Why are we allowed to cancel fractions in limits?

Question

For example:

$$\lim_{x\to 1} \frac{x^4-1}{x-1}$$

We could expand and simplify like so:

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

In this case we divided out $x-1$ on top and bottom even though technically, at $x=1$, we have $\frac{0}{0}$ that we're just tossing aside.

But what allows us to do this?

Because in the definition of that limit the $x$ is quantified on a set in which $x\neq1$ (notice the $\mathbf{0<}|x-1|<\delta$ in the definition). Since when $x\neq1$ the two functions, after and before cancelling, are equal, then the two limits are equal. — orole, Jan 30 '18 at 23:46
Ah so I'm not really dividing out $0/0$ since we're not actually reaching $x=1$? — user525966, Jan 30 '18 at 23:48
Yes, for limit there is that $0<$ in the definition. If the $0<$ is not there it results in another concept, continuity. — orole, Jan 30 '18 at 23:49
$\lim_\limits{x\to a} f(x) g(x) = \lim_\limits{x\to a} f(x) \lim_\limits{x\to a} g(x)$ and $\lim_\limits{x\to 1} \frac {x-1}{x-1} = 1$ — Doug M, Jan 31 '18 at 00:06
@DougM that doesn't actually answer the OP question. It can still be asked why can we divide by $x -1$ to dertmine what $\lim \frac {x-1}{x-1}$ is. — fleablood, Jan 31 '18 at 00:57
@fleablood I didn't say you can divide one factor in the numerator by one factor in the denominator (although effectively you can). I was trying to suggest that can separate these "troublesome" factors and evaluate them independently. At some point you need a definition of exactly what is a limit. — Doug M, Jan 31 '18 at 01:21
@DougM I don't follow, do you mean that $f(x) = x-1$ and $g(x) = \frac{1}{x-1}$ so we can compute the limits separately? $\lim_{x \to 1}(x-1) \cdot \lim_{x \to 1} \frac{1}{x-1} = 0 \cdot \frac{1}{0}$ though — user525966, Jan 31 '18 at 01:25
no, I mean $f(x) = (x^3+ x^2 + x + 1)$ and $g(x) = \frac {x-1}{x-1}$ then $\lim_\limits{x\to 1} \frac {x^4-1}{x-1} = \lim_\limits{x\to 1} f(x) \lim_\limits{x\to 1} g(x), f(x)$ is a polynomial, hence continuous. Leaving it to show that $|g(x) -1| < \epsilon$ everywhere that $g(x)$ is defined. — Doug M, Jan 31 '18 at 01:34
@DougM So ... why can we say $\lim_{x\to 1} \frac {x-1}{x-1} = 1$? Magic? God told us? Because we can divide by $x-1$ when $x \ne 1$? Just shut up and accept it? There's nothing here that answers the question of the OP. — fleablood, Jan 31 '18 at 04:26
@fleablood I thought it was largely "Because we can divide by $x−1$ when $x≠1$" is it not? Have I misunderstood? — user525966, Jan 31 '18 at 04:34
@fleablood $|\frac {x-1}{x-1} - 1| < \epsilon$ for all $x$ in the domain. — Doug M, Jan 31 '18 at 04:55
Well, yes, tha's the reason why and is the answer to the question. But your original comment does not address that nor adds anything to the issue. Your original comment entirely begs the question. The question was "why can we claim $\lim\frac{x^4-1}{x-1}=\lim x^3+x^2+x+1$? Isn't that dividing by 0?" and your comment $\lim\frac{x^4-1}{x-1}=\lim (x^3+x^2+x+1)\lim\frac {x-1}{x-1}$ does't answer that AT ALL. The questions STILL "why can we claim $\lim\frac{x-1}{x-1} = 1$? Isn't that dividing by 0?" Isolating the "troublesome" factors doesn't answer the question; it merely delays the inevitable. — fleablood, Jan 31 '18 at 15:08
@fleablood: DougM's comment simplifies the question, and makes it somewhat easier to apply the definition of a limit. You're right that it's not a full answer to the question, but presumably that's why DougM posted it as a comment and not as an answer to the question! — psmears, Jan 31 '18 at 21:24

user · Accepted Answer · 2018-01-31T17:54:39.103

51

Simply because we are dealing with values $x\neq 1$ in this case, thus for algebraic rule we are allowed to cancel out

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

Remember indeed that by the definition of limit we are demanding that $$\forall \varepsilon>0 \quad \exists \delta>0 \quad \text{such that}\quad \color{green}{\forall x\neq1}\quad|x-1|<\delta \implies|f(x)-L|<\varepsilon$$

