This is probably a trivial question however i cannot find the correct information online.

When simplifying mappings from $\mathbb{R}$ to $\mathbb{R}$ such as:


Why is it okay to simplify this to simply just $x$? if $x=1$ do we not have the singularity $$\frac{0(1)}{0}$$ Thank you

  • 2
    $\begingroup$ Yes, one should specify that equality only holds if $x\ne 1$. $\endgroup$ – André Nicolas Apr 23 '16 at 23:13
  • $\begingroup$ Strictly speaking, $x \mapsto \frac{x(x-1)}{(x-1)}$ does not determine a mapping from $\mathbb{R}$ to $\mathbb{R}$. If you have a mapping from $\mathbb{R} \setminus \{ 1 \}$ to $\mathbb{R}$ it is OK to simplify, when you remember that the mapping is from $\mathbb{R} \setminus \{ 1 \}$ to $\mathbb{R}$, also after the simplification. $\endgroup$ – Jeppe Stig Nielsen Apr 23 '16 at 23:18
  • $\begingroup$ In addition to the point @AndréNicolas made, you also have $\lim_{x\to 1} \frac{x(x-1)}{x-1} = 1$, which agrees with the value of the function $x$ at $x=1$. This tells us that this singularity is removable. $\endgroup$ – Nicholas Stull Apr 23 '16 at 23:19
  • $\begingroup$ As FUNCTIONS, $x$ and $\frac {x(x-1)}{x-1}$ are not equal - they don't even have the same domain. IF you restrict the domain of "$x$" to $\mathbb R \setminus \{1\}$ then they become equal. If you mean them as "rational functions" in the sense of ALGEBRA, in the field of "rational functions" (the quotient field of the ring of polynomials in one variable), they are equal - but that is not a set of "functions" in the set-theoretic sense of the word "function." So it all depends on your CONTEXT for the question. $\endgroup$ – mathguy Apr 23 '16 at 23:57