# limit

• January 18th 2009, 05:48 AM
Nacho
limit
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}
$

thanks
• January 18th 2009, 06:27 AM
Isomorphism
Quote:

Originally Posted by Nacho
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}
$

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 :)
• January 18th 2009, 08:26 AM
Jester
Quote:

Originally Posted by Nacho
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?
• January 18th 2009, 08:41 AM
Nacho
Quote:

Originally Posted by Isomorphism
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 :)

$
L = \mathop {\lim }\limits_{x \to \infty } x\left( {1 - x\ln \left( {\frac{{x + 1}}
{x}} \right)} \right) = \mathop {\lim }\limits_{x \to \infty } x - x^2 \ln \left( {1 + \frac{1}
{x}} \right) =
$
$
\mathop {\lim }\limits_{x \to \infty } x - x^2 \sum\limits_{n = 0}^\infty {\frac{{\left( { - 1} \right)^n \left( {\frac{1}
{x}} \right)^{n + 1} }}
{{n + 1}}}
$

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

thanks
• January 18th 2009, 08:47 AM
Jester
Quote:

Originally Posted by Nacho
$
L = \mathop {\lim }\limits_{x \to \infty } x\left( {1 - x\ln \left( {\frac{{x + 1}}
{x}} \right)} \right) = \mathop {\lim }\limits_{x \to \infty } x - x^2 \ln \left( {1 + \frac{1}
{x}} \right) =
$
$
\mathop {\lim }\limits_{x \to \infty } x - x^2 \sum\limits_{n = 0}^\infty {\frac{{\left( { - 1} \right)^n \left( {\frac{1}
{x}} \right)^{n + 1} }}
{{n + 1}}}
$

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

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
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