# Limit Question - Explanation

The limit of $f(x) = x$, as $x$ tends to zero is zero.

What's the limit of the function $\dfrac{x^2}{x}$ as $x$ tends to zero? and

What's the limit of the function f(x) = (modulus of x)/x ?

I am interested in this question from a teaching perspective. How would you explain and contrast these three cases in an intuitive manner to a student?(i.e something like " if you take values of $x$ arbitrary close to zero then .....or using the epsilon-delta definition in an intuitive way.) I know that the epsilon-delta definition is not exactly intuitive, but I guess, a reasonable degree of intuition should be possible for such simple cases.

I am not a teacher, but I think you understand a concept properly only when you are able to explain it clearly to someone else, so....

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This has already been discussed quite a bit, but since my understanding of "explain ... in an intuitive manner to a student" seems to be quite different from the discussion so far, I thought I'd offer my approach to this.

The basic idea is to write $x^2$ as $x \cdot x$, which allows us to see that $x^2$ is a small number times $x$ (when $x$ is small). For instance:

When $x = \frac{1}{10},$ then $x \cdot x = \frac{1}{10} x,$ and hence $x^2$ is $10$% of $x.$

When $x = \frac{1}{100},$ then $x \cdot x = \frac{1}{100} x,$ and hence $x^2$ is $1$% of $x.$

When $x = \frac{1}{1000},$ then $x \cdot x = \frac{1}{1000} x,$ and hence $x^2$ is $0.1$% of $x.$

When $x = \frac{1}{10000},$ then $x \cdot x = \frac{1}{10000} x,$ and hence $x^2$ is $0.01$% of $x.$

As $x$ gets smaller and smaller, the values of $x^2$ become a smaller and smaller percent of the values of $x.$ Algebrically, we can see this easily from $\frac{x^2}{x} = x,$ but I think most anyone who is initially puzzled over what is going on intuitively will probably not be helped much by someone immediately pointing out the identity $\frac{x^2}{x} = x.$ However, with the examples above in sight, such a student can now see that $\frac{x^2}{x} = x$ efficiently summarizes the results seen in these examples.

Naturally, at some point we should point out what we really mean by "smaller and smaller". For example, the values of $1 + \frac{1}{n}$ for $n=1,\;2,\ldots$ get smaller and smaller. But, for an initial investigation into what Nikhil Panikkar asked about, these are things that can be looked at later (along with the idea of having to use arbitrary sequences approaching $0,$ the use of epsilon-delta neighborhood language, etc.).

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I liked your explanation, but I was thinking of explaining it intuitively in terms of the epsilon delta definition. I understand that the epsilon delta definition is not intuitive, but I guess a reasonable degree of intuition should be possible - atleast for such a simple problem. My main issue is that the epsilon delta definition doesn't seem to explain discontinuties well. As I pointed before in a comment, the differing signs of right hand and left hand limits seem intuitively reasonable in the case of a continuous function, but not in the case of a function with discontinuities. – Nikhil Panikkar Feb 20 '13 at 14:21
@Nikhil Panikkar: I've seen some great graphical explanations, but I don't know how to do something like that in a post here (and also don't really have the time, even if I did know). However, here are some things I've known about or found that could be of help: Eric Schechter's essay on funnels (probably too advanced for here, however), the epsilon delta game, and this mathoverflow post. – Dave L. Renfro Feb 20 '13 at 14:32
Thank you, for pointing me to these links. I'll read and let you know. – Nikhil Panikkar Feb 21 '13 at 3:07
@Nikhil Panikkar: This morning I came across something (by accident) that should be included with the references I gave on 20 February 2013: Leonard Gillman, Rigor in calculus, Notices of the American Mathematical Society 44 #8 (September 1997), 932-934. In particular, see Figure 3 and the discussion that relates to it. – Dave L. Renfro Apr 19 '13 at 18:12

I think a concept that this question is trying to elucidate is that the limit depends on values of a function near a point rather than at a point. Specifically: $\frac{x^2}{x}$ is equal to $x$ everywhere except at $x = 0$, where the former function is undefined. But for the purposes of taking a limit as $x \rightarrow 0$, the behavior at $x = 0$ doesn't matter.

