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This might be a very stupid question. I was trying to evalueate the limit of the function as x tends to 1 of (x-1)/(x^2-1). This function clearly has a hole at that point because it becomes 0/0 but after simplifying it, the function becomes 1/x+1 and now the hole magically disappears. And the limit is clearly 1/2. But now this function has a hole at x=-1. Arent both of thease function the same? Why does this happen?

Dong ky
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    Consider $f(x) = \frac xx$. This is almost the same as the function $g(x) = 1$. Indeed, $f(x) = g(x)$ for all $x$ except $x=0$. The difference is that $g$ is defined everywhere, whereas $f$ is not defined at $x=0$. They are not the same function. When you simplify, make a side note that the domain still excludes the "hole". – Théophile Aug 18 '22 at 17:04
  • For more detailed discussion, see this question. – Théophile Aug 18 '22 at 17:07
  • So every time we perform an alegebraic operation to a function , It becomes a different function? – Dong ky Aug 18 '22 at 17:07
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    The hole at $x=-1$ was always there, because $x^2-1=(x+1)(x-1)$. The hole doesn't "move". The first hole at $x=1$ is removable and the second hole at $x=-1$ is always there, whether you're talking about $(x-1)/(x^2-1)$ or $1/(x+1)$. – march Aug 18 '22 at 17:07
  • Not every time, because, for instance, $(x+1)(x-1) = x^2-1$ for every possible $x$. You do have to be careful about the domains of functions when you are composing functions that have restricted domains. – march Aug 18 '22 at 17:10
  • @Dongky What you're missing is the domain. The function "$\frac{x-1}{x^2-1}$, where $x \in \Bbb R \setminus {-1,1}$" is identical to the function "$\frac 1{x+1}$, where $x \in \Bbb R \setminus {-1,1}$". Note that despite the simplification, the domain did not change. – Théophile Aug 18 '22 at 17:11
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    Got it! Thank you! – Dong ky Aug 18 '22 at 17:12
  • We do want to replace $\frac{x-1}{x^2-1}:\mathbb{R}\setminus{\pm1}\to\mathbb{R}$ by the new function $\frac{1}{x+1}:\mathbb{R}\to\mathbb{R}$, rather than the function $\frac{1}{x+1}:\mathbb{R}\setminus{\pm1}\to\mathbb{R}$, since our knowledge that the former is continuous allows us to replaced the limit computation by function evaluation. We could detal with the latter function, sure, but definitely evaluating the function is easier. – plop Aug 18 '22 at 17:27
  • @user85667 - who is this "we", and why do they and you care so much about Dong ky's function? While it is commonly useful to remove removable singularities, there are times when it is important instead to recognize that the function is not defined there and let it be. – Paul Sinclair Aug 19 '22 at 17:38
  • @PaulSinclair Not interested. – plop Aug 19 '22 at 17:47

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