I came across this expression, which I was asked to simplify and then choose the number that would make the expression undefined:


I simplified the expression to $\frac{z+1}{2z-3}$, and then chose $z = \frac{3}{2}$ since this would evaluate to a zero in the denominator. However, it turns out that $z = 0$ also makes it undefined, but I don't understand why. If we replace $z$ with $0$ we get $\frac{-1}{3}$. So my question is why can't the value of $z$ equal to $0$ in this case?

Any explanation will be appreciated.

  • 1
    $\begingroup$ The limit exists, but the function is not defined, when $z=0$. In a graph at $z=0$ there would be an infinitely small "hole". When using a graph to determine a function's behavior, holes are worthless, and so we call them "removable discontinuities." $\endgroup$ – David P Jul 16 '16 at 21:08
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    $\begingroup$ "If we replace $z$ with $0$ we get $-\frac{1}{3}$" - well, which expression did you replace $z$ with $0$ in? Certainly not in the original expression, but that's exactly the expression under consideration. $\endgroup$ – arctic tern Jul 16 '16 at 21:13
  • $\begingroup$ To clarify previous answers, when cancelling out $z/z = 1$ then we assume $z \neq 0$, otherwise $0/0$ is undefined (thus not equal to 1). Thus you have to consider the cases $z =0$ and $z \neq 0$, and you only considered the latter. $\endgroup$ – user305860 Jul 16 '16 at 21:49
  • $\begingroup$ There are some contexts where $x/x$ means something that is defined at $x=0$, and other contexts where $x/x$ means something that is undefined at $x=0$. In a context where the expression means "plug in the value and then perform the indicated arithmetic operations", $x/x$ is undefined at $x=0$ (and, thus, $x/x$ and $1$ are not equivalent expressions). $\endgroup$ – user14972 Jul 18 '16 at 13:25

In performing your simplification, you are writing $$ \frac{17z^3 + 17z^2}{34z^3 - 51z^2} = \frac{17z^2}{17z^2} \cdot \frac{z + 1}{2z - 3}$$ and then cancelling the numerator and denominator of the first factor. However, if $z = 0$, then the first factor becomes $\frac{0}{0}$, which is undefined. You can only cancel common factors if they are non-zero.

  • $\begingroup$ As suggested in @DavidP's comment, your calculation of $\frac{-1}{3}$ gives the limit of the rational function as $z \rightarrow 0$. If $z \rightarrow 0$, but is not equal to $0$, then you can perform the cancellation, and then what remains ($\frac{z+1}{2z-3}$) tends to $\frac{-1}{3}$. $\endgroup$ – Shagnik Jul 16 '16 at 21:11

$$\frac{17z^3+17z^2}{34z^3-51z^2} = \frac{z^3+z^2}{2z^3-3z^2} =\frac{z^2(z+1)}{z^2(2z-3)} $$

So here you should look at the denominator; equate it to zero and see what roots it has, the solutions will be the numbers that make the expression undefined:


$$ z_{1} =0 , z_{2} = 3/2 $$

So as you can see a $0$ will certainly make the expression undefined. In this kind of problems it's always important to never cancel out variables, as you are bound to loss roots in the process.


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