¿Cómo calcular $\lim\limits_{x \to 0}\frac1{x} + \frac{\ln(1-x)}{x^2}$?

Question

¿Cómo calcular el siguiente límite?

$$\lim_{x \to 0}\frac1{x} + \frac{\ln(1-x)}{x^2}$$

Answer 1

Tienes\begin{eqnarray*} \lim_{x\to 0} \frac{1}{x}+\frac{\ln(1-x)}{x^2} &=& \lim_{x\to 0} \frac{x+\ln(1-x)}{x^2}, \end{eqnarray *} Observe entonces que % $ $$ \lim_{x\to 0} x+\ln(1-x)=0,\: \lim_{x\to 0} x^2= 0, $por la L'Hospital regla $$\begin{align*} \lim_{x\to 0} \frac{1}{x}+\frac{\ln(1-x)}{x^2}&= \lim_{x\to 0} \frac{\frac{d}{dx}(x+\ln(1-x))}{\frac{d}{dx}x^2}\\ &= \lim_{x\to 0} \frac{1-\frac{1}{1-x}}{2x}\\ &= \lim_{x\to 0}\frac{\frac{1-x - 1}{1-x}}{2x}\\ &= \lim_{x\to 0} \frac{\frac{x}{x-1}}{2x}\\ &= \lim_{x\to 0}\frac{1}{2(x-1)}\\ &= -\frac{1}{2}. \end{align*} $$

Answer 2

con la serie $$\log(1-x)=-\sum_{n=1}^{\infty}\frac{x^n}{n}$$ for $-1\leq x < 1$ tenemos $$\frac{1}{x}+\frac{\log(1-x)}{x^2}=-\frac{1}{x^2}\sum_{n=2}^{\infty}\frac{x^n}{n}$$ so the limit as $x\to0$ is $-\frac{1}{2}$