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$$\lim_{x\to 0^{+}}\frac{\ln(x)}{x}$$

I know that $\lim_{x\to 0^{+}}\ln(x)=-\infty$ and $\lim_{x\to 0^{+}}\frac{1}{x}=\infty$ can I say anything about $-\infty\cdot\infty$ or it is intermediate of L'hopital?

Clement C.
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gbox
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    When $x$ is close to $0$ but positive, $\ln x$ is large negative, $1/x$ is large positive, so the product is large negative. – André Nicolas Jul 14 '16 at 17:12

5 Answers5

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We have, as $x \to 0^+$, $$ \frac{\ln(x)}{x}=\frac{1}{x} \times \ln(x) \to +\infty \times (-\infty)=-\infty. $$

Olivier Oloa
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    @gbox This goes as the usual rule of signs, the product of two numbers with different signs is negative, if the two numbers tend to $\infty$, the sign rule still applies. – Olivier Oloa Jul 14 '16 at 17:19
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Another way to evaluate the limit is to note that $\log(x)\le x-1$ for all $x>0$, which I showed in THIS ANSWER using only the limit definition of the exponential function and Bernoulli's Inequality. Therefore,

$$\begin{align} \frac{\log(x)}{x}&\le \frac{x-1}{x}\\\\ &=1-\frac1x\\\\ & \to -\infty\,\,\left(\text{as}\,\,x\to 0^+\right) \end{align}$$

Community
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Mark Viola
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  • $log(x)\leq x-1$ is by taylor expansion? – gbox Jul 14 '16 at 17:20
  • @gbox my favourite way of seeing it is by noting that $y = x+1$ is tangent to $e^x$ at $x = 0$, and it's always below its graph, so for any $x \in \Bbb R$, $e^x \ge x + 1$ (which after manipulation gives the desired inequality) –  Jul 14 '16 at 17:50
  • @gbox I've edited with a link to an answer I posted which establishes the inequalities $$\frac{x-1}{x}\le \log(x)\le x-1$$ from only the limit definition of the exponential function and Bernoulli's Inequality. – Mark Viola Jul 14 '16 at 17:50
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Here is a formal proof which uses the definition of an infinite limit directly:

We have $$\lim_{x\to 0^+}\frac{\ln x}{x}=-\infty$$ if for any $k>0$ one can find a $\delta>0$ such that for all $x\in(0,\delta)$ $$\frac{\ln x}{x}\leq-k.\quad (*)$$

Since $\ln x$ is concave it lies everywhere below the tangent at $x=1$, i.e. $$\ln x\leq x-1$$ for all $x>0$. Therefore $$\frac{\ln x}{x}\leq 1-\frac{1}{x}$$ for all $x>0$.

Thus as long as$$1-\frac{1}{x}\leq -k,$$ then $(*)$ will also be true. So one can take $\delta=\frac{1}{1+k}$ (which satisfies the above inequality as an equality).

smcc
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The expression is negative for $0<x<1$, so the limit is $-\infty$.

As you pointed out, $ln(x)$ tends to $-\infty$ and $\frac{1}{x}$ tends to $\infty$. So, the limit cannot be a finite real number.

You cannot apply L'Hospital here, because the limit of $\frac{1}{x}$ , $x$ tending to $0$ from above, does not exist either.

Peter
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"I know that $\lim_\limits{x\to0+}\ln(x)=-\infty,$ and $\lim_\limits{x\to0+}1x= \infty$ can I say anything about $−\infty\cdot \infty$"

Lets parse this. What dies $\lim_\limits{x\to0+}\ln(x)=-\infty$ really mean?

You learned the $\epsilon - \delta$ definition of limit, but it probably didn't stick the first time. So, back to definitions.

$\forall M>0, \exists \delta > 0$ such that $0<x<\delta \implies \ln x < -M$

Pick an number that you think bounds $\ln x$ and I can find a value of $x$ in the neighborhood of $0$ that busts your bounds.

Now we add that we know something similar about $\frac 1x$ -- that I can find a value of $x$ such that $\frac 1x > M$

As long as $\frac 1x$ and $-\ln x$ are also greater than 1, then there exists an $x$ where $\frac {ln x}{x} < -M$

Mathing it up a little bit.

$\forall M>0, \exists \delta > 0$ such that $0<x<\delta \implies \frac {\ln x}{x} < -M$

$\lim_\limits{x\to 0+} \frac {\ln x}{x} = -\infty$

Doug M
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