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
$\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