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Formally speaking, can $\frac{x}{x}$ be defined for $x=0$? Normally this division turns to be 1 but we also know that we can only divide it when the denominator is different from zero. In this case, the numerator will also be zero but still the denominator will be zero at this same point.

We will have $\frac{0}{0}$ and we know that zero in the numerator turns everything to zero. But zero in the denominator is undefined.

So, formally, can this division be made without taking extra careful?

FELIPE_RIBAS
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3 Answers3

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$$\frac xx = 1, \text{ for all } x \neq 0$$

So the function is defined for all $x$ in the domain of the function, and the domain of the function (the domain of values for which $\frac xx$ is defined) is $\mathbb R - \{0\}$.

So yes, extra care must be taken (and this is true for all rational functions, and many other functions, too) to first determine the domain of the function. Once you've done that and made that domain explicit (in this case the domain being all real numbers except $0$), you can then manipulate the function, e.g., simplifying or canceling like terms.

amWhy
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$\frac{a}{b}=a\cdot \frac{1}{b}$... so if $\frac{1}{b}$ is undefined then so is $\frac{a}{b}$.

JP McCarthy
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By common convention, without further context $\,f(x) = x/x\,$ is not defined at $\,x = 0,\,$ since evaluating at $\,x = 0\,$ leads to division by $\,0.\,$ However, further contextual information may reveal that this discontinuity at $\,x = 0\,$ is an artifact of the contsruction of $\,f.\,$ For example, if we obtained that expression for $\,f\,$ by solving the equation $\,x f = x\,$ and we know in our context that the solution must be continuous (e.g. we may know a priori that $\,f(x)\,$ must be a polynomial in $\,x),\,$ then the unique solution is $\,f(x) = 1.\,$ In this case the discontinuity at $\,x = 0\,$ is an artifact of the algebra involved in the solution process - it was introduced by dividing by $\,x.\,$ In algebra, such removable discontinuities are often removed by cancellation, and this is a powerful algebraic method of removing such singularities, e.g. see this answer where it is used to algebraically define polynomial derivatives, and to obtain a slick proof of Sylvester's determinant identity.

In summary, the denotation of the expression $\,x/x\,$ depends upon the ambient context and, as such, it may or may not be defined and/or continuous at $\,x = 0.\,$ Lacking any further context, the normal convention is that it is not defined at $\,x = 0.$

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Bill Dubuque
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  • So let's say that in my original function the domain is all real numbers but through my algebric method I found x/x. This will cause a limitation when x=0 so even though my algebric method can help me find the correct answer, this answer should be limited to all real numbers except zero, right? So formally I should explicitly demonstrate by another method that the particular case (x=0) is also included in the solution, am I right? – FELIPE_RIBAS Mar 20 '14 at 16:23
  • @user2110874 It depends on the context. If you know a priori that the solution for $,f,$ must be continuous then the unique solution is $,f(x) = 1.,$ But absent such further knowledge it is not defined at $,x = 0.$ – Bill Dubuque Mar 20 '14 at 16:37