Graph of $\quad\frac{x^3-8}{x^2-4}$.

I was using google graphs to find the graph of $$\frac{x^3-8}{x^2-4}$$ and it gave me:

Why is $x=2$ defined as $3$? I know that it is supposed to tend to 3. But where is the asymptote???

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That's not a "dot" at $x=2$: it's a "hole". In a rational function, when there is a common factor $(x - a)$ in the numerator and denominator, it "cancels", but only in the sense that the function remains undefined at $x = a$, but a finite limit for the function will exist as $x$ approaches $a$. Vertical asymptotes will occur at the values of $x$ at which the limit doesn't exist. – RecklessReckoner May 2 '13 at 2:55
Most of what you said makes sense, but that is not a hole. Google graphs has this feature where where can "trace". That dot is the tracing ball. – Hele May 2 '13 at 2:56
Nevertheless, that is an undefined point. I don't know what sort of representation Google uses for that. – RecklessReckoner May 2 '13 at 2:57
Have a look : Click Me! – Hele May 2 '13 at 2:59
Well, it is regrettable that Google's plotter has such a deficiency... This shows once again that one shouldn't rely on computers too much on mathematical matters: one need to have some awareness oneself about special situations. – RecklessReckoner May 2 '13 at 3:01

Because there is a removable singularity at $x = 2$, there will be no asymptote.

You're correct that the function is not defined at $x = 2$. Consider the point $(2, 3)$ to be a hole in the graph.

Note that in the numerator, $$(x-2)(x^2 + 2x + 4) = x^3 - 8,$$ and in the denominator $$(x-2)(x+ 2) = x^2 - 4$$

When we simplify by canceling (while recognizing $x\neq 2$), we end with the rational function $$\frac{x^2 + 2x + 4}{x+2}$$

We can confirm that the "hole" at $x = 2$ is a removable singularity by confirming that its limit exists: $$\lim_{x \to 2} \frac{x^2 + 2x + 4}{x+2} = 3$$

In contrast, however, we do see, that there is an asymptote at $x = -2$. We can know this without graphing by evaluating the limit of the function as $x$ approaches $-2$ from the left and from the right:

$$\lim_{x \to -2^-} \frac{x^2 + 2x + 4}{x+2} \to -\infty$$

$$\lim_{x \to -2^+} \frac{x^2 + 2x + 4}{x+2} \to +\infty$$

Hence, there exists a vertical asymptote at $x = -2$.

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+1 for the detailed explanation. – Hele May 2 '13 at 3:10
Thanks, Hele. Glad to help! – amWhy May 2 '13 at 3:11
For my dear tutor here. +1 – Babak S. May 2 '13 at 4:37
Hello, @Babak! did you sleep well? – amWhy May 2 '13 at 4:38
Yes @amWhy. It was deep and dreamy and empty of any Maths. :-) – Babak S. May 2 '13 at 4:39

There is no asymptote at $x=2$. Note that $$\frac{x^3-8}{x^2-4}=\frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)}.$$ For $x\ne 2$, we can cancel the $x-2$.

So near $x=2$ our function is very well behaved, it has a nice limit. The singularity at $2$ is called a removable singularity. If we define a new function $g(x)$ by $g(x)$ equal to our given expression when $x\ne 2$, and $g(2)= 3$, the function $g(x)$ is very nice everywhere except at $x=-2$. (The singularity at $x=-2$ is not removable.)

Many pieces of graphing software completely ignore removable singularities. At least Alpha had the decency to put a dot there.

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Some of the older graphing calculators which had lower resolution used to leave a "gap" (no "pixel" darkened) on the display for the function. – RecklessReckoner May 2 '13 at 2:59
Nope, my TI-84 does'nt show anything. – Hele May 2 '13 at 3:01
Pity, because it really ought to. Some resident graphing software on personal computers doesn't show anything either, but will indicate a problem with the point when you place the cursor there (if it is permitted to do so accurately enough) by declaring there is no defined value there -- some even make an audible "click"! – RecklessReckoner May 2 '13 at 3:04

There will only be an asymptote if the limit is infinity. in your case, for all points $x\neq 2$: $$\frac{x^3-8}{x^2-4} = \frac{x^2+2x+4}{x+2}$$ Which has no discontinuities!

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