# Why does simplifying a function give it another limit [duplicate]

$$\lim_{x\to 1} \frac{x^3 - 1}{x^2 + 2x -3}$$

This does obviously not evaluate since the denominator equals $0$. The solution is to:

$$\lim_{x\to 1} \frac{(x-1)(x^2+x+1)}{(x-1)(x+3)}$$ $$\lim_{x\to 1} \frac{x^2 + x + 1}{x + 3}$$ $$\frac{1+1+1}{1+3} = \frac{3}{4}$$

My question: what is actually happening? How can simplifying a function give it another limit? Is it a complete other function and if so why would it be relevant to our original question?

• The limits are not different. The first form is just a indeterminate limit of the kind $\frac00$, this mean that you need some manipulation to get the value for this limit. Jun 12, 2016 at 0:55
• This might surprise you, but 0/0 can be made to equal any number. In reality, you are not changing the limit. Instead, you are limiting the infinite range that 0/0 returns to the fnite range of a single real value. In essence, the first limit you obtained simply stated "this expression could be equal to literally anything". That is why it is called indeterminate; it has no apparent value. Jun 12, 2016 at 0:59
• @TheGreatDuck: More accurate would be to say "any real number can be the limit of the form 0/0". Your phrasing makes it sound like $\lim_{x \to 0} \frac{x}{x}$ can be set equal to anything you want.
– user14972
Jun 12, 2016 at 6:13
• @lastresort: Sorry, being Euler does not make one correct or even close to correct. Jun 12, 2016 at 9:23
• @TheGreatDuck: Look I've had enough of this. I've helped you as much as I could, but now you're just pretending to know mathematics that you don't seem to. If you're willing to learn, I'm willing to teach (and learn), but if not I'm going to spend my time with other people. That very thing cited by lastresort is utter nonsense, and I don't care who wrote it. Jun 12, 2016 at 15:49

Perhaps you (and I think many beginners in calculus) have the notion that limit of a function is evaluated by plugging the value of the variable. Thus for the function $$f(x) = \frac{x^{3} - 1}{x^{2} + 2x - 3}\tag{1}$$ you say that its limit as $x \to 1$ can not be evaluated by plugging $x = 1$ because denominator vanishes.

It is very important to understand that limit of a function at a point is something completely different from the value of the function at that point. A limit of a function at a point has nothing to do with the value of the function at that point, but it has everything to do with the values of the function near that point.

However, there is a catch!! There are many functions seen commonly in calculus for which the limit at a point turns out to be the same as their value at that point and hence for such functions it is possible to evaluate the limit just by plugging. I think it is this behavior of some functions combined with the fact I described in last paragraph which creates utter confusion in the mind of a beginner. At one stage I mentioned that limit is not same as value of a function and next I mention that for some functions the limit is same as its value. Really confusing!

The only way to sort out this confusion is to learn to identify at least some basic types of functions which have this nice property that their limit at a point is same as their value at that point. Such functions are called continuous functions. Using a series of theorems on limits it can be proved in a step by step manner that any function made up of algebraic (which includes polynomials and rational functions), trigonometric (direct and inverse), logarithmic and exponential functions using a finite number of arithmetic operations and compositions is continuous wherever it is defined. The type of function described in last sentence is called an elementary function.

The function $f(x)$ given by equation $(1)$ is an elementary function and we need to calculate its limit as $x \to 1$. From what we have mentioned in last paragraph it is clear that we can evaluate its limit simply by plugging provided it is defined at the point under consideration. The trouble is that $f(x)$ is not defined at $x = 1$. Then we use the fact mentioned in the beginning that $\lim_{x \to 1}f(x)$ has nothing to do with its value at $x = 1$. The limit operation $\lim_{x \to 1}$ ensures that $x \neq 1$ and we can now use any sort of transformation on $f(x)$ under the assumption that $x \neq 1$ and try to simplify it in the form of an elementary function which is perhaps defined at $x = 1$.

Here we are lucky and by cancelling the factor $x - 1$ from numerator and denominator we reach another function $$g(x) = \frac{x^{2} + x + 1}{x + 3}\tag{2}$$ Note that both $f$ and $g$ are different functions ($f$ is not defined at $x = 1$ whereas $g$ is defined there), but $f(x) = g(x)$ as long as $x \neq 1$. Hence as far the limit operation $\lim_{x \to 1}$ is concerned both $f(x), g(x)$ have same behavior. And now we see that $g(x)$ is also an elementary function and it is defined at $x = 1$ and hence its limit as $x \to 1$ is same as its value $g(1) = 3/4$. Thus limit of $f(x)$ as $x \to 1$ is also $3/4$.

• I think yours is the only answer that really tackles the confusion and its source. Jun 12, 2016 at 6:22
• @user21820: Thanks man! BTW when I was a beginner in calculus long back, then even instructors told that evaluating limits is easy: try plugging, if it does not work try to transform the expression somehow so that plugging works. Actually that is the practical method by which limits are evaluated in step by step manner. Unfortunately most teachers and books don't mention how this technique works and how it does not contradict the definition of limits. I have tried to sort this out and I am happy that you liked my answer. Jun 12, 2016 at 7:11
• When I was taking my first calculus course in university, I gave a trick question to my professor: Find $\lim_{x \to 0} \dfrac{|x+1|+|x-1|-2}{|x+1|+|x-1|-2}$. His first (instinctive) reaction was to cancel them away and get $1$. =D Jun 12, 2016 at 7:14
• @ritwiksinha: Its a trick question. Limit does not exist. Jun 12, 2016 at 9:06
• @ritwiksinha: Nope. Both the numerator and denominator are $0$ not just when $x = 0$. For a (real) limit to be defined at a point the expression has to be defined for an open interval minus the point. Jun 12, 2016 at 10:09

