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I can sort of understand why the left limit of $f(x)=\frac{|x|}{x}$ is $-1$ and the right one is $1$ but why isn't the limit undefined? Won't the function ultimately assume $0/0$?

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    Are you saying that the limit should be $0/0$? that doesn't make sense. And when you are talking about limits, you never actually reach that value, x is just going infinitely close to $0$ – 2017-02-24
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    Yes I know it never reaches zero but the part about limits that confuses me is why do we substitute X with a certain value when we compute limits when it never really reaches that value... – 2017-02-24
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    Only time you do that with limits is when you know the function is continuous at that point. Then substituting that value to find limit is justified – 2017-02-24
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    @user3423386: You know the _definition_ of "limit", right? It does not say anything about substituting $x$ for the value it approaches. Doing so is a shortcut to finding the limit that works **if** (but only if) you already know the function you're taking a limit of is _continuous_ a that point. In this case, that function isn't even defined at $x=0$, so the substitution has no right to work (and doesn't). – 2017-02-24
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    If $f$ is a continuous function, then $\lim_{x\to a}f(x) = f(\lim_{x\to a}x)$. We know (because hopefully we have previously proven) that $\lim_{x\to a}x = a$. This is the macinery behind "just inserting the limit value", and in my opinion it pays off to be aware of this every time you do it. It is an easy way to avoid many dumb mistakes when things get more complicated. – 2017-02-24
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    I see now. Thank you for the clarification. And yes the first thing I noticed about the function is that it wasn't defined at zero. I was just a bit confused that's all. The approaching value can be substituted to find the limit only if the function is continuous at that point. It's clear to me now. Thanks guys! – 2017-02-24
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    "I can sort of understand why the left limit is -1 and the right one is 1 but why isn't the limit defined?". Because the left limit is different then the right limit. "Won't the function ultimately assume 0/0?" Is 0/0 an actual thing? – 2017-02-24
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    No it isn't... I just worded it really badly. – 2017-02-24
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    "Why do we substitute x with a certain value when we calculate limits when it never really reaches that value" Because we are lazy cheaters. Seriously. We have proven that *if* a function is continuous, and *if* a function is defined at x, then lim f (x)=f (x). So why bother to correctly calculate the limit when we know it will be f (x) anyway? But if the function *isn't* continuous or it is not defined at x, we can't do that. |x|/x is a perfect example. It's not continuous at x=0, it "jumps" from -1 to 1, and at 0 it is 0/0 which is meaningless garbage. So we can not do lim f (0)=f (0) – 2017-02-24

2 Answers 2

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Let $\epsilon > 0$, $f(x) = |x|/x$. Set $\delta >0$. Then

  • $\forall x \in (-\delta,0), |f(x) -(-1)| = ||x|/x+1| = |(-x)/x+1| = 0 < \epsilon$
  • $\forall x \in (0,\delta), |f(x) - 1| = ||x|/x-1| = |x/x-1| = 0 < \epsilon$

Hence, $\lim\limits_{t\to0^-} f(x) = -1$ and $\lim\limits_{t\to0^+} f(x) = 1$. Since they aren't equal to each other, the limit is undefined.

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The limit is defined as the common value of the left- and right-hand limits, and doesn't exist if they differ (or if at least one of them doesn't exist). The absolute-value function is a famous example of a function with a turning point where its derivative is undefined instead of being zero.