I was working on a few problems from James Stewart's Calculus book (seventh edition) and I found the following:


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

Since there's a $|x|$ on the limit and knowing that $|x| = -x$ for any value less than zero, we have

$$\lim_{x \to 0^-} \left( \frac{1}{x} - \frac{1}{|x|} \right) = \lim_{x \to 0^-} \frac{2}{x}$$

So far so good. Continuing,

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

since the denominator becomes smaller and smaller. When checking the textbook's answer I've found the following:

Short answer

Long answer

Am I missing something or should the limit really be $- \infty$ ?

  • 2
    $\begingroup$ Saying the limit is $-\infty$ is (roughly) the same thing as saying the limit doesn't exist. $\endgroup$ – Zain Patel Jul 5 '16 at 10:48
  • 1
    $\begingroup$ You may want to read this: math.stackexchange.com/questions/1782077/… $\endgroup$ – Zain Patel Jul 5 '16 at 10:49
  • $\begingroup$ @ZainPatel got ya, thank you, best regards! $\endgroup$ – bru1987 Jul 5 '16 at 10:52
  • $\begingroup$ Saying the limit doesn't exists isn't the same as saying the limit is $\;-\infty\;$ . Perhaps one could say "the limit doesn't exist finitely". $\endgroup$ – user351910 Jul 5 '16 at 11:13
  • $\begingroup$ Both answers are correct, but your answer is better (since it gives information about why the limit does not exist). $\endgroup$ – user84413 Jul 5 '16 at 17:13

Saying that the limit is equal to $\infty$ is a mathematical shorthand (amongst some mathematicians, at least) for:

Given any real number $M$, there is a real $\delta$ (depending on $M$) such that $\frac{1}{x^2} > M $ for all $x$ satisfying $0 <x <|\delta|$.

It is usually advised that beginners avoid using $\infty$ since it leads to careless or wrong manipulations of the symbol all too often.


bru, I think that the problem is just about terminology. Your derivation is correct, but it is likely that what Stewart is claiming is (I guess) that a limit that goes to $-\infty$ or to $+\infty$ on only one side (as in this example, where the limit is only from the left), is "non existing". Otherwise, the statement "the limit does not exist becuase the denomiator approaches $0$ while the numerator does not" would be simply nonsense.

  • $\begingroup$ I see friend, thank you for explaining. I appreciate, best regards! $\endgroup$ – bru1987 Jul 5 '16 at 11:50
  • $\begingroup$ @bru1987 thank you. If you like my answer you can check on it, or upvote it. Thank you again. $\endgroup$ – RandomGuy Jul 5 '16 at 11:52

I agree that the limit is negative infinity. I could understand "does not exist" if the question was asking for a two-sided limit, but it clearly isn't. It's asking for the left-hand limit. The text differentiates between specifically identifying ± infinite limits and non-existent limits, so it should be consistent in the solution manual. I just burned an hour trying to figure out if this was some strange concept that I missed... and I'm the teacher! Glad I found this thread!


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