# Limit calculation and discontinuity

Having a function, which has a polynomial in the denominator like:

$$\lim_{x \to 2}\,\dfrac{x+3}{x-2}$$

We see there is a discontinuity at x=2, because it sets the denominator to 0. But multiplying by its conjugate we get:

$$\dfrac{x+3}{x-2}\cdot\dfrac{x-2}{x-2} = \dfrac{(x+3)(x-2)}{x^2-4}$$ And now there is no discontinuity in $x=2$ anymore, but should not that function be an equivalent of the original one? we are just multiplying by $1$.

UPDATE:

Apologies, I chose a very poor example.

Consider this one:

$$\lim_{x \to -2}\,\dfrac{x+2}{x^3+8}$$

When doing f(-2) you get 0/0. By doing factor, cancel and plug in:

$$\lim_{x \to -2}\, \dfrac{(x+2)}{(x+2)(x^2-2x+4)}$$

$$\lim_{x \to -2}\, \dfrac{1}{x^2-2x+4}$$

Then f(-2)=1/12, it is not 0/0 anymore. So my question is that how come that two functions that should be equivalent, have different discontinuit?

• Why don't you calculate $\lim_{x \to 2^-}\,\dfrac{x+3}{x-2}$ and $\lim_{x \to 2^+}\,\dfrac{x+3}{x-2}$? Thus, you could see what happens in $x = 2$. Commented Oct 21, 2014 at 16:01

The functions $f(x)=\dfrac{x+2}{x^3+8}$ and $g(x)=\dfrac{1}{x^2-2x+4}$ are not equivalent, since $x=-2$ is not in the domain of $f$, but is in the domain of $g$.

When you're looking for the limit, you are interested in the behaviour as $x$ approaches $-2$, and both functions have the same behaviour in this case. That's why it's allowed to algebraically manipulate the rational function as far as limits are concerned.

• I read your message yesterday, and I am still trying to land it in my head and make peace with it. It makes a lot of sense, but the fact they are not equivalent, breaks my little understanding of functions and polynomials (and this may be a good thing). Commented Oct 22, 2014 at 9:49
• No problem, a new concept needs some time. I'm glad I could help. Consider clicking the checkbox if you thought my answer was helpful. :) Commented Oct 22, 2014 at 9:55

What is $x^2 - 4$ at $x = 2$? You still have a denominator of zero.

As $x \to 2$, the limit of $\dfrac {x+3}{x -2}$ does not exist.

$$\lim_{x\to 2^-} \dfrac{x+3}{x-2} = -\infty$$

$$\lim_{x\to 2^+} \dfrac{x+3}{x-2} = +\infty$$

Edit: Note that the functions are equivalent everywhere except at $x=-2$. The original (second function in the post) is not defined at $x = -2$, whereas the subsequent is defined there. Since they have different domains on which each is defined, they are not strictly equivalent.

Recall that when finding the limit as $x\to -2$, we are not interested in the behavior of the function at $x = -2$, only as $x$ approaches $-2$, and for this purpose (finding the limiting behavior of the function as $x\to -2)$, the functions are equivalent.

• Sorry, I chose a very bad example for the post. Please take a look to the update. Commented Oct 21, 2014 at 16:22

The mistake you are making is that $\lim_{x\to -2}\frac{x+2}{x^3+8} = \lim_{x\to -2}\frac{x+2}{(x+2)(x^2-2x+4)} = \frac{0}{0}$ is not necessarily discontinuous. It is a so called indeterminant form. Typically, you have to do something algebraic to "get to" the actual limit, similar to what you did by factoring and cancelling. So indeed $$\lim_{x\to -2}\frac{x+2}{x^3+8} = \lim_{x\to2} \frac{1}{x^2-2x+4} = \frac{1}{12}.$$

Multplying by the conjugate quantity does not take away the discontinuity ($2$ is still a root of the denominator $x^2-4$) at $x=2$. What you can say is :

$$\lim \limits_{\substack{x \to 2 \\[1mm] x < 2}} \frac{x+2}{x-2} = - \infty$$

and

$$\lim \limits_{\substack{x \to 2 \\[1mm] x > 2}} \frac{x+3}{x-2} = + \infty.$$

Edit : The example you gave when you edited you post could be generalized : let $P$ and $Q$ be two polynomial functions defined on $\mathbb{R}$ and $a \in \mathbb{R}$ such that : $a$ is a root of $P$ and $Q$ with multiplicity $1$. Since $a$ has multiplicity $1$, there exist a polynomial function $P_{1}$ (resp. $Q_{1}$) such that $\forall x \in \mathbb{R}, \, P(x) = (x-a)P_{1}(x)$ and $P_{1}(a) \neq 0$ (resp. $\forall x \in \mathbb{R}, \, Q(x) = (x-a)Q_{1}(x)$ and $Q_{1}(a) \neq 0$). The function

$$x \, \longmapsto \, \frac{P(x)}{Q(x)}$$

is not defined at $a=0$ but note that :

\begin{align*} \frac{P(x)}{Q(x)} &= {} \frac{ \require{cancel} \cancel{\color{blue}{(x-a)}}P_{1}(x) }{ \require{cancel} \cancel{\color{blue}{(x-a)}}Q_{1}(x) } \\[2mm] &= \frac{P_{1}(x)}{Q_{1}(x)} \\ \end{align*}

As a consequence :

$$\lim \limits_{x \to a} \frac{P(x)}{Q(x)} = \lim \limits_{x \to a} \frac{P_{1}(x)}{Q_{1}(x)} = \frac{P_{1}(a)}{Q_{1}(a)}.$$

In your example : $a=-2$, $P(x) = x+2$, $P_{1}(x) = 1$ and $Q(x) = x^{3} + 8$, $Q_{1}(x) = x^{2}-2x+4$.

• Sorry, I chose a very bad example. Please take a look to the update. Commented Oct 21, 2014 at 16:22
• @NullOrEmpty : I edited my answer accordingly. Commented Oct 21, 2014 at 20:41