(x2-2.5x+1.5)/(x-1.5)
www06.wolframalpha.com
www06.wolframalpha.com
let a=1.5 for brevity and let f(x) = (x-a)(x-1)/(x-a).
Relabel f(x)= N(x)/D(x) (N for numerator, D for denominator)
N(x) and D(x) are analytic and so tend to zero as x--> a, so f(x) tends to the indeterminate form 0/0.
In that case, according to L'hopital's rule, f(a) = N'(a) / D'(a).
N'(x) = (x-a) + (x-1) by the product rule. D'(x) = 1.
Plugging in x=a:
f(a) = (a-1) / 1 = .5
Which is what Alpha implies.
See also: http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule
L'Hopital's rule says that the limit of the function exists, not the function itself. A removable singularity is still a singularity, just one that's more well-behaved than a pole (or gods forbid, an essential singularity). While the function can be analytically extended to be defined at the singularity, that's an extension of the function, rather than the function itself...
... sorta. Since extensions of analytic function are unique, we usually treat the extension and the function as the same thing. Whether this is revealing a deeper truth about the function, or sloppy notation, is up to the individual practitioner, and the context. Certainly, the definition of the function that has the removable singularity is sufficient to define the function that has the singularity removed. For analysis they're basically the same and we usually extend as far as we can. But for topology, a single point separates open and closed sets, which are worlds apart, and the extended function is a very different beast from the original one. This just goes to show how limits identify the points that are most important to a (structured) set.
However, for the purposes of Wolfram|alpha, I don't fault them for making that mistake. Removable singularities can be safely extended over for the purposes of computational mathematics.
Yes, the inequality simplifies, but when x=1.5 you are multiplying both sides by 0 which is not a permitted operation.
In this case nothing goes wrong if you blindly simplify and use the simpler inequality. Mathematicians go ballistic when you do that because there are other occasions where similar operations create nonsense. The usual proof that 1=2 is an example.
Trying to show why this sort of thing is wrong is like trying to show that a fallacious proof of a true theorem is wrong. Counter-examples are hard to produce because the theorem is true.
And that's why it's a general rule. If you are dividing by something you have to preclude the possibility that the denominator is 0.
I'm not sure what the point of the post is, or why the discussions here are using so many words I don't understand.
((1.5 - 1.5) * (1.5 - 1))/(1.5 - 1.5)
(1.5 - 1)
Did you get the same answer? If not then there must either a bug in your calculator/spreadsheet or in wolfram alpha.
0.5 * 0 / 0 = 0.5
Excel or your calculator will freak out, but it's still true.
What I should have said is that if you intentionally craft the circumstances of an equation such that you are dividing by zero, it is your premise that is nonsense, NOT the equation itself.
http://www.wolframalpha.com/input/?i=xy/y+where+y%3D0+and+x+...
Whereas it gets this completely fine: http://www06.wolframalpha.com/input/?i=60%25+of+%2420
It seems to forget what % means..
edit:
And then http://www.wolframalpha.com/input/?i=60000%20percent%20squar...
Performing "obvious" completions of incompletely defined functions is assuming you know the context, and sometimes you don't.
This is a sloppy engineering solution that is not guaranteed to be right in all contexts. Without the context, the correct solution is to say that the question is ill-defined/undefined when x=1.5, and all values of x such that x>1 and x.ne.1.0 satisfy the given inequality.
Okay: what did we learn from that?
rofl: http://www06.wolframalpha.com/input/?i=422340980293840234098...
full name: 42 undecillion, 234 decillion, 98 nonillion, 29 octillion, 384 septillion, 23 sextillion, 409 quintillion, 823 quadrillion, 409 trillion, 823 billion, 94 million, 823 thousand and 411...