By simplifying, you are removing the point discontinuity, also appropriately called the removable discontinuity, at $x = 1$. • It appears that the graph does not represent the problem correctly. The red point does not correspond to $(1, 3/4)$. Jun 12, 2016 at 11:29
• @ParamanandSingh, fixed.
– cr3
Jun 12, 2016 at 18:48
• I don't think this really answers the question of why or how. Jan 6, 2018 at 21:36

It's not that the simplified function has a different limit, it's that the limit of the original expression cannot be found by evaluation. In other words, write $$f(x) = \frac{x^{3} - 1}{x^{2} + 2x - 3},\qquad g(x) = \frac{x^{2} + x + 1}{x + 3} = \begin{cases} f(x) & x \neq 1, \\ \frac{3}{4} & x = 1. \end{cases}$$ Since $g$ is a quotient of polynomials with denominator non-zero at $x = 1$, its limit at $1$ can be found by evaluation. The limit of $f$ at $1$ has the same value, since $f(x) = g(x)$ for all $x \neq 1$.

I think the important thing to understand is that the limit does not necessarily have anything to do with the actual value of the function at the point, or whether it even has a value at all. It's possible that the function does have a value when you plug the number into the expression, but that's still not the value of the limit (such a function would not be continuous). When you do the division you are actually changing the function from one that has no value at the point to one that does have a value, and the relationship between the functions is that the new one's value is equal to the limit of both of them.

• +1 I always appreciate an answer which is eloquent, instructive, and doesn't sacrifice correctness to achieve it.
– Ovi
Sep 14, 2018 at 7:05

Consider $$\lim_{x=2}{x^2-4\over x-2}.$$ We factor and simplify this to get $${x^2-4\over x-2}=x+2.$$

In a elementary algebra class, we think of these two expressions as being equal. But they are not. Both sides will produce the same value for every $x$ except $x=2$. The left-hand side is undefined for $x=2$ while the right-hand side produces $4$.

One good way of understanding the limit is to consider the graphs. Of course, the graph of $y=x+2$ is a line with slope $1$ and $y$-intercept $(0,2)$. On the other hand, the graph of $y=(x^2-4)/(x-2)$ is the same line with a hole at $(2,4)$.

• so... is algebra all a lie? Are the two functions not equal? Jan 8, 2018 at 4:24
• @jwan They're equal as algebraic expressions but not as functions. In general everything involving polynomials is still valid when you substitute except when it gives you division by 0. Sep 22, 2018 at 22:28

It doesn't get a new limit, it actually just didn't have a value before (because it was $0/0$), but it still had that same limit. The original function and the new function are actually different functions that agree everywhere except $x=1$, for which the first has no value. This is a consequence of the fact that when broken down, a function is just a rule for turning $x$'s into $y$'s.

What's happening is the following: The given expression $$\Phi(x)$$ is undefined at $$x=1$$, but for all $$x\ne1$$ it is equal to some simpler expression $$\Psi(x)$$, which is defined and continuous at $$x=1$$. It follows that $$\lim_{x\to1}\Phi(x)=\lim_{x\to1}\Psi(x)=\Psi(1)={3\over4}\ .$$

The point is your limit is so-called indeterminate form $0/0$. If you have information that bouth funxtions $f(x)$ and $g(x)$ have limits $0$ in some point you actually have no information about limit of $\frac{f(x)}{g(x)}$ in this point.

Maybe more simple exmple will make this fact clearer. It is obvious that $$\lim_{x\to 0}\frac{3x}{4x} = \frac{3}{4}$$ as $\frac{3x}{4x} = \frac{3}{4}$. But formally $\lim_{x\to 0}3x = \lim_{x\to 0}4x = 0$, so this is indeteminate form $0/0$.

Such problems are dominated by what we call as indeterminacy, that is when you have at the same time zero at the numerator and the denominator, or infinity at both again when you are trying to compute such limits (if you had zero at the denominator but a non zero number at the numerator, the result would be infinity. But right now you can't know the answer). It is not that the function itself is changed, rather that you changed its' form in a way that you are "lifting the indeterminacy". L'Hôpital's rule is a really useful theorem that it enables you easily to lift such indeterminacies, thus calculating the limit.

There is more to functions that mere formulae. Relations between $y$ and $x$ through $f(x)$ can be defined in many ways. For instance, for the identity function $f(x)=x$, you can write $y= f(x)$ (explicit form).

You can also find $(x,y)$ pairs such that $|y-x|=0$ (implicit form). You can use the Cauchy functional equation ($f(u+v) = f(v)+f(v)$) with additional constraints such as continuity and a fixed point like $f(1)=1$.

Implicit, explicit, functional definitions are more or less well defined and handy to use.

What you are trying to do is to find a somewhat natural extension to a not-so-well defined function. Imagine I pose $y=f(x)= \frac{x}{x}$. It is not defined for $x=0$. But morally, since the function is constant for all other $x$, one might want to "extend" it at $0$, in a natural way, and the limit is one of the many possible natural ways, based on continuity. Then, you could draw the graph without lifting the pen at $0$.

A similar question is discussed in Why is $y=\frac{x+y}{x}$ the same as $y=\frac{x}{x−1}$, with explicit and implicit forms? They are not the same, but are closely related.

The simplification you use here is practice for more complicated cases. For instance, you can define or extend $\sin(x)/x$ at $0$. $\sin(x)$ is not a polynomial, but locally it is very to one ($x-x^3/6$), and then some simplifications may apply.