We repeat the phrase "approach zero" regularly, but what exactly does it mean to say "$\delta t$ approaches zero"?

P.S. If this a stupid question, then please forgive me.

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    $\begingroup$ It is very far from a stupid question. Actually, it is the very heart of differential calculus. Nevertheless, it would be easier to answer if you specified an example where you found it. The exact meaning can vary slightly depending on the context. $\endgroup$
    – ajotatxe
    Commented Feb 25, 2018 at 9:24
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    $\begingroup$ Very good question. $\endgroup$
    – user517784
    Commented Feb 25, 2018 at 9:26
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    $\begingroup$ This expression is only one part of a larger phrase, such as "the limit as $\delta t$ approaches $a$ of $f$ is equal to $L$." It is the full larger phrase which must be defined. The snippet "$\delta t$ approaches $a$" has no precise meaning on its own. You might be interested in reading about the $\epsilon-\delta$ definition of the statement $\lim_{x \to a} f(x) = L$. (By the way, here $f$ is a function defined on a subset of $\mathbb R$ and $a$ is a limit point of the domain of $f$.) $\endgroup$
    – littleO
    Commented Feb 25, 2018 at 10:09
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    $\begingroup$ I thought it was-approach0.xyz/search $\endgroup$
    – Soham
    Commented Feb 25, 2018 at 16:56
  • $\begingroup$ @Soham This was posted before that search engine came into being... But amazing, I too thought the same... $\endgroup$ Commented Dec 9, 2023 at 19:09

7 Answers 7


Your question isn't stupid, it's the heart of calculus.

An introductory step from algebra to calculus is in the context of slope. Algebra allows us to find an average slope, while calculus allows us to find the instantaneous slope. In other words, algebra gives us the slope of a line, while calculus gives us the slope of a point.

Slope of a point? Yes, but let's stay with lines for now. The slope of a line is given by the following function. We divide the change in $y$ by the change in $x$:

$$m=\frac{y_2 - y_1}{x_2 - x_1}$$

However, if the line is curved, say $f(x)=x^2$, then each point on that line has a different slope. If we are asked to find the slope when $x=5$, then we can approximate it by finding the average slope between $x=4$ and $x=6$:


But this isn't the correct answer. If we wanted to get closer to the correct answer, we would choose values that are closer to 5:


The pattern to notice is that the more accurate our answer becomes, the smaller the difference between $x_2$ and $x_1$. In fact, the correct answer will be found when the difference is zero. However, when we go to write this down, we have a problem:


Specifically, we can't divide by zero. We've gone as far as algebra can take us, and we need a new way to talk about math. We need calculus. In algebra, we saw that we get closer and closer to the correct answer. In calculus, this is called the "limit". We get closer and closer to the limit as the divisor gets closer and closer to zero. The divisor "approaches zero".

Finally, we have "What is the limit as x approaches zero?"

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    $\begingroup$ Specifically, we can't divide by zero. In the case that the nominator is zero, the claim is simply wrong. It is the solution of 0*m = 0 and this in fact solvable by every finite member of R (just to exclude infinities added in some extended versions of R). This case is called indeterminate and is often solved by looking which real number is approached when the term approaches zero (not only in calculus). But some functions cannot be "fixed" this way and remain undefined when 0/0 is approached. $\endgroup$ Commented Feb 26, 2018 at 2:11
  • $\begingroup$ @ThorstenS. Yes, I've been thinking that there's something wrong with this. I was more focused on how the idea "approaching zero" happens. It's marked as the answer, so maybe I shouldn't edit it? $\endgroup$
    – Carl
    Commented Feb 26, 2018 at 2:30
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    $\begingroup$ You are free to edit your answer. $\endgroup$ Commented Feb 26, 2018 at 2:40
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    $\begingroup$ It's not the slope of a point (a point has no slope), it is the slope at the point. Different functions may have different slopes at the same point. $\endgroup$
    – celtschk
    Commented Feb 26, 2018 at 7:50
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    $\begingroup$ You need to do something about $y1, y2, x1, x2$. It's probably subscripts, in which case you need to write y_1 y_2 x_1 and x_2, but just in case you actually meant exponents, I've let them be for now. $\endgroup$
    – Arthur
    Commented Feb 26, 2018 at 9:17

