I ran into a scenario when practicing L'Hôpital's rule which yielded -infinity/0. I broke this down into $-1 \cdot \infty \cdot \frac 1 0$, which I assumed equaled $-1\cdot\infty\cdot\infty$, which simplified to $-1\cdot\infty$ which equals negative infinity. So where did I go wrong with my logic as Wolfram Alpha claims the answer is positive infinity?

Here is the limit problem:

$$\lim_{x\to0}\frac{\ln(\sin x)}{\ln(\cos x)}$$

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    $\begingroup$ It doesn't. It doesn't really work that way. What is the exact limit problem? $\endgroup$ – Tyler Oct 13 '13 at 19:03
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    $\begingroup$ $\frac{1}{0}$ is NOT equal $\infty$. You cannot divide by zero. $\endgroup$ – EricAm Oct 13 '13 at 19:14
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    $\begingroup$ l'Hôpital's rule does not apply, as Skylion indicated. $\endgroup$ – dfeuer Oct 13 '13 at 19:14
  • $\begingroup$ Wolfram Alpha says -infinity/0 = infinity though... $\endgroup$ – Skylion Oct 13 '13 at 19:18
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    $\begingroup$ I doubt that Wolfram Alpha says, word for word, that $-\infty/0=\infty$. That is your interpretation of what it ways. $\endgroup$ – André Nicolas Oct 13 '13 at 19:26

Wolfram Alpha does not say that $-\infty/-0 = \infty$ exactly, it says that this is equal to complex infinity. What's happening is that Wolfram Alpha is coming up with an interpretation for your inputs that makes the input sensible. Specifically, you can't divide infinity by zero in the context of real or complex numbers, but you can do this in the context of the Riemann sphere, which is usually treated as the union of the complex numbers with a single point at $\infty$.

  • $\begingroup$ Beat me by few seconds :) $\endgroup$ – N. S. Oct 13 '13 at 19:39
  • $\begingroup$ Hmm. @N.S., somehow I like Thomas' explanation and interpretation better. With complex logarithm we run into branch selection problems here. For example we could choose a branch of $\ln\cos z$ that approaches $2\pi i$ as $z\to0$. There's no way of making $\ln\sin z$ continuous in any punctured neighborhood of the origin, so I'm unhappy about treating this as a complex limit. $\endgroup$ – Jyrki Lahtonen Oct 13 '13 at 20:08
  • $\begingroup$ @JyrkiLahtonen Did you check the wolfram alpha link in the question ;) ? The question and the Wolframalpha link are about $-\infty/0=\infty$ not about that particular limit ;) $\endgroup$ – N. S. Oct 13 '13 at 20:11
  • $\begingroup$ Oh!? I didn't check that, sorry. Strange, but that does make your (and Aaron's) interpretation plausible. $\endgroup$ – Jyrki Lahtonen Oct 13 '13 at 20:14

As has already been pointed out in the comments above, $-\infty / 0$ is not a well defined expression. We are not allowed to divide by $0$ and we als have to remember that $\infty$ is not a number. That said, you can still find the limit. In asking that, we are just interested in what happens to the values of $f(x) = \frac{\ln(\sin(x))}{\ln(\cos(x))}$ when as $x$ approaches $0$.

First, I assume that $x$ is approaching $0$ from the right: $$ \lim_{x\to 0^+} \frac{\ln(\sin(x))}{\ln(\cos(x))} $$

since when $x$ is close to $0$ and negative, then $\sin(x)$ is negative and then $\ln(\sin(x))$ isn't defined.

So as $x$ approaches $0$ from the right, then $\sin(x)$ approaches $0$ from the right. Then $\ln(x)$ approaches $-\infty$.

Likewise, $\cos(x)$ approaches $1$ from the left, so $\ln(\cos(x))$ approaches $\ln(1) = 0$ from the left: $$ \ln(\cos(x)) \to 0^- \quad\text{as}\quad x\to 0^+ $$

And here then is the crucial part. As $x$ gets closer and closer to $x$ while being positive, we have just nooted that $\ln(\cos(x)$ is negative while $\ln(\sin(x))$ is negative. So for values of $x$ very close to $0$ but positive, $f(x)$ is something negative divided by something negative, hence positive.

