quote Originally Posted by Isomorphism View Post
    You can use the power series of \ln (1 +t) = t - \frac{t^2}{2} + \frac{t^3}{3} - \frac{t^4}{4} + ..., with some justification. And then rearrange and simplify to take limits
    <br />
L = \mathop {\lim }\limits_{x \to \infty } x\left( {1 - x\ln \left( {\frac{{x + 1}}<br />
{x}} \right)} \right) = \mathop {\lim }\limits_{x \to \infty } x - x^2 \ln \left( {1 + \frac{1}<br />
{x}} \right) = <br />
<br />
\mathop {\lim }\limits_{x \to \infty } x - x^2 \sum\limits_{n = 0}^\infty  {\frac{{\left( { - 1} \right)^n \left( {\frac{1}<br />
{x}} \right)^{n + 1} }}<br />
{{n + 1}}} <br />

    <br />
L = \mathop {\lim }\limits_{x \to \infty } x - x + \frac{1}<br />
{2} - \frac{1}<br />
{{3x}} + \frac{1}<br />
{{4x^2 }} + ... = 0 + \frac{1}<br />
{2} - 0 + 0 - ... = \frac{1}<br />
{2}<br />

    thanks
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  • #5
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    Quote Originally Posted by Nacho View Post
    <br />
L = \mathop {\lim }\limits_{x \to \infty } x\left( {1 - x\ln \left( {\frac{{x + 1}}<br />
{x}} \right)} \right) = \mathop {\lim }\limits_{x \to \infty } x - x^2 \ln \left( {1 + \frac{1}<br />
{x}} \right) = <br />
<br />
\mathop {\lim }\limits_{x \to \infty } x - x^2 \sum\limits_{n = 0}^\infty {\frac{{\left( { - 1} \right)^n \left( {\frac{1}<br />
{x}} \right)^{n + 1} }}<br />
{{n + 1}}} <br />

    <br />
L = \mathop {\lim }\limits_{x \to \infty } x - x + \frac{1}<br />
{2} - \frac{1}<br />
{{3x}} + \frac{1}<br />
{{4x^2 }} + ... = 0 + \frac{1}<br />
{2} - 0 + 0 - ... = \frac{1}<br />
{2}<br />

    thanks
    Just an added note. If you let x = \frac{1}{t} then your limit becomes

    \lim_{t \rightarrow 0} \frac{t - \ln (1+t)}{t^2}.

    Using the Taylor series given by Isomorphism gives the result really quickly.
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  • #6
    Member Nacho's Avatar
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    Quote Originally Posted by danny arrigo View Post
    Just an added note. If you let x = \frac{1}{t} then your limit becomes

    \lim_{t \rightarrow 0} \frac{t - \ln (1+t)}{t^2}.

    Using the Taylor series given by Isomorphism gives the result really quickly.
    ooh! that limit is more easy to do than the mine

    thanks
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