I know that $\lim_{x\to0}\frac{x}{x}=$ 1.
But in my text book, it is written that it is $\infty$ and even $\frac10=\infty$.
But how is it possible? And are they both same? What is the difference between indeterminate and undefined?
And if they are $\infty$ then what is the reason? And why can't we define something for these numbers? Textbook fault??

Now see that book says when $x=0$, $t=\infty$. If it is wrong, why did they just write it?

  • 1
    $\begingroup$ You have given a good example ($\frac xx$) for which the limit is 1. But you could have looked at $\frac {x^2}{x}$ (limit $0$) or $\frac {2x}{x}$ (limit $2$) or $\frac {x}{x^2}$ (limit $\infty$), and so on. The fact that we can "easily" get all these different values suggests that it will be hard to declare a single value for $\frac 00$ $\endgroup$ – lulu Aug 2 '15 at 12:52
  • $\begingroup$ So? Is it perfectly infinite? $\endgroup$ – Aditya Agarwal Aug 2 '15 at 12:55
  • $\begingroup$ I don't know what "perfectly infinite" means. I am arguing that it isn't a well defined number at all. In certain specific contexts (like the limit you gave, or the examples I gave) we can give it a meaning but that meaning remains specific to that context. $\endgroup$ – lulu Aug 2 '15 at 12:57
  • $\begingroup$ My book says that (while addressing an integral problem)- $x=t^{-1}$, so when $x=0$ then, $t=\infty$ So is it right? $\endgroup$ – Aditya Agarwal Aug 2 '15 at 13:00
  • 2
    $\begingroup$ @AdityaAgarwal No it is blatantly wrong. At least wrong in the context of calculus. $\endgroup$ – user103816 Aug 2 '15 at 13:03

The limit $\displaystyle \lim_{x \rightarrow 0 } \frac{x}{x}=1$, as you said. It is not infinity. Why, because the fraction is simplified to $1$. So no matter where $x$ tends the limit will always be $1$.

As for undefined it means that something is not defined. For example the function $f(x)=\frac{1}{x}$ is not defined at $x_0=0$, that is it is undefined at $x_0=0$.

Indeterminate, on the other hand means other thing. Consider the limit:

$$\lim_{n \rightarrow +\infty} \left(1-\frac{1}{n} \right)^n$$

You can see immediately that it is of the form $1^{\infty}$, which is a very well known indeterminate form. Some could speculate here that this limit would be $1$. No, it is not. It equals $e^{-1}$.

Now, let's take a look at another limit. For example:

$$\lim_{x\rightarrow 1^{+}} \frac{\sqrt{1-x^2}}{1-x^2}$$

Again we see that as $x \rightarrow 1^+$ both the nominator and the denominator are zero. This is an indeterminate form $\frac{0}{0}$. However you can sub $1-x^2 =u$ and change variables. Then $u \rightarrow 0$ and the limit is expressed as:

$$\lim_{x \rightarrow 0^+} \frac{\sqrt{u}}{u}$$

which clearly is $+\infty$.

So, indeterminate form means that you can not estimate the limit right away, that is plug in the value and calculate. Undefined means that something is not defined. So, there terms are not the same.

  • $\begingroup$ Good explanation. (+1) $\endgroup$ – user103816 Aug 2 '15 at 13:04

We like to think about continuous functions. Everything to us is continuous, and the basics of physics and calculus, where limits came from, dealt mainly with continuous functions.

Why does that matter? Because a continuous function is exactly a function $f$ such that $f(x_0)=\lim_{x\to x_0}f(x)$.

So when we want to calculate, for example, $\lim_{x\to 2}x+1$, we know we can just put the limit into the function. Namely, $2+1$ and get $3$. This works great, $\lim_{x\to 2}\sqrt x=\sqrt2$ and $\lim_{x\to 0}e^x=1$.

But then we run into functions which are not continuous. Like $\frac1x$. Then when we calculate $\lim_{x\to 0}\frac1x$, we are tempted to replace $x$ by $0$, and obtain $\frac10$. But wait, we can't divide by $0$. So here we need to be more careful and investigate the limit by its definition.

$\lim_{x\to x_0}f(x)=L$ (meaning it exists and it equals to $L$), if for every small enough interval around $L$, we can find some other small enough interval around $x_0$, such that whenever $x$ is in that interval near $x_0$, $f(x)$ is in the interval near by $L$.

If no such $L$ exists, then the limit does not exist. We are sometimes being informal and say that the limit is $\infty$ or $-\infty$ if the function becomes eventually larger (or smaller) than any other real number.

Non-existence of limits

So we are back to the case of $\frac1x$ and $x\to 0$. The limit does not exist, since there is no real number which satisfies the definition. The limit is not $\infty$ either, since by approaching $0$ only from the positive side, we obtain increasingly larger positive real numbers; and by approaching from the negative side, we will grow further and further negative.

So the limit does not exist at all. Again, this can be mitigated by saying that you are not interested in the sign of the limit, and you sort of "glue" the two points together. Then the limit is $\infty$. But to say that without additional explanation is just bad pedagogy.

Indeterminate forms

So what are indeterminate forms? Our instinct to just plug in the limits into the function is strong, and we sometimes get meaningful values. But sometimes we get something like $\frac00$ or $1^\infty$ or even $0-0$.

These case cannot be "easily determined", and they need further investigation. For example, $\lim_{x\to 0}x-0$ has the form $0-0$, but it is clear why the limit is indeed $0$.

On the other hand, $\lim_{x\to 0}\frac{x^2}{x}$ is also $\frac00$, but simplification shows that it is in fact again $\frac1x$ so the limit does not exist in the strong sense of the word.

To sum up

Limits have a particular definition. And while it's very easy to forget it and assume our universe is entirely made of continuous functions, this is not true. So we can't quite plug in the limit into the equation and get it over with. Sometimes, it is clear why there is no limit, since the function does not "stabilize" into some value.

But in other times, plugging the limit into the equation will produce a situation where we cannot determine immediately whether or not there is a limit. Then we need to simplify the function, or investigate other routes for a solution (e.g. L'Hospital's rule).

  • $\begingroup$ I know all this, but then why does my textbook say this (see the edit) $\endgroup$ – Aditya Agarwal Aug 2 '15 at 13:38
  • $\begingroup$ "What is the difference between indeterminate and undefined?" $\endgroup$ – Asaf Karagila Aug 2 '15 at 13:42
  • $\begingroup$ Okay, 1 part answered. (y) Thank you so much! $\endgroup$ – Aditya Agarwal Aug 2 '15 at 13:45
  • $\begingroup$ Is $\frac10=\frac00=\frac{a}0$ for any constant $a$? $\endgroup$ – Aditya Agarwal Aug 2 '15 at 13:46
  • $\begingroup$ @AdityaAgarwal No, $a \over 0$ is undefined for any $a$, that is we do not assign any meaning to it. Since it is mathematically meaningless, any expression involving it is also meaningless. Now equality just means that the thing on the left side can be replaced by the thing on the right side of equality in every possible case; perhaps in this sense you could say two meaningless things are equal :-). $\endgroup$ – user103816 Aug 3 '15 at 1:19

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