# Justification for evaluation of limits?

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)}##?

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.

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Mark44
Mentor
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##
No, this is incorrect. The limit here does not exist, because the left-side limit and the right-side limit are not equal.
Mr Davis 97 said:
, 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)}##?

HallsofIvy
Homework Helper
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)}##?
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.

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.

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).