Imagine the following game, played by two players:

Both players are given a number, the same number (let's say in our case it is zero) and they are taking turns trying to find a number that is closer than the one found by the previous player. After 100 tries, if neither of them has quit yet, a coin is flipped to decide the winner. $$\begin{align} \text{Player A}&:\text{10}\\ \text{Player B}&:\text{1}\\ \text{Player A}&:\text{0.5}\\ \text{Player B}&:\frac{1}{3}\\ \text{Player A}&:-\frac{1}{\pi}\\ \text{Player B}&:\frac{1}{8}\\ \text{Player A}&:\frac{1}{16}\\ \text{Player B}&:-\frac{1}{1,000}\\ \text{Player A}&:-\frac{1}{1,000,000}\\ \text{Player B}&:\frac{1}{10,000,000}\\ \text{Player A}&:\frac{1}{10^{15}}\\ \text{Player B}&:\frac{1}{10^{3,026}}\\ \end{align}$$ They soon realize that this can last forever. For instance, if one player chooses a number, let's say $a$, the other player can choose the number $\frac{a}{2}$ which is always closer to zero than $a$.

If we perform this procedure of approximating zero with whatever accuracy you want, in such a way that you can find a number, $a$, arbitrarily close to it - we say "$a$ approaches zero".

So, in terms of real numbers, getting arbitrarily close to a number in terms of distance (i.e. the absolute value of real numbers) is considered approaching a number with another. However, you may alter the way you think two numbers are close, and then the situation gets messy - see, for such cases, courses like Real Analysis or Topology.

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    $\begingroup$ You might add that "approaching zero" can also be understood as approaching from positive direction only, depending on the context (e.g. negative times usually don't make sense). Also there's a small typo ("whicever"). $\endgroup$
    – Jasper
    Commented Feb 25, 2018 at 11:17
  • $\begingroup$ Thanks for the remarks! :) $\endgroup$ Commented Feb 25, 2018 at 11:20
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    $\begingroup$ I'm not sure I like the example; I think it glosses over the issue of having arbitrarily small accuracy; why it's specifically $0$ that is being approached rather than anything else. For example, with these choices, the players could just as well be playing the "find numbers closer to $-1$" game, but badly. Or maybe they were playing the "find numbers closer to $10^{-10000}$ game? $\endgroup$
    – user14972
    Commented Feb 25, 2018 at 14:52
  • $\begingroup$ Well, in my writing I assumed that both players choose their numbers in a "greedy" way, which results to an absolutely monotonous sequence. However, you are right that this sequence may converge to whichever number, even $e$, however, I feel that using monotonicity with respect to absolute values and clearly stating which is the limit may make the concept of convergence a little more clear. I know it is a mere a example and, of course, it cannot cover all the cases. It may even cause the misconception that convergence has to be monotonous. (Continued...) $\endgroup$ Commented Feb 25, 2018 at 16:18
  • $\begingroup$ (...Continued) Isn't it true, however, that, for every sequence, any finite numbers of its elements, do not define its limit? So, $\frac{1}{n}$ is not obvious that converges to $0$ if one just writes down it first terms, as well as the sequence mentioned does not. But, the players' understanding that this may take forever is what may help find notion of convergence in this example. Thank you for your comment and you observations! They helped a lot clarify some things! :) $\endgroup$ Commented Feb 25, 2018 at 16:23

The limit definition might help here. We say that a function $f$ "approaches zero", if for every $\epsilon > 0$, there is a $\delta > 0$ such that, if $x>\delta$, $|f(x)| < \epsilon$. Think of it this way: You can make the "outputs" of the function as close as you like to zero by choosing large enough "inputs".


