# limit

• Jan 18th 2009, 05:48 AM
Nacho
limit
prove without l´hopital that:

$\displaystyle \mathop {\lim }\limits_{x \to \infty } x\left( {1 - x\ln \left( {\frac{{x + 1}} {x}} \right)} \right) = \frac{1} {2}$

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

Originally Posted by Nacho
prove without l´hopital that:

$\displaystyle \mathop {\lim }\limits_{x \to \infty } x\left( {1 - x\ln \left( {\frac{{x + 1}} {x}} \right)} \right) = \frac{1} {2}$

thanks

You can use the power series of $\displaystyle \ln (1 +t) = t - \frac{t^2}{2} + \frac{t^3}{3} - \frac{t^4}{4} + ...$, with some justification. And then rearrange and simplify to take limits :)
• Jan 18th 2009, 08:26 AM
Jester
Quote:

Originally Posted by Nacho
prove without l´hopital that:

$\displaystyle \mathop {\lim }\limits_{x \to \infty } x\left( {1 - x\ln \left( {\frac{{x + 1}} {x}} \right)} \right) = \frac{1} {2}$

thanks

Can I use Calculus but not L'Hopital's rule?
• Jan 18th 2009, 08:41 AM
Nacho
Quote:

Originally Posted by Isomorphism
You can use the power series of $\displaystyle \ln (1 +t) = t - \frac{t^2}{2} + \frac{t^3}{3} - \frac{t^4}{4} + ...$, with some justification. And then rearrange and simplify to take limits :)

$\displaystyle 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) = $$\displaystyle \mathop {\lim }\limits_{x \to \infty } x - x^2 \sum\limits_{n = 0}^\infty {\frac{{\left( { - 1} \right)^n \left( {\frac{1} {x}} \right)^{n + 1} }} {{n + 1}}} \displaystyle L = \mathop {\lim }\limits_{x \to \infty } x - x + \frac{1} {2} - \frac{1} {{3x}} + \frac{1} {{4x^2 }} + ... = 0 + \frac{1} {2} - 0 + 0 - ... = \frac{1} {2} thanks • Jan 18th 2009, 08:47 AM Jester Quote: Originally Posted by Nacho \displaystyle L = \mathop {\lim }\limits_{x \to \infty } x\left( {1 - x\ln \left( {\frac{{x + 1}} {x}} \right)} \right) = \mathop {\lim }\limits_{x \to \infty } x - x^2 \ln \left( {1 + \frac{1} {x}} \right) =$$\displaystyle \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}}}$

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

thanks

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

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

Using the Taylor series given by Isomorphism gives the result really quickly.
• Jan 18th 2009, 09:15 AM
Nacho
Quote:

Originally Posted by danny arrigo
Just an added note. If you let $\displaystyle x = \frac{1}{t}$ then your limit becomes

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

Using the Taylor series given by Isomorphism gives the result really quickly.

ooh! that limit is more easy to do than the mine :)

thanks