Evaluating limit of $\lim_{x \to 0^+} x \ln(x)$

Question

Problem.

Evaluate the limit $$\lim_{x \to 0^+} x \ln(x)$$

I split the limit up in a product of $x$ and $\ln(x)$ and get $0 \cdot -\infty$ which sounds a bit shady...Can I simply say that this is obviously $0$ because $0 \cdot a = 0, \forall a \in \mathbb{R}$...?

For some reason I only think of using l'hospital when I've got a fraction. But thanks for the laugh, heh — user3200098, Feb 28 '14 at 16:08
Isn't there a function for using l'Hospital? I'm tired of writing $\frac{d}{dx}$ over both the numerator and the denominator every step. — user3200098, Feb 28 '14 at 16:21
@user3200098 THere is a way. One time calculate your limit but raise $x$ to the power $n$ and then to the Frenchman way. The limit is still zero. From now on, call this a Standard Limit. You will never have to calculate it again. — imranfat, Feb 28 '14 at 16:37

score 6 · Accepted Answer · answered Feb 28 '14 at 16:09

6

Hint: $$ \lim_{x \to 0^+} x\ln(x) =\lim_{x \to 0^+} \frac{\ln(x)}{1/x} $$

answered Feb 28 '14 at 16:09

Ben Grossmann

225,327

Does this mean that ∴ the limit is 0? – Anonymous Pi Mar 01 '14 at 03:23
@AnonymousPi Do you know L'Hopital's? – MT_ Mar 01 '14 at 03:34
Emmmm... no. I know what is a limit, and if you tell me the result of one is something, I understand, but I really don't have much experience calculating them. – Anonymous Pi Mar 01 '14 at 03:36
@AnonymousPi here's an explanation of L'Hôpital's rule, if you're interested. L'Hôpital's rule allows you to simplify the limit of an indeterminate fraction by taking the derivative of the numerator and denominator. What I've done here is put the limit in the form of an indeterminate fraction, so that we could directly apply L'Hôpital's rule. – Ben Grossmann Mar 01 '14 at 16:55

score 6 · Answer 2 · edited Feb 28 '14 at 17:57

6

Another way to think about it: let $x=e^{-t}$; then the limit becomes:

$$\lim_{t\to \infty}\;\; -t \, e^{-t}$$

Maybe that's clearer.

edited Feb 28 '14 at 17:57

recursion.ninja

1,430

answered Feb 28 '14 at 16:11

Ron Gordon

138,521

score 3 · Answer 3 · edited Apr 13 '17 at 12:21

As shown in this answer, for all $x\in\mathbb{R}$, $$ x\lt1+x\le e^x\tag{1} $$ If $u\gt0$, applying $(1)$ to $u/2$ and squaring gives $$ u^2/4\lt e^u\tag{2} $$ Multiplying $(2)$ by $\dfrac4{ue^u}$ yields $$ 0\lt\frac{u}{e^u}\lt\frac4u\tag{3} $$ Thus, applying $(3)$ and the Squeeze Theorem, $$ \begin{align} \lim_{x\to0^+}x\log(x) &=\lim_{x\to0^+}\log(x)e^{\log(x)}\\ &=\lim_{u\to\infty}-ue^{-u}\\ &=-\lim_{u\to\infty}\frac{u}{e^u}\\[3pt] &=0\tag{4} \end{align} $$

score 2 · Answer 4 · answered Feb 28 '14 at 16:10

2

$$\lim_{x\to 0} x\log x=\lim_{x\to 0}\frac{\log x}{\frac1x}\stackrel{l\,'H}=\lim_{x\to 0}\frac{\frac1x}{-\frac1{x^2}}=\ldots$$

answered Feb 28 '14 at 16:10

DonAntonio

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alex · Answer 5 · 2014-02-28T16:25:51.857

1

Set $t=\frac{1}{x}$, then

$$\lim_{x \to 0^+} x \ln(x)=\lim_{t\rightarrow +\infty}\frac{\log(\frac{1}{t})}{t}=\lim_{t\rightarrow +\infty}\frac{-\log(t)}{t}=0.$$

edited Feb 28 '14 at 16:25

answered Feb 28 '14 at 16:15

alex

795

@DonAntonio Sorry, a typo. Thanks for pointing out. – alex Feb 28 '14 at 16:25

Evaluating limit of $\lim_{x \to 0^+} x \ln(x)$

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