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How do I prove (without L'Hôpital's rule) that $$\lim_{x\to0^+} x\cdot\ln x = 0$$.

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 $-\infty$, right ?

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

So when $x\to0^+$, I conclude that $x\cdot\ln x\ge-1$.

J.G.
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Kasper
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4 Answers4

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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: $\ln x < \ln 2 + \sqrt{x} =o(x)$.)

Ofir
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    Is it always true that you can rewrite $\lim_{x→0^+}f(x) = \lim_{x→∞}f(1/x)$? – Kasper Dec 13 '12 at 22:48
  • @Kasper, that's a great question. My intuition is no, and the reason is that there are tricky situations that arise when moving from a one-sided limit to a two-sided limit. I would say, however, that it may be true that $$\lim_{x \to 0}f(x)=\lim_{x \to \infty}f\left(\frac{1}{x}\right).$$ However, I am uncertain. – 000 Dec 13 '12 at 22:55
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    @Kasper - The answer is "Yes". Try to prove it using the definition of one-sided limit at 0 and limit at infinity. Show that if one limit exists, the other must equal it. – Ofir Dec 13 '12 at 23:03
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    @Limitless - There's no problem because the limit at infinity is not two-sided. Actually, your proposed equality is false, because in the LHS you evaluate $f$ on both negative and positive values, and at the RHS you evaluate $f$ only on positive values. Take $f(x)=\frac{1}{x}$ to reach a contradiction. – Ofir Dec 13 '12 at 23:05
  • @Ofir Let $s_n>0$ and $\lim_{n→∞} s_n = 0$. By definition $$\lim_{x→0^+}f(x)=\lim_{n→∞}f(s_n)$$. Let $t_n=1/s_n$. Then $\lim_{n→∞}t_n=+∞$. So $$\lim_{x→0^+}f(x)=\lim_{n→∞}f(s_n)=\lim_{n→∞}f(1/t_n)=\lim_{x→∞}f(1/x)$$

    Do you mean something like this ?

     – Kasper Dec 13 '12 at 23:21
  • @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) $. – Ofir Dec 13 '12 at 23:46
  • You could make another "rigorous" argument that the limit \lim_{x \rightarrow \infty} \ln x / x = 0 by using L'Hopitals Rule. – G. Khanna Jun 26 '21 at 04:12
  • @G.Khanna L'Hopital's rule was excluded. – Anne Bauval Mar 05 '24 at 13:14
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$$x \ln x=-x⋅\int_x^1 \frac{1}{t}dt$$

Now, let $0<a$ 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 $0<a<1$, so picking $a=\frac{1}{2}$ makes the proof much cleaner....

N. S.
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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.

Ayman Hourieh
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Another method is to use Taylor's series: $\ln(1-y)=-y-\frac{y^2}{2}-\frac{y^3}{3}-\ldots$ $$\therefore \lim_{x\to 0}x\cdot \ln x$$ $$=\lim_{x\to 0}x\cdot \lim_{x\to 0}\ln x$$ $$=\lim_{x\to 0}x\cdot \lim_{x\to 0}\ln (1-(1-x))$$ $$=\lim_{x\to 0}x\cdot \lim_{x\to 0}\left(-(1-x)-\frac{(1-x)^2}{2}-\frac{(1-x)^3}{3}-\ldots\right)$$ $$=0\cdot \left(-1-\frac{1}{2}-\frac{1}{3}-\ldots\right)$$ $$=0$$ Proved.

Harish Chandra Rajpoot
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  • This answer is wrong from the very beginning. $\lim_{x\to 0}x\cdot \lim_{x\to 0}\ln x$, as well as $0\cdot \left(-1-\frac{1}{2}-\frac{1}{3}-\ldots\right)$, are the indeterminate form $0\cdot(-\infty)$. – Anne Bauval Mar 05 '24 at 13:18