# Why must polynomial division be done prior to taking the limit?

Suppose I wish to evaluate the following, $$\mathop {\lim }\limits_{x \to 2} \left( {{{{x^2} - 4} \over {x - 2}}} \right)$$ If I just substitute two into $x$, it can't be done because the answer would be undefined (division by zero).

But, if I complete the polynomial division, that I hate to do because I'm all thumbs at it, $$\mathop {\lim }\limits_{x \to 2} \left( {{{{x^2} - 4} \over {x - 2}}} \right) = \mathop {\lim }\limits_{x \to 2} \left( {x + 2} \right) = 4$$

Please tell me what's going on here?

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– Zircht Jun 15 '14 at 23:43
Thank you kindly Zircht, it would have taken me ages to find that post :) – Mike Jun 15 '14 at 23:46
David's answer is good, but I would add this: The main reason for introducing limits in differential calculus is to deal with the ones where you get $0/0$ if you just substitute the value for the variable. The reason is that that's what you get in the definition of the derivative: $\lim\limits_{h\to0}\dfrac{f(x+h)-f(x)} h$. If you plug in $0$ for $h$, you get $0/0$. ${}\qquad{}$ – Michael Hardy Jun 16 '14 at 1:23
...and an addition to @MichaelHardy's addition: the question asked in the OP is in fact nothing other than the derivative of $x^2$ at $x=2$. – David Jun 16 '14 at 5:37

The point is that the functions $$\frac{x^2-4}{x-2}\quad\hbox{and}\quad x+2$$ are equal except at $x=2$, where the second is defined and the first is not. If you look closely at the definition of a limit as $x\to a$, you will see that it is carefully framed in such a way that the value of the function (if any) when $x=a$ is irrelevant. Therefore the two functions above have the same limit as $x\to2$. However, as you have noted, you cannot just substitute $x=2$ in the first as it is undefined. On the other hand, the second function is defined at $x=2$, and better still, it is continuous at $x=2$, because it is a polynomial. Therefore, using the definition of continuity, $$\lim_{x\to2}(x+2)=2+2=4\ .$$ And finally, as already noted, $$\lim_{x\to2}\frac{x^2-4}{x-2}=\lim_{x\to2}(x+2)=4\ .$$