Longit644

Tính giới hạn sau:
$$\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{1}{x} - \dfrac{1}{{\ln \left( {x + 1} \right)}}} \right),\,\,x > - 1$$

Do $ ln(x+1) \rightarrow 0$ khi $x \rightarrow 0$ Áp dụng khai triển Taylor :$ ln(x+1)=x-\dfrac{1}{x^2} + o(x^2)$ $ \lim_{x \to 0} (\dfrac{1}{x}-\dfrac{1}{ln(x+1)})= \lim_{x \to 0} \dfrac{ln(x+1)-x}{xln(x+1)} = \lim_{x \to 0} \dfrac{x-\dfrac{x^2}{2} + o(x^2)-x}{x(x+o(x))} = \lim_{x \to 0} \dfrac{-\dfrac{x^2}{2} + o(x^2)}{x(x+o(x))} =-\dfrac{1}{2}$