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?

Answer 1

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 $f(x)=x^2$

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

Answer 2

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$0$...

Answer 3

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 $a.$

Answer 4

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.

Answer 5

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.

Answer 6

Well, the simple answer is: Because limits never depend on values in single points, so we are allowed to modify the limited function in finitely many points without changing the limit.

• Either the point is the one in which we are taking the limit; but we know that it's excluded.
• Or it's any other point, and than it's "far away" from the limit point and can be ignored altogehter.

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.

Answer 7

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: $$\begin{align}\lim_{x\to 1} \frac{x-1}{x-1} = \lim_{x\to 1} 1&=1;\\ \lim_{x\to 1} (x^3 + x^2 + x + 1) &= 4;\\ \lim_{x\to 1} \frac{x^4-1}{x-1}=\lim_{x\to 1} \frac{(x-1)(x^3 + x^2 + x + 1)}{x-1} &= \\ \lim_{x\to 1} \frac{x-1}{x-1}\cdot \lim_{x\to 1} (x^3 + x^2 + x + 1) &= 1\cdot 4=4.\end{align}$$

See also: Limit Theorems.