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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?

TZakrevskiy
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Dima Railguner
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  • In the question marked as duplicate, André Nicolas wrote I do not believe this should be closed, since it describes an interesting aproach to the problem that is absent elsewhere. I agree with him. Despite their similar titles, the questions are not strictly the same. – Siméon Jan 08 '14 at 12:24

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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 $0$ .

Siméon
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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 $$x \ln x = (x-1) \ln x + \ln x = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}(x-1)^{n+1} + \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}(x-1)^n = (x-1) + \sum_{n=1}^\infty \left( \frac{(-1)^{n+1}}{n} + \frac{(-1)^{n+2}}{n+1}\right) (x-1)^{n+1} \to -1 + \sum_{n=1}^\infty \left( \frac{(-1)^{n+1+n+1}}{n} + \frac{(-1)^{n+2+n+1}}{n+1} \right) = -1 + \sum_{n=1}^\infty \left( \frac{1}{n} - \frac{1}{n+1} \right) = -1+1 = 0$$

Ulrik
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If we can use the fact that $$\frac{\ln(u)}{u}\to 0\ \ \text{($u\to\infty$)},$$ consider $x=1/u.$

mathlove
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    The fact assumed here ($\lim\limits_{u \to \infty}\dfrac{\log u}{u} = 0$)can be proved without using L'Hospital via the inequality $\log u = \int_{1}^{u}dt/t < \int_{1}^{u}t^{a - 1},dt$ where $a > 0$. Then we get $\log u < \dfrac{u^{a} - 1}{a} < \dfrac{u^{a}}{a}$ for $a > 0$ and $u > 1$. This means that $0 < \dfrac{\log u}{u} < \dfrac{u^{a - 1}}{a}$. Choosing any $a$ with $0 < a < 1$ our job is done when we take limits as $u \to \infty$. – Paramanand Singh Jan 08 '14 at 11:06
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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?

Emanuele Paolini
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  • How would you prove $;2^t>t^2;$ without derivatives? – DonAntonio Jan 08 '14 at 12:55
  • You can prove the inequality $2^{n-1}>n^2$ for $n$ sufficiently large by induction on natural numbers. Then it depends on how $a^x$ has been defined... it is enough to know that it is an increasing function. – Emanuele Paolini Jan 10 '14 at 09:08