Also, the bottom is close to $0$ (small number) while the top is large (and negative). That makes the whole thing very large and positive

Therefore The whole limit is positive $\infty$.

If you want a precise proof and you are willing to accept that $$ \ln(\sin(x)) \to -\infty \quad\text{as}\quad x\to 0^+ $$ and $$ \ln(\cos(x)) \to 0^- \quad\text{as}\quad x\to 0^+ $$ then let $N>0$ be given. We want to find a $\delta > 0$ such that if $0<x<\delta$ then $f(x) > N$. Now pick a $\delta_1$ such that $\ln(\sin(x)) < -N$ when ever $0<x<\delta_1$.

Pick a $\delta_2$ such that $0 > \ln(\cos(x)) > -1$ whenever $0<x<\delta_2$. Then for $\delta = \min\{\delta_1, \delta_2\}$ you will have $f(x) > N$ for $0<x< \delta$.

  • $\begingroup$ Would you mathematically prove that though? I get it how you can estimate that the limit either through checking points or through the graph, but how would you prove that algebraically? Also why does Wolfram say -infinity/0 = + infinity? $\endgroup$ – Skylion Oct 13 '13 at 19:17
  • $\begingroup$ @Skylion: You want an $\epsilon-\delta$ type of proof? $\endgroup$ – Thomas Oct 13 '13 at 19:18
  • $\begingroup$ I don't care too much about the limit, I just want to know why Wolfram is evaluating -infinity/0 as +infinity $\endgroup$ – Skylion Oct 13 '13 at 19:20
  • $\begingroup$ @Skylion: I tried to explain things a bit better. If you want something different, you might have to go to the definition of a limit. $\endgroup$ – Thomas Oct 13 '13 at 19:26
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    $\begingroup$ This is a good answer @Skylion you should care, it is not evaluating what you think it is. Limits are not substitutions. $\endgroup$ – Alec Teal Oct 13 '13 at 19:42

$$ \lim_{x\to 0^{+}}{\ln\left(\sin\left(x\right)\right) \over \ln\left(\cos\left(x\right)\right)} = \lim_{x\to 0^{+}}{\ln\left(x\right) \over \ln\left(1 - x^{2}/2\right)} = 2\lim_{x\to 0^{+}}{1 \over x^{2}}\,\left[-\ln\left(x\right)\right] = {\large +\,\infty} $$

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    $\begingroup$ You should at least give some justification for the sawn-off Taylor-series substitution. $\endgroup$ – AlexR Oct 13 '13 at 19:38
  • $\begingroup$ @AlexR It's quite obvious: Just divide and multiply and you get some "1" limits. I imagine the OP and everybody else knows that. $\endgroup$ – Felix Marin Oct 13 '13 at 20:41
  • $\begingroup$ Still you silently use the properties of $\ln$ AND the two Taylor series. Not that obvious to the OP I guess when looking at the question itself. $\endgroup$ – AlexR Oct 14 '13 at 8:27
  • $\begingroup$ Is it obvious? I'm assuming we're not allowed to use L'Hospital's rule. To prove $\lim_{x\to 0^+} \frac{\ln(\sin(x))}{\ln(x)}=1$, see this question. To prove $\lim_{x\to 0} \frac{\ln(\cos (x))}{\ln\left(1-\frac{x^2}{2}\right)}=1$, where I used $x\to 0$ instead of $x\to 0^+$ because $\cos (x)$, $\frac{x^2}{2}$ are both even functions, I've tried a few of the methods given in the link I gave, and none of them worked. $\endgroup$ – user263326 Sep 17 '17 at 18:22
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    $\begingroup$ Also, look at the series that WolframAlpha finds for $\ln\left((\cos x) / \left(1-\frac{x^2}{2}\right)\right)$ wolframalpha.com/input/?i=series+ln((cos+x)%2F(1-((x%5E2)%2F2. We can't immediately find, e. g., $\frac{x^{4}}{24\ln\left(1-\frac{x^2}{2}\right)}$, because it gives an indeterminate form $\frac{0}{0}$. This is another method used in an answer. $\endgroup$ – user263326 Sep 17 '17 at 18:23

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