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?