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How do I prove (without L'Hôpital's rule) that $\lim_{x→0} x⋅\ln x = 0$.

I'm trying to get some intuitive sense for this, but it's quite hard. It's like trying to prove that $x$ goes faster to $\ 0 \ $ then $\ln x$ goes to $-∞$, right ?

I tried this, for $x∈(0,1)$ $x \ln x=x⋅-∫_x^1 \frac{1}{t}dt>x⋅\frac{(1-x)}{-x}=-1+x$

So when $x→0$, I conclude that $x⋅\ln x≥-1$.

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    I don't know how to implement this, but how about squeeze theorem? It was the first thing that came to mind.2012-12-13

3 Answers 3

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Since $\ln(\frac{1}{x})=-\ln x$, $\lim_{x\to 0^{+}} x \cdot \ln x = \lim_{x\to \infty} \frac{1}{x} \cdot\ln \frac{1}{x} = - \lim_{x\to \infty} \frac{\ln x}{x}$

But $\ln x$ is "slower" than $x$ (since $e^x$ is faster than $x$), so this limit is zero.

(Rigorously: for positive $x$, $2 e^x > x^2$ (by comparing power series of both sides). Plug $\sqrt{x}$ instead of $x$ and take logarithm of both sides: $\ln x < \ln 2 + \sqrt{x} =o(x)$.)

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    @Kasper - yes, you got the gist of it, but some comments are in order. First of all, if you know that both limits exist and you just want to demonstrate equality, you could have taken a *specific* $s_n$. Secondly, I would have done what you did in the following more detailed way: Suppose $\lim_{x \to 0^{+}} f(x)$ exists and equals $L$. $\forall t_n \to \infty$ (notice the "$\forall$"!), consider $s_n = \frac{1}{t_n} \to 0$. Since $\lim f(\frac{1}{t_n}) = \lim f(s_n) = L$, it follows that $\lim_{x \to \infty} f(\frac{1}{x}) = L = \lim_{x \to 0^{+}} f(x) $.2012-12-13
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$x \ln x=-x⋅\int_x^1 \frac{1}{t}dt$

Now, let $0 be arbitrary. Then, for all $x \in (0,1)$ we have

$\frac{1}{t^{1-a}}\leq \frac{1}{t} \leq \frac{1}{t^{1+a}}$

Thus

$-x⋅\int_x^1 \frac{1}{t^{1+a}}dt \leq -x \ln(x) \leq -x⋅\int_x^1 \frac{1}{t^{1-a}}dt$ $-x⋅\frac{x^a-1}{a} \leq -x \ln(x) \leq -x⋅\frac{x^{-a}-1}{-a}⋅$

now let $x \to 0$.

P.S. The argument works directly with any $0, so picking $a=\frac{1}{2}$ makes the proof much cleaner....

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Make the change $x = e^{-y}$: $ \lim_{x \to 0} x \ln x = - \lim_{y \to \infty} y e^{-y} $

Show that for $y > 0$: $ e^y > \frac{y^2}{2!} $

And use the squeeze theorem.