$$\lim_{x\rightarrow 0}\frac{\ln (x+1)}{x}=0$$
$$\lim_{x\rightarrow 0}\frac{\log _\alpha {(x+1)}}{x}=\log _\alpha {e}$$
$$\lim_{x\rightarrow 0}(1+x)^\frac{1}{x}=e$$
$$\lim_{x\rightarrow 0}\frac{(\alpha ^x-1)}{x}=\ln \alpha$$
$$\lim_{x\rightarrow 0}x^x=1$$
2010-12-09 • Просмотров [ 1417 ]
