Reduce a fraction

Question

Given the function

$f(x)=\frac{x^2-5}{x+\sqrt{5}}$

If I draw this function in maple, I will get a line. How can that be true? I should expect a line except in area of $x = -\sqrt{5}$, where $f(x) \rightarrow \infty$ or $f(x) \rightarrow -\infty $. Of course Maple has factorized the numerator and reduced.

$\frac{x^2-5}{x+\sqrt{5}} = x-\sqrt{5},\ x\neq -\sqrt{5}$

My question is now, how can we ever reduce such a fraction with only condition $x\neq -\sqrt{5}$, when it is "$-\sqrt{5}$ and around it".

Answer 1

A function doesn't go to infinity every time it's denominator goes to zero. If the numerator also goes to zero, then "anything" can happen. In this case, the graph of your function looks exactly like the graph of $x - \sqrt{5}$ except that there's a tiny hole in the line at $x=-\sqrt{5}$ where the function is undefined. Maple won't see the hole. Bottom line is that you have everything right, but you expected something different.

Answer 2

Note that $x^2 - 5 = x^2 - \sqrt 5^2 = (x-\sqrt5)(x+\sqrt 5)$, so we have $$ f(x) = \frac{x^2 - 5}{x + \sqrt 5} = \frac{(x-\sqrt 5)(x + \sqrt 5)}{x+\sqrt 5} = \frac{x-\sqrt 5}{1} = x-\sqrt 5 $$ However, this manipulation does not suddenly make $f$ defined when $x = -\sqrt 5$. The only thing it does is that it simplifies the process of studying how $f$ behaves very close to $x = -\sqrt 5$, namely, it behaves exactly like the function $g(x) = x-\sqrt 5$, which is defined on the whole of the real line.

Answer 3

The arithmetical expression $$\frac{x^2-5}{x+\sqrt5}$$ is not defined for $x=-\sqrt 5$.

To define a function, we must specify the domain. A definition like $$f(x)=\frac{x^2-5}{x+\sqrt5}$$ makes no sense because the domain is not specified. Think that $x$ could be a sheep, or a chair.

On the other hand, a definition like $$f(x)=\frac{x^2-5}{x+\sqrt5}\text{ for }x\in\Bbb R$$ doesn't either make sense, because the expression has no sense for an element of the domain (namely, $-\sqrt 5$).

This definition is correct: $$f(x)=\frac{x^2-5}{x+\sqrt5}\text{ for }x\in\Bbb R\setminus\{-\sqrt 5\}$$

This definition is also correct, and defines the same function: $$f(x)=x-\sqrt 5\text{ for }x\in\Bbb R\setminus\{-\sqrt 5\}$$

Lastly, another correct definition, but this time it does not define the same function, because the domain is not the same:

$$g(x)=x-\sqrt 5\text{ for }x\in\Bbb R$$