Why is it legit to evaluate $\lim_{x\rightarrow 1} \frac{(x-1)(x+1)}{x-1}$ by cancelling common factors?

I haven't ever taken an analysis course, so maybe that's where I would really learn this, but I've always wondered why it's okay to do this when evaluating a limit. I guess it's the case that there is a theorem which says that the limit of a rational function as $x\rightarrow a$ is equal to the limit of that function in lowest terms as $x\rightarrow a$, regardless of division by zero. Is there a name for that theorem? Or does it follow from some other property of limits or rational functions?

Solution 1:

The important thing to remember about the limit process in something like this is that whenever we evaluate something like $\lim_{x\rightarrow 1} f(x)$, we don't require the function $f(x)$ to be defined at $x=1$, so we look at the values of $f(x)$ as $x$ approaches 1, but taking values which are not equal to 1.

In your example, it means that as $x$ approaches 1 without actually equalling 1, the factor may be cancelled because it is definitely not zero.

I don't know of any general result, though, and have always done things like this by examining each case on its merits, by asking if we can be sure that cancellation is allowable in a particular case.

Solution 2:

The limit is about letting $x$ "approach" the value of $1$, not setting $x=1$.

For this reason you are allowed to cancel since the denominator will never be $0$ for these values of $x$.

Solution 3:

One relatively subtle reason that this is true for rational functions is that polynomial functions (so in particular, the denominator of your rational function) have what are known as isolated zeroes: because a polynomial has only finitely many zeroes (this is a form of the Fundamental Theorem of Algebra), no two zeroes can be too close to each other. In particular, around each zero there's a disc of finite radius in which no other zeroes can occur. This means that a rational function is well-defined in some neighborhood of each of the zeroes of its denominator.

For an example of what can go wrong when you step away from rational functions, consider instead the function $\displaystyle f(x) =\frac{x\sin(\frac{1}{x})}{\sin(\frac{1}{x})}$: then can you see why $\displaystyle\lim_{x\rightarrow 0} f(x)$ is undefined?

Solution 4:

Your expression $f(x)$ is not definied at $x=1$, therefore the limit represents the value of $f(x_n)$ for each sequence $x_n\rightarrow1$ with the property that $x_n \neq 1$ for all $n$. But when $x_n \neq 1$ you can of course cancel the two terms as they are not zero.

Solution 5:

It is a theorem that if the limits below are exist and finite, $$ \lim_{x\to x_{_{0}}} f(x)=A $$ and $$ \lim_{x\to x_{_{0}}} g(x)=B, $$ then $$ \lim_{x\to x_{_{0}}} f(x)g(x) $$ exists and equals $AB$.

In your case $$ \lim_{x\to 1} \frac{x-1}{x-1}=1 $$ and $$ \lim_{x\to 1} x+1=2, $$ so your limit exists and equals $2$. To emphasize the idea $$ \lim_{x\to 1} \frac{x-1}{x-1}=1=g(x), $$ where the function $g(x)$ is defined for $x\neq 1$ as the constant $1$. This is the reason why we can cancel the $\frac{x-1}{x-1}$ to $1$.