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Prove $\lim\limits_{x\to 0^+} \frac{\ln x}{x} = -\infty$

I've seen the following proof but I think it's invalid:
$$\lim\limits_{x\to 0^+} \frac{\ln x}{x} = \lim\limits_{x\to 0^+}\ln x \cdot \lim\limits_{x\to 0^+} \frac{1}{x} = -\infty \cdot \infty = -\infty$$

Is this proof valid? If not, I'd be glad for an alternative.

Thanks.

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  • $\begingroup$ Use comparative increasing $\endgroup$
    – Maman
    Aug 10, 2014 at 15:03
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    $\begingroup$ The two sided limit does not exist. Do you mean from $0^-$? $\endgroup$
    – Jeel Shah
    Aug 10, 2014 at 15:03
  • $\begingroup$ Please change $+$ by $\times$ in the middle! $\endgroup$
    – Hamou
    Aug 10, 2014 at 15:05

3 Answers 3

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Let $y=\frac1x$, then as $x \rightarrow 0^+,y \rightarrow \infty$

$\lim\limits_{x\to 0^+} \frac{\ln x}{x}=\lim\limits_{y\to \infty} y \ln{\frac1y}=\lim\limits_{y\to \infty}- y \ln{y}$

As both $y$ and $\ln(y)$ increase to infinity, therfore $\lim\limits_{y\to \infty}- y \ln{y}=-\infty$

Hence $\lim\limits_{x\to 0^+} \frac{\ln x}{x}=-\infty$

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  • $\begingroup$ If I understand correctly, you used: $\ln\frac{1}{y} = -\ln y$. $\endgroup$ Aug 10, 2014 at 15:19
  • $\begingroup$ Or more genrealy, $\ln(a/b) = \ln(a)-\ln(b)$. Thanks! $\endgroup$ Aug 10, 2014 at 15:21
  • $\begingroup$ or you can look at it in a way $\ln \frac1{y}=\ln (y^{-1})=-\ln y$. $\endgroup$ Aug 11, 2014 at 17:54
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set $\log n = t, \ n = e^t$ and consider two cases: $$ \lim_{t \to \infty} \frac{t}{e^t} = 0 $$ $$ \lim_{t \to - \infty} \frac{t}{e^t} = - \lim_{v \to \infty}e^v v = - \lim_{v \to \infty} e ^{\log v + v} = - \infty $$ to see that the one-sided limit doesn't exist (here set $-t = v$).

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From an $M,\delta$ perspective, the proof is essentially correct. Let's generalize slightly:

Suppose $f$ and $g$ are two functions for which $\lim\limits_{x \to 0^+} f(x) = -\infty$, $\lim\limits_{x \to 0^+} g(x) = +\infty$. Then, $\lim\limits_{x \to 0^+} f(x)\cdot g(x) = - \infty$.

Proof: Consider $f$ and $g$ with the desired properties. We will prove $\lim\limits_{x \to 0^+} f(x)\cdot g(x) = - \infty$ by showing that for any $M$, there exists a $\delta > 0$ such that for all $x$, $0 < x < \delta$, we have that $f(x)g(x) < M$.

Consider real $M$, arbitrary. Since $\lim\limits_{x \to 0^+} f(x) = -\infty$, there exists a $\delta_1 > 0$ such that for all $x$ with $0 < x < \delta_1$, $f(x) < -1$. Also, since $\lim\limits_{x \to 0^+} g(x) = +\infty$, there exists a $\delta_M > 0$ such that for all $x$ with $0 < x < \delta_M$, $g(x) > \max\{-M, 0\}$. Now, set $\delta = \min\{\delta_1, \delta_M\}$, and consider $x$ with $0 < x < \delta$. There are two cases to consider. If $M \ge 0$, then $f(x)g(x) < 0 \le M$, as required. Alternatively, if $M < 0$, then $-M > 0$, so we have

$$f(x)\cdot g(x) < -1 \cdot g(x) < -1 \cdot -M \le M.$$

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  • $\begingroup$ Very nice. thank you! $\delta,M$ feels more regorus in a way... $\endgroup$ Aug 10, 2014 at 19:50

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