# Why are we allowed to cancel fractions in limits?

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 0:06
• @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 0:57

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 $f(x)=x^2$

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

• Where did you get that definition? Can you provide a link? – Stefan Pochmann Jan 31 '18 at 11:37
• 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? – gimusi Jan 31 '18 at 11:52
• @StefanPochmann en.wikipedia.org/wiki/(ε,_δ)-definition_of_limit – mbomb007 Jan 31 '18 at 17:47

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 $a.$

• This doesn't seem to address the question; it doesn't say why the cancellation is allowed. – user2357112 Jan 31 '18 at 8:42
• @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
• 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 5 '18 at 22:17

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.

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$0$...

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.

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}

However, you of course cannot do crazy stuff like claiming that $$\lim_{x\to0-} \frac{x\sqrt{x}}{\sqrt{x}} = \lim_{x\to0-} x = 0$$ because here you modify the domain of the function in more than finitely many points.