Similarly, you might ask about the limit as $x \rightarrow 0$ of a function:

$$f(x) = \left\{\begin{array}{cc} 50 & x = 0 \\ x & x \neq 0 \end{array}\right.$$

Of course, this functions isn't equal to $x$, but when it comes to taking a limit that doesn't matter.

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Ok, I understand, that a function doesn't need to be defined at a point for it have a limit at that point. But how do you evaluate that limit and explain it intuitively to a student? Specifically, what is the limit of the function x^2/x as tends to zero? and how do you contrast this with the first case? – Nikhil Panikkar Feb 20 '13 at 6:16
So what you're saying is that the limit of x^2/x is equal to zero as x tends to zero. But do you explain it intuitively? For here it seems that we are arbitrarily assigning a value to the limit. Can we for example, claim that we can make f(x) as close to zero as we wish by taking x closer and closer to zero? – Nikhil Panikkar Feb 20 '13 at 6:24
To make my comment above, more clear, if f(x) is a continuous at a point, then we can claim to make f(x), as close to its value at that point,as we wish, by taking x closer and closer to that point. But in the case of a discontinuous function, how do we choose a value, to which f(x) can be made arbitrarily close? – Nikhil Panikkar Feb 20 '13 at 6:39
To make my question, even more precise, consider a function f(x), whose value is constant everywhere except at one particular point(where it is undefined). Then you can say that the limit of f(x) as x approaches that point, is the constant value, for as long as x is not equal to zero, there is a unique value(the constant), which the function assumes. One such function is x/x. But for a function which assumes different values at points other than those of discontinuity (such as x^2/x), how would you choose a unique value? (and explain it intuitively)? – Nikhil Panikkar Feb 20 '13 at 6:52

Let me start my answer rigorously.

Lemma: Let $x_0$ be a point and $f,g$ be two functions and supposed that at a punctuated neighborhood of $x_0$ the functions equals. Then $\lim_{x\to x_0} f(x)=\lim_{x\to x_0} g(x)$.

Thus, since $x^2/x$ and $x$ equals in the punctured neighborhood $0<|x|<1$ of $x_0=0$, they have the same limit.

The intuition that should be explained to a student is that

1) limit of a function is local (depends on the values only near the point $x_0$)

2) the limit does not depend on the function at the actual points (so depends only on values in the punctured neighborhood...)

I think a student may well understand these two points intuitively, and hence be able to understand why $x$ and $x^2/x$ have the same limit as $x\to 0$.

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Ok, I understand. I have a related question : Generally students taking an introductory course in calculus, think of limits in terms of the epsilon-delta definition(i.e making f(x) as near as we wish to a particular value by making x approach nearer and nearer to another value.). The second condition stated by you is generally covered by delta being greater than zero in the definiton.But what about the first ? – Nikhil Panikkar Feb 20 '13 at 11:37
A related question is - Generally the epsilon delta definition is used as a proof writing tool i.e to prove that an already given limit exists.Can epsilon delta definitions be used to evaluate limits? – Nikhil Panikkar Feb 20 '13 at 11:40
One doubt, that I am having now - When x approaches zero from the left, x^2/x takes negative values(if you are actually plugging in numbers). And when x approaches zero from the right, x^2/x takes positive values. So does the limit exist? With f(x) = x, one has the luxury that the function is defined at x = 0, but what about this function? I think the answer to this question has to do with how you define 'local'.... is it? – Nikhil Panikkar Feb 20 '13 at 12:14
Regarding my fisrt comment to your answer, I guess, it is given by the absoute value of x - a being less than delta. But the questions raised in my other two comments stand. – Nikhil Panikkar Feb 20 '13 at 12:22
$x$ also takes negative values, I don't see what is your problem in this. A property P holds locally about $x_0$, means: there exists a (punctured) neighborhood of $x_0$ in which property P holds. – Lior B-S Feb 20 '13 at 15:02