Let's say we have a function, for example, $$ f(x) = \frac{x-1}{x^2+2x-3}, $$ and we want to now what is $$ \lim_{x \to 1} f(x). $$ The result is $\frac{1}{4}$.

So there exists a limit as $x \to 1$.

My teacher says that the limit at $x=1$ doesn't exist. How is that? I don't understand it. We know that a limit exists when the one sided limits are the same result.

Thank you!

  • $\begingroup$ Try factoring the denominator. This is probably a homework problem, so look at the numerator for a hint. $\endgroup$ – copper.hat Oct 16 '14 at 22:35
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    $\begingroup$ @copper.hat: I completely disagree! In the definition of "$\lim_{x\to 1} f(x)$", $f(1)$ doesn't need to be defined (in fact, in most interesting cases it isn't), and even if it is defined, that value shouldn't be taken into account. Adding "$x \neq 1$" doesn't make any difference whatsoever. $\endgroup$ – Hans Lundmark Oct 17 '14 at 7:13
  • $\begingroup$ @HansLundmark: I stand corrected. $\endgroup$ – copper.hat Oct 17 '14 at 15:03

It's possible that your teacher was pointing out the fact that the function doesn't exist at $x = 1$. That's different from saying that the limit doesn't exist as $x \to 1$. Notice that by factoring, $$ f(x) = \frac{x-1}{x^2 + 2x - 3} = \frac{x-1}{(x-1)(x+3)} $$

As long as we are considering $x \ne 1$, the last expression simplifies: $$ \frac{x-1}{(x-1)(x+3)} = \frac{1}{x+3}. $$ In other words, for any $x$ other than exactly $1$, $$ f(x) = \frac{1}{x+3}. $$ This helps understand what happens as $x$ gets ever closer to $1$: $f(x)$ gets ever closer to $$ \frac{1}{(1) + 3} = \frac{1}{4}. $$

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    $\begingroup$ The word 'exactly' unnecessary. $\endgroup$ – CiaPan Oct 17 '14 at 8:12
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    $\begingroup$ @CiaPan: What about the word 'is'? $\endgroup$ – Nikolaj-K Oct 17 '14 at 13:43
  • $\begingroup$ @NikolajK that depends on what 'is' is. $\endgroup$ – Yakk Oct 17 '14 at 14:53
  • $\begingroup$ @Yakk: It's equivalent to equivalence in univalent foundations. $\endgroup$ – Nikolaj-K Oct 17 '14 at 16:37

Your teacher is not correct. There are two easy ways to do this problem. The first is by factoring the denomiator:

$$\lim_{x\to 1}\frac{x-1}{(x-1)(x+3)}=\lim_{x\to 1}\frac{1}{x+3}=\frac{1}{4}$$

The second is by using L'Hospital's rule, which is a useful identity in limits. By L'Hospital's rule, we know that

$$\lim_{x\to 1}\frac{x-1}{x^2+2x-3}=\lim_{x\to 1}\frac{1}{2x+2}=\frac{1}{4}$$

This limit exists, because it is simply a discontinuity in the function, but it is a discontinuity at a single point. When we have certain indeterminate forms in limits, we may apply L'Hospital's rule, which allows us to better compute the limit.

L'Hospital's rule states that, in certain indeterminate forms:

$$\lim_{x\to n}\frac{f(x)}{g(x)}=\lim_{x\to n}\frac{f'(x)}{g'(x)}$$

It is worth clearing up a particular misconception that seems to have arisen. For $$f(x)=\frac{x-1}{x^2+2x-3}$$ we cannot compute the value of $f(1)$, because this results in an indeterminate form. However, the limit exists, because there is simply a local discontinuity at a single point in an otherwise continuous function.

  • $\begingroup$ A third alternative, to whoever may care, is to note that the limit if $\dfrac 1{f'(1)}$, where $f(x)=x^2+2x-3$. $\endgroup$ – Git Gud Oct 16 '14 at 23:57
  • $\begingroup$ To justify the use of l'Hospital, you also need to point out that $g' \neq 0$ in a punctured neighbourhood of the point in question. $\endgroup$ – Hans Lundmark Oct 17 '14 at 7:14
  • $\begingroup$ A hard requirement of the usability of l'Hôpital's rule is that the resulting "right-hand-size" is exists. But I guess that's what Hans is saying in more fancy terms? $\endgroup$ – rubenvb Oct 17 '14 at 8:00
  • $\begingroup$ @rubenvb: No, it's not enough that $\lim f/g$ is of type $0/0$ and that $\lim f'/g'$ exists. There are examples which show this. You really need $g'$ to be nonzero near the point. (Those examples are a bit artificial to be honest, but still...) $\endgroup$ – Hans Lundmark Oct 17 '14 at 15:10
  • $\begingroup$ @Hans $g'$ can be zero if $f'$ is zero, because you just have another L'Hospital's case and take $f''/g''$ Otherwise, just compute the limit normally. $\endgroup$ – Aza Oct 17 '14 at 16:13

I don't know what your teacher means exactly. Limits are defined when $x$ tends to some number, or infinity. Not when $x$ is this number.

The value that the function takes at the limit poit is irrelevant (to compute the limit). In fact, in most high school limit exercises, the function is not defined at the point that $x$ tends to.


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