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Why does the following limit not exist?

$\lim \limits_{x \to 0^-} \left(\frac{1}{x} - \frac{1}{|x|}\right)$

I understand that the equation simplifies to $\frac{2}{x}$, but I don't understand how the absolute value affects the result of the limit, when otherwise, the limit would be negative infinity.

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  • $\begingroup$ Without the absolute value the limit would be $0$ $\endgroup$ – Elliot G Dec 5 '19 at 21:31
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    $\begingroup$ The limit is $-\infty$. It's the two sided limit that doesn't exist. $\endgroup$ – Matt Samuel Dec 5 '19 at 21:32
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We have that the limit doesn't exist finite but the limit exists and it is equal to $-\infty$, indeed since $x<0$ we have $|x|=-x$ and therefore

$$\lim \limits_{x \to 0^-} \left(\frac{1}{x} - \frac{1}{|x|}\right)=\lim \limits_{x \to 0^-} \left(\frac{1}{x} + \frac{1}{x}\right)=\lim \limits_{x \to 0^-} \frac{2}{x} =-\infty$$

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  • $\begingroup$ As I commented to another user, I have Stewart's Calculus, and the textbook says the limit does not exist. $\endgroup$ – sui Dec 5 '19 at 21:49
  • $\begingroup$ @sui The limit doesn't exists finite is ok but in this case the limit exists. Maybe there is a typo with $x\to0^-$? $\endgroup$ – user Dec 5 '19 at 21:52
  • $\begingroup$ @sui, would Stewart say that $\lim_{x\to0}1/x^2$ does exist? If so, can you point to a specific example in Stewart where he explicitly says "the limit exists" when in fact it's infinite? (I don't have a copy of Stewart handy, but I'm sure other people do.) I've always regarded the formula $\lim f(x)=\infty$ as saying "the limit does not exist because..." with the definition of infinite limit given as the reason. $\endgroup$ – Barry Cipra Dec 5 '19 at 23:12
  • $\begingroup$ @BarryCipra I say that the limit exist when it is finite or positive infinite or negative infinite and I say that it doesn't exists otherwise. $\endgroup$ – user Dec 5 '19 at 23:16
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    $\begingroup$ @user, I agree, it's a question of convention and/or exact definition. That's why I asked the OP to look to how Stewart treats things. The OP seems to be saying that Stewart regards one particular infinite limit as not existing; either Stewart is consistent in regarding infinite limits as non-existent (which would agree with my point of view), or there is an error in the book. $\endgroup$ – Barry Cipra Dec 5 '19 at 23:37
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The limit is -$\infty$ from the left; the limit from both sides does not exist. Look at this desmos graph for reference:

https://www.desmos.com/calculator/5lr2i9i9l4

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  • $\begingroup$ I'm confused. I'm going through my old textbook to brush up on Calculus, and the solution says the limit does not exist because the denominator approaches 0, while the numerator doesn't? $\endgroup$ – sui Dec 5 '19 at 21:44
  • $\begingroup$ Perhaps the verbiage in your textbook is assuming that if the limit does not equal a real number, then they just denote it as not existing? As in if the limit goes to plus or minus infinity, then the limit itself does not exist. This would not be a very awful way of describing a limit though, since a limit approaching plus or minus infinity is useful information in calculus. Maybe you could look to see if the author specifies that a limit like this is just denoted as not existing somewhere in the textbook? It certainly approaches negative infinity. $\endgroup$ – Ty Jensen Dec 5 '19 at 21:48
  • $\begingroup$ There's an entire section on infinite limits, so I don't think that's it. It's Stewart's Calculus, which afaik is well-regarded, which is why I doubted my own answer instead of the textbook. $\endgroup$ – sui Dec 5 '19 at 21:51
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    $\begingroup$ Beats me! From here I would tell you to contact the author for an explanation... but sadly James Stewart passed in 2014 according to my research. If I were you I would just understand that this is likely a mistake and the limit in fact approaches negative infinity. $\endgroup$ – Ty Jensen Dec 5 '19 at 21:58

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