I'm having trouble understanding what really happens when we evaluate the limit of a certain function f(x) as x approaches a certain value.

For ex, if we have lim x-->2 $\frac{x^2 + x -6}{(x-2)}$ we can't just plug in 2 and evaluate that because f(x) is undefined when x=2. So we factor out the numerator and find that lim x-->2 $\frac{x^2 + x -6}{(x-2)} = \frac{(x+3)(x-2)}{(x-2)}$

Here is where there seems to be an inconsistency as far as I understand what's going on. We are ok with canceling out the (x+2) terms because we're not saying that x=2, we're saying x approaches 2, so (x-2) in the denominator =/= 0 and we can simplify.

However, then it seems we just plug 2 into (x+3) and say that lim x-->2 $\frac{x^2 + x -6}{(x-2)} = 5$.

That's very confusing to me because we go from not being ok with plugging in the value 2, instead imagining that we're getting closer and closer to it from both ends, to just plugging in 2 and saying (2+3) = 5.

I understand that there's no more problem with using 2 once we got rid of (x-2) in the denominator, but what happened to just approaching x?

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    $\begingroup$ We are still just looking at $x+3$, and letting $x$ approach $2$. The difference is that it is clear what $x+3$ approaches as $x$ approaches $2$, while it was not clear what $\frac{x^2+x-6}{x=2}$ approaches as $x$ approaches $2$. $\endgroup$ – André Nicolas Jun 11 '14 at 22:52
  • $\begingroup$ @AndréNicolas thank you, that answers perfectly what I was asking; now I feel I actually know what's going on throughout the whole evaluation process. $\endgroup$ – jeremy radcliff Jun 11 '14 at 23:00
  • $\begingroup$ You are welcome. There are other manipulations that change an expression from one whose limit is not clear to one whose limit is clear. $\endgroup$ – André Nicolas Jun 11 '14 at 23:05
  • $\begingroup$ @AndréNicolas When people talk about continuity and how $\frac{x^2 + x -6}{(x-2)}$ isn't continuous but $(x+3)$ is, which makes it ok to evaluate the latter at 2 exactly; aren't those two functions different functions? Is it ok to basically "turn" the former into $(x+3)$?. It seems to me you were saying that we still evaluate using values that approach 2 for x, just that after simplification it becomes obvious what (x+3) is as x approaches 2, but what's truly going on is that we're still approaching 2 and not just plugging in the exact value 2. $\endgroup$ – jeremy radcliff Jun 11 '14 at 23:16
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    $\begingroup$ Yes, we are always approaching. And the two functions are not the same. The first is not defined at $x=2$, while the second is. However, the two functions agree at all $x\ne 2$. $\endgroup$ – André Nicolas Jun 11 '14 at 23:22


$$\frac{x^2 + x - 6}{x - 2} = x + 3$$

for all $x \ne 2$, it follows (from the definition of the limit, since we exclude the possibility that $x = 2$)

$$\lim_{x \to 2} \frac{x^2 + x - 6}{x - 2} = \lim_{x \to 2} x + 3$$

More generally, if $f$ and $g$ are functions which agree everywhere on an interval containing $a$, except possible the point $a$ itself, then

$$\lim_{x \to a} f(x) = \lim_{x \to a} g(x)$$

Now to actually prove that $\lim_{x \to 2} x + 3 = 5$, one could again proceed directly from the definition and use an $\epsilon-\delta$ argument. Alternatively, one could note that this function agrees (on its domain) with a continuous function which is defined everywhere, and continuity means that we can just evaluate at the relevant point.

So in short, we know that our complicated not-always-defined function agrees everywhere (except at a single point) with a nice continuous function, which justifies just evaluating the nice function.


If the function is continuous the value of the limit is the value of the function; so, you can just plug in the value. When the function is not continuous (or not even defined) then we need to do something else (like cancel a factor).


With limits you are never concerned with the behavior of the function at the limit point, only the behavior around the limit point.

We say that a function is continuous at $x = a$ if $\lim_{x\to a} f(x) = f(a)$. If a function does not have this property ($\lim_{x\to a} f(x) \neq f(a)$) then the function is discontinuous at $x=a$.

Your first function $f(x) = \dfrac{x^2+x-6}{x-2}$ is not continuous at $x =2$ because it has a discontinuity at $x = 2$. We cannot calculate the limit by means of "plugging in" because $\lim_{x\to a} f(z) \neq f(a)$. On the other hand, $f(x) = x+3$ is continuous so we can state that $\lim_{x\to 2} x+3 = 2+3 = 5$.

In many, but not all cases, discontinuities arise because of a division by zero.

For reference, all polynomial functions are continuous. Functions that are the division of two polynomials are continuous wherever the denominator is not zero. You can use this fact to evaluate many limits by just plugging the value into the equation.


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