Spivak Calculus Limit Intuition Clarification This is from page 94 of Spivak's "Calculus" 4th edition. He builds up the definition of a limit from examples, but I am confused about this paragraph:
Let $f(x) = \frac{1}{x}$ (for $x \neq 0$)

To show in general that $f$ approaches $1/a$ near $a$ for any $a$ we proceed in basically the same way, except that, again, we have to be a little more careful in formulating our initial stipulation. It's not good enough simply to require that $|x-a|$ should be less than 1, or any other particular number, because if $a$ is close to 0 this would allow values of $x$ that are negative (not to mention the embarrassing possibility that $x=0$, so that $f(x)$ isn't even defined!).

Why would negative values of $x$ be bad in this example? Couldn't the value of $a$ be negative, since $f(x)$ is defined for $x < 0$? 
 A: You are correct that $f(x)$ is defined for $x<0$ (though it is not defined for $x=0$), but it could be a problem that $x<0$ is close to $a>0$ but $\dfrac1x$ is not close to $\dfrac1a$, when studying $\lim\limits_{x\to a}f(x)$.
A: To understand why negative x values would be bad you should consider the graph of 1/x . 
In particular you should focus on the two different behaviors of the function when it approaches 0 from the left (from the negative side) or from the right (from the positive side).
If you want to show that $\lim_{x\to a} f(x) = 1/a$ you should be careful when choosing your interval on the x axis when $|a|$ is small: if your $(a-\delta, a+\delta)$ open interval centered at $a$ (for $\delta > 0$) is too big, you will also take into account values of the function in the "wrong" side (positive or negative) of the graph, where $f$ has a different behavior and therefore won't describe "what the function is doing" around the point $a$
The above holds for every function that has a discontinuity in its domain.
