limit

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  • January 18th 2009, 08:41 AM
    Nacho
    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
  • January 18th 2009, 08:47 AM
    Jester
    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.
  • January 18th 2009, 09:15 AM
    Nacho
    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