Is it possible to prove that $\lim_{x\to 0} {ln(x)x} = 0$ without L'Hospital's rule? Could somebody give me the answer:
Is it possible to prove that $\lim_{x\to 0} {\ln(x)x} = 0$ without L'Hospital's rule?

One can also use the following power series, which converge for $-1 < x \leq 1$:
$$\ln(1+x) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} x^n$$
Then $x \ln x$ becomes
$$x \ln x = (x-1) \ln x + \ln x = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}(x-1)^{n+1} + \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}(x-1)^n = (x-1) + \sum_{n=1}^\infty \left( \frac{(-1)^{n+1}}{n} + \frac{(-1)^{n+2}}{n+1}\right) (x-1)^{n+1} \to -1 + \sum_{n=1}^\infty \left( \frac{(-1)^{n+1+n+1}}{n} + \frac{(-1)^{n+2+n+1}}{n+1} \right) = -1 + \sum_{n=1}^\infty \left( \frac{1}{n} - \frac{1}{n+1} \right) = -1+1 = 0$$
