GMB/12/2008

[Virgil Nicula]

Sa se determine \( L\equiv \lim_{x\to\infty}x\cdot\left(1-\frac {\ln x}{x}\right)^x\ . \)

[Marius Mainea]

Prin logaritmare si aplicarea regulii lui L'Hospital obtinem

\( \lim_{x\to\infty}(\ln x+x\ln (1-\frac{\ln x}{x}))=\lim_{x\to\infty}\frac{\frac{\ln x}{x}+\ln (1-\frac{\ln x}{x})}{\frac{1}{x}}=\lim_{x\to\infty}\frac{\frac{1-\ln x}{x^2}+\frac{x}{x-\ln x}\frac{\ln x-1}{x^2}}{\frac{1}{x^2}}=\lim_{x\to\infty}1-\ln x+\frac{x(\ln x-1)}{x-\ln x}=\lim_{x\to\infty}1+\frac{\ln^2x-x}{x-\ln x}=\lim_{x\to\infty}1+\frac{\frac{\ln^2x}{x}-1}{1-\frac{\ln x}{x}}=1+\frac{0-1}{1-0}=0 \)

Asadar limita initiala este \( e^0=1 \)

[Virgil Nicula]

Sa se determine \( L\equiv \lim_{x\to\infty}x\cdot\left(1-\frac {\ln x}{x}\right)^x\ . \)

\( L\equiv\lim_{x\to\infty}\ x\cdot\left(1-\frac {\ln x}{x}\right)^x\ \stackrel{\left(x:=\frac 1x\right)}{\ =\ }\ \lim_{x\to 0}\ \frac {(1+x\ln x)^{\frac 1x}}{x}=e^l\ , \) unde \( l=\lim_{x\to 0}\ \frac {\ln (1+x\ln x)-x\ln x}{x}\ \stackrel {(l^{\prim}H)}{\ =\ } \)

\( \lim_{x\to 0}\ \left[\frac {1+\ln x}{1+x\ln x}-(1+\ln x)\right]=-\lim_{x\to 0}\ \frac {x\ln x(1+\ln x)}{1+x\ln x}=0\ . \) In concluzie, \( \lim_{x\to\infty}x\cdot\left(1-\frac {\ln x}{x}\right)^x=1\ . \)

Comentariu. Am folosit limita remarcabila \( \underline {\overline {\left\|\ \lim_{x\to 0}\ x\ln^{\alpha} x=0\ ,\ \alpha > 0\ \right\|}}\ . \) Iata o demonstratie a acesteia.

Folosim substitutia \( x:=x^{\alpha}\ \ :\ \ \lim_{x\to 0}\ x\ln^{\alpha} x=\lim_{x\to 0}\ x^{\alpha}\left(\ln x^{\alpha}\right)^{\alpha}=\alpha ^{\alpha}\cdot\lim_{x\to 0}\ \left(x\ln x\right)^{\alpha}=0 \)

deoarece \( \lim_{x\to 0}\ x\ln x\ \stackrel{\left(x:=\frac 1x\right)}{\ =\ }\ -\lim_{x\to\infty}\ \frac {\ln x}{x}=-0\ . \) Mai ramane sa aratam de ce \( \lim_{x\to\infty}\ \frac {\ln x}{x}=0\ \ldots\ . \)