Răspuns:
[tex]ln(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}x^n =x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots [/tex]
[tex]x-ln(1+x) = x - x + \frac{x^2}{2} - \frac{x^3}{3}+ \frac{x^4}{4}-\cdots = \frac{x^2}{2} - \frac{x^3}{3} + \frac{x^4}{4} - \cdots [/tex]
[tex] \lim_{x\to 0} \frac{x-ln(1+x)}{x^2}=\lim_{x\to 0} (\frac{1}{2}-\frac{x}{3} + \frac{x^2}{4} - \cdots) = \frac{1}{2}[/tex]
