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?
 A: Using the integral formula $\ln(x)=\int_1^x \frac{dt}{t}$ and $\ln(y^2)=2\ln(y)$, we have for all $x \in (0,1)$,
$$
|x\ln(x)| =2\left|x\ln(x^{1/2})\right| \leq 2\int_{\sqrt{x}}^1\frac{x}{t}\,dt\leq 2 \int_{\sqrt{x}}^1\frac{x}{\sqrt{x}}\,dt \leq 2\sqrt{x}(1-\sqrt{x}).
$$
The conclusion follows by squeezing since the limit of the r.h.s. as $x\to 0^+$ is $0$ .
A: 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
$$x \ln x = (x-1) \ln x + \ln x = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}(x-1)^{n+1} + \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}(x-1)^n = (x-1) + \sum_{n=1}^\infty \left( \frac{(-1)^{n+1}}{n} + \frac{(-1)^{n+2}}{n+1}\right) (x-1)^{n+1} \to -1 + \sum_{n=1}^\infty \left( \frac{(-1)^{n+1+n+1}}{n} + \frac{(-1)^{n+2+n+1}}{n+1} \right) = -1 + \sum_{n=1}^\infty \left( \frac{1}{n} - \frac{1}{n+1} \right) = -1+1 = 0$$
A: If we can use the fact that
$$\frac{\ln(u)}{u}\to 0\ \ \text{($u\to\infty$)},$$
consider $x=1/u.$
A: You can prove that $e^t > 2^t > t^2$ for $t$ sufficiently large. Hence you can prove that
$$
\lim_{t\to +\infty} \frac{t}{e^t} = 0
$$
without using derivatives.
Let $x=e^{-t}$ and you find
$$
0 = \lim_{t\to +\infty} \frac{t}{e^t} = \lim_{x\to 0} -x \log x.
$$
The point, however, is: how the function $\log x$ has been defined?
