# Are equivalent rational polynomials still equivalent if not all of their limits at corresponding points are equal?

Apologies if I've used terms incorrectly. What I mean:

1. Far as I know, the limit of $x-5$ exists for all real numbers.

2. $\displaystyle\frac{(x-5)(x+5)}{x+5} = x-5$

3. But the limit of the left-hand side of $(2)$ doesn't exist (or isn't defined?) at $x = -5$, whereas the limit of the right-hand side $x-5$ does exist at $x = -5$.

So, if the left and right-hand parts of $(2)$ are equivalent, but only one has a limit at $x = -5$, does that make them, I don't know, equivalent but not interchangeable? Or is it not even a useful question to ask?

• the limit of 2 does exist for both left and right hand side. – Arashium Feb 21 '15 at 19:54
• Limit means that the function approaches the value,$-5$ in this case but it doesn't take the value $-5$.It wouldn't make sense if it took the value $-5$ since the expression is not defined there. – kingW3 Feb 21 '15 at 19:59

Your language is a bit imprecise. It would be much better if you just used limit notation. Also, note that when you evaluate a limit, you're not at all interested in the behavior of the function at the point, but in a neighborhood of it. This is paramount, as you can evaluate limits towards point which aren't in the function's domain.

To summarize:

1. $\lim\limits_{x \to \alpha} x - 5 \in \mathbb R\,~\forall \alpha \in \mathbb R$
2. Note that $$\frac{(x-5)(x+5)}{x+5} = x-5\quad\forall x \in \mathbb R \backslash\{5\}$$ so it wouldn't even make sense to consider the point $x = -5$ for the left-hand side, because it isn't even defined there. You can, however, write \begin{align}&\lim_{x \to -5} \frac{(x-5)(x+5)}{x+5} = -10\\ &\lim_{x \to -5} x - 5 = -10 \end{align}

Why we aren't interested in the exact point $x = -5$? If you recall the definition of a limit, $$\lim_{x \to \alpha}f(x) = l \iff \forall\varepsilon > 0,\,\exists \delta\mbox{ s.t. }0 < |x - \alpha| < \delta \implies |f(x) - l| < \epsilon,$$ you'll notice the part $0 < |x - \alpha| < \delta$. That expression is equivalent to $x \in (-\delta,~\alpha)\cup(\alpha,~\delta)$: the $x$ never takes the value $\alpha$. This way we can define limit for discontinuous functions and extend the definition at infinity, for example.

Notice:

In definition of limit:

$$\lim_{x \to a} f(x) = L \iff (\forall \varepsilon > 0)(\exists \ \delta > 0) (\forall x \in D)(0 < |x - a | < \delta \ \Rightarrow \ |f(x) - L| < \varepsilon)$$

$0 < |x - a |$ means that $x\neq a$ so in $lim_{x\to -5} f(x)$, $x$ is never equal to $(-5)$. $x$ can be very close to $(-5)$ as much as you want but not exactly equal to $(-5)$