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Take the equation $\displaystyle y=\frac{x^2+x-6}{x^2-4}$, which is formed from simplifying $\displaystyle y= \frac{(x+3)(x-2)}{(x+2)(x-2)}$.

If we divide out $(x-2)$, from the numerator, we get the function $\displaystyle y = \frac{x+3}{x+2}$ which is defined for all real numbers except where $x = -2$, where there is an infinite discontinuity.

However, if we look at the original function, wouldn't 2 be a removable discontinuity? Multiplying the equation by $\displaystyle \frac{x-2}{x-2}$ doesn't change the equation, except where $x=2$, because the denominator is now 0. I understand that the limit exists, yet Wolfram|Alpha seems to say that not only the limit, but the point actually exists. We couldn't have divided $(x-2)$ out if the denominator was equal to 0. Have I gone horribly wrong somewhere?

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6  
Alpha automatically removes removable singularities (at least when it spots it ;-)) –  Fabian May 5 '11 at 16:54
2  
It's actually common practice in some of the advanced parts of mathematics to have functions with removable discontinuities be defined at those "undefined" points by taking limits. –  J. M. May 5 '11 at 23:37

2 Answers 2

The point does not exist, as the division by zero is not defined. Wolfram Alpha just gives you the limit as you can see in this example too. However the limit exists so you can divide $(x-2)$ from the numerator like you said but in this case you have always that $x \neq 2$ because you consider the limit so canceling the term is a valid step to get the function without the disconuity.

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Wolfram Alpha is just sloppy. If you put in x/x at x=0 you get x/x= 1 at x=0. It may be that the practice of ignoring this is encouraged by practical applications because , when we get an expression, The cancelling parts were introduced spuriously to begin with. What actual circumstances would realistically have such canceling parts?

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