# Justification for evaluation of limits?

1. Feb 12, 2015

### Mr Davis 97

I am curious about the process of evaluating a limit. Firstly, I know that if a function $f(x)$ is continuous then one can usually just plug in the the number that $x$ is approaching in the limit, since criteria for a continuous function is that $\lim_{x \to a}f(x) = f(a)$. However, what if we have the function $f(x) = \frac{(x-1)^2}{(x+1)}$, then if we try to evaluate the limit $\lim_{x \to -1}\frac{(x-1)^2}{(x+1)}$, we do so by cancelling the linear factors to get $f(x) = x - 1$, which then gives $\lim_{x \to -1}(x - 1) = -2$. But what is the justification that we're allowed to cancel out the factors when evaluating the limit? This produces an entirely new function, since $f(x) = x - 1$ is obviously different than $f(x) = \frac{(x-1)^2}{(x+1)}$. Why are we allowed to manipulate one function into another in this way when evaluating limits? What guarantees that $\lim_{x \to -1}(x - 1) = \lim_{x \to -1}\frac{(x-1)^2}{(x+1)}$?

2. Feb 13, 2015

### Cruz Martinez

Both functions are exactly the same function written differently except for x=-1 which is when one of them becomes undefined.
The algebraic manipulations you used to rewrite the initial function are perfectly valid ones for all x not equal to -1.
But this is ok since when you evaluate a limit you want the value to which f(x) gets closer and closer, not the actual f(a).

I guess to say it another way, you're replacing your old function with another function which is continuous at x=a, as oppossed to the original one, and which is also equal to the original except maybe at x=a, so that you can evaluate the limit by substitution. Both functions will have the same limit as x approaches a, if the limit exists, because both functions are identical everywhere except at x=a.

Edit: I didnt see how you manipulated the function when i replied, what linear factor did you cancel?
In any case, my argument remains unchanged except of course for the fact that i now think the algebraic manipulations of your example are not valid.

Last edited: Feb 13, 2015
3. Feb 13, 2015

### Staff: Mentor

No, this is incorrect. The limit here does not exist, because the left-side limit and the right-side limit are not equal.

4. Feb 17, 2015

### HallsofIvy

I presume the real difficulty you see is that the denominator goes to 0 at x= 1. There is no problem otherwise: 6/2 is equal to 3

One very important, but often overlooked, property of limits is this: "If f(x)= g(x) for all x other than a, then $\lim_{x\to a} f(x)= \lim_{x\to a} g(x)$".
That follows from the definition of "limit":
$\lim_{x\to a} f(x)= L$ if and only if, for any $\epsilon> 0$, there exist $0< \delta$ such that if $0< |x- a|< \delta$ then $|f(x)- L|< \epsilon$.

Note that $$0< |x- a|$$. What happens at x= a is irrelevant to the limit. In the case $$\frac{x^2- 1}{x- 1}$$ is equal to x+ 1 for all x except x= 1 but that is enough.

5. Mar 3, 2015

### AURUM

draw a graph for both the initial and the final functions and see for yourself if they are different plus there is an error in your question.

6. Mar 4, 2015

### HallsofIvy

If you mean "draw graphs for both $y= (x^2- 1)/(x- 1)$ and $y= x+ 1$", if you do it correctly you will see that they area NOT the same! The graph of $y= x+ 1$ is a straight line through (0, 1) with slope 1. The graph of $y= (x^2- 1)/(x- 1)$ is that same straight line except there is a "hole" at (1, 2).