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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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    $\begingroup$ Related: math.stackexchange.com/q/462199/121880 $\endgroup$
    – user121880
    Jun 15, 2014 at 23:43
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    $\begingroup$ Thank you kindly Zircht, it would have taken me ages to find that post :) $\endgroup$
    – Michael Lee
    Jun 15, 2014 at 23:46
  • $\begingroup$ 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 $0/0$. ${}\qquad{}$ $\endgroup$
    – Michael Hardy
    Jun 16, 2014 at 1:23
  • $\begingroup$ ...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 $x=2$. $\endgroup$
    – David
    Jun 16, 2014 at 5:37

2 Answers 2

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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, $$\lim_{x\to2}\frac{x^2-4}{x-2}=\lim_{x\to2}(x+2)=4\ .$$

This kind of problem is superficially very simple, but as you can see, there is quite a lot behind it if you want to fully understand what is going on.

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  • $\begingroup$ You write eloquently, kindest regards. $\endgroup$
    – Michael Lee
    Jun 16, 2014 at 0:05
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Do not think of it as algebraic simplification. Think of the problem first. The original expression is a discontinuous function thus we are not able to solve for it. So we need to find an equivalent expression which is not discontinuous to make it easy for us to solve. Here simple algebraic operation gives us another function which is equivalent in the range we care about (between 0 and 2) and is continuous at 2, so we can solve for it.

In other cases you may have to use trigonometric, logarithmic, or other operations to find an expression that can be solved.

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