Note also that the same cancellation is used to prove the basic derivatives case, for example for $f(x)=x^2$

$$\lim_{x\to x_0}\frac{x^2-x_0^2}{x-x_0}=\lim_{x\to x_0}\frac{\color{red}{(x-x_0)}(x+x_0)}{\color{red}{x-x_0}}=\lim_{x\to x_0}(x+x_0)=2x_0$$

edited Jan 31 '18 at 17:54

answered Jan 30 '18 at 23:52

user

154,566

Where did you get that definition? Can you provide a link? – Stefan Pochmann Jan 31 '18 at 11:37
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Usually it is stated as $0<|x-x_0|<\delta$ which is equivalent to $|x-x_0|<\delta$ and $x\neq x_0$, some doubt on it? – user Jan 31 '18 at 11:52
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@StefanPochmann https://en.wikipedia.org/wiki/(ε,_δ)-definition_of_limit – mbomb007 Jan 31 '18 at 17:47
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I've seen very often people make a distinction between $\lim_{x \rightarrow 1}f(x)$ and $\lim_{\substack{x \rightarrow 1 \ x \neq 1}} f(x)$, but in this case the $x-1$ in the denominator makes it pretty clear which definition is used. – Stef May 09 '23 at 23:34

score 14 · Answer 2 · answered Jan 31 '18 at 00:25

14

Proposition 1: If $f(x) = g(x)$ whenever $x\ne a,$ then $\lim\limits_{x\,\to\,a} f(x) = \lim\limits_{x\,\to\,a} g(x).$

Proposition 2: After the cancelation, the resulting function is continuous at $a,$ so the limit can be found by plugging in $a.$

answered Jan 31 '18 at 00:25

Michael Hardy

1

1

This doesn't seem to address the question; it doesn't say why the cancellation is allowed. – user2357112 Jan 31 '18 at 08:42
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@user2357112 : It appears to me that that's exactly what it does say. Maybe you need to clarify the question further if this does not address it. – Michael Hardy Jan 31 '18 at 13:23
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When I state Proposition 1 in class I say, "Limits don't see the point!" – Matthew Leingang Jan 31 '18 at 21:06
@user2357112 : At this point perhaps I should tell you that you may be missing something if you don't think this explains why the cancellation is allowed, but I probably won't be able to explain what until you clarify further. – Michael Hardy Feb 05 '18 at 22:17

score 10 · Answer 3 · answered Jan 31 '18 at 00:53

You are correct. At the point $x=1$ the expression is undefined/behaves badly and has no value.

But limits aren't about functions at the point $x = 1$. They are about functions near the point $x = 1$. In fact, they are specifically about when $x \ne 1$ (but is close to $1$).

$\lim_{x\to a} f(x) = K$ means if $x$ is NEAR $a$ then $f(x)$ is NEAR $K$.

And if $x$ is near $a$ then $x$ isn't $a$ and it is perfectly fine to divide by $x -a$ when $x \ne a$.

Now your hackles should be raised when you hear something like "$\frac {x^4 -1}{x-1}$ is near $4$ when $x$ is near $1$" and ask yourself what can "near" possibly mean in precise mathematical terms.

That's a question for another time.

score 6 · Answer 4 · edited Jan 31 '18 at 13:21

You never actually reach $1$... $x$ gets closer and closer to $1$ without ever being $1$...
Therefore, you can divide by $x-1$; it's never $0$... See limits.

Consider the function $f(x)=\begin{cases} 1 \text{ when } x=0 \\ \frac1x \text{ when } x\not= 0\end{cases} \cdots$

Study the limiting behavior of $f$ at $0$... Notice it has nothing to do with $f$'s value, $1$, at$0$...

score 3 · Answer 5 · answered Jan 31 '18 at 11:58

The functions defined by the expressions

$$\frac{(x-1)(x^3 + x^2 + x + 1)}{x-1} \quad\text{and}\quad x^3 + x^2 + x + 1$$

are not the same (because they are defined on different domains), but they agree outside of $x=1$. And the limit $\lim_{x\to 1}$ does not care about the value (if existent) at $x=1$, but only about values close to $1$.

Conclusion: Since the limit only sees the parts of these function in which they agree, it cannot distinguish between the two expressions (even though they are differnt from your perspective), and has to give the same result for both.

score 1 · Answer 6 · answered Jun 10 '18 at 05:48

Algebraic Limit Theorem: Let the limits exist: $$\lim_\limits{x\to a} f(x)=L \quad \text{and} \quad \lim_\limits{x\to a} g(x)=M.$$ Then: $$\begin{align}&1) \ \lim_\limits{x\to a} (f(x)\pm g(x))=\lim_\limits{x\to a} f(x)\pm \lim_\limits{x\to a} g(x)=L\pm M;\\ &2) \ \lim_\limits{x\to a} (f(x)\cdot g(x))=\lim_\limits{x\to a} f(x)\cdot \lim_\limits{x\to a} g(x)=L\cdot M;\\ &3) \ \lim_\limits{x\to a} (f(x)/ g(x))=\lim_\limits{x\to a} f(x)/ \lim_\limits{x\to a} g(x)=L/M; \quad (\text{provided:} \lim_\limits{x\to a} g(x)=M\ne 0). \\ \end{align}$$ Note that: $$\begin{align}\lim_{x\to 1} \frac{x-1}{x-1} = \lim_{x\to 1} 1&=1;\\ \lim_{x\to 1} (x^3 + x^2 + x + 1) &= 4;\\ \lim_{x\to 1} \frac{x^4-1}{x-1}=\lim_{x\to 1} \frac{(x-1)(x^3 + x^2 + x + 1)}{x-1} &= \\ \lim_{x\to 1} \frac{x-1}{x-1}\cdot \lim_{x\to 1} (x^3 + x^2 + x + 1) &= 1\cdot 4=4.\end{align}$$

Why are we allowed to cancel fractions in limits?

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