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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. –  000 Dec 13 '12 at 22:35
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3 Answers

up vote 3 down vote accepted

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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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
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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....

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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.

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