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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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    Alpha automatically removes removable singularities (at least when it spots it ;-))2011-05-05
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    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.2011-05-05

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