The phrase "$x$ approaches zero" is a colloquial way of evoking a precise abstract thought figure as follows:

(i) If $x$ is an independent variable of some function $f$ it means that we are going to look whether $\lim_{x\to0} f(x)$ exists and if yes will try to determine its value, according to the rules defined in the books.

(ii) If $x$ is a dependent variable, i.e., $x=f(t)$, and we are envisaging a limiting process $t\to \tau$ for some proper or improper accumulation point $\tau$ of the domain of $f$, then "$x$ approaches zero" means $\lim_{t\to\tau} f(t)=0$, whereby the latter has to be expanded according to the rules given in the books.


Your question is far from a stupid question. In as recent as the $1960$'s, mathematician Abraham Robinson pioneered what it means for a value to approach $0$, so don't worry, loads of professionals still ponder this question ;)

There are different answers to this question depending on which branch of mathematics you are in. For example, if you are working in nonstandard analysis, for a value to "approach zero" means it essentially "becomes" an infinitesimal. To illustrate this idea, we say an infinitesimal in nonstandard analysis is a value which is so small, that two real numbers who are an infinitesimal distance apart cannot be distinguished from each other. A good example is $0.999\ldots$ and $1$.

Although, I'm assuming, given your tag, that you're referring specifically to calculus. In which case, consider the following. Let $a$ be a constant. On the real number line, the distance between $0$ and a constant is that constant. So using this idea, we can say that for a constant $a$ to approach $0$, the distance between $a$ and $0$ gets less and less, but $a$ and $0$ will never be exactly on top of each other. We can couple this idea with the implications of a value approaching $0$. For example, if we had some function, $g$, such that $$g(a)=\frac{1}{a}$$then for $a$ to approach $0$, would mean that $g(a)$ would approach (very) large values. Ultimately, trying to understand what it means for single value to approach $0$ can be slightly mind boggling, so as a result, we usually look instead at what a value approaching $0$ implies, which has lead to some beautiful tools in mathematics such as epsilon-delta proofs and the classic derivative, both of which you will, if you continue studying calculus, come across in due time.


Definition: $\delta t$ means the change in time, or any other variable being considered.

Then $\delta t\to0$ simply means that you are forcing two points in time closer and closer together. This is an extremely important concept because it forms the fundamentals of calculus.

The gradient of a line is given by $$\frac{\delta y}{\delta x}=\frac{y_2-y_1}{x_2-x_1}$$ and we can apply this to curves as well. We have $$\frac{f(x)-f(a)}{x-a}$$ at some point $x=a$ and this is known as the average rate of change. But to find the rate of change (or slope) exactly at the point $(a,f(a))$, we let $x=a+t$ and take the limit as $t\to0$. $$\lim_{t\to0}\frac{f(x)-f(a)}{x-a}=\lim_{t\to0}\frac{f(a+t)-f(a)}{t}$$ This is known as the instantaneous rate of change, or more commonly, the derivative.

Example: Take the function $f(x)=\sin x$ and we want to find its slope at $x=\pi$. Then the definition of the derivative is $$\begin{align}f'(\pi)=\lim_{t\to0}\frac{f(\pi+t)-f(\pi)}{t}&=\lim_{t\to0}\frac{\sin(\pi+t)-\sin\pi}{t}\\&=\lim_{t\to0}\frac{\sin\pi\cos t+\sin t\cos\pi}{t}\\&=-\lim_{t\to0}\frac{\sin t}{t}\end{align}$$ which is a very famous limit and it evaluates to $1$ (see here). Hence $f'(\pi)=-1$.


You mean when you take the derivative of a function? Then it just means taking the limit $t \to 0$, like in $\lim_{t\to0}\frac{f(x+t)-f(x)}{t}$.

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    $\begingroup$ Yes you got his point but your answer doesn't explain "approaches zero". $\endgroup$
    – user517784
    Commented Feb 25, 2018 at 9:28
  • $\begingroup$ also it talks about $t\to 0$ while it could have been better to use $dt\to 0$. $\endgroup$
    – Surb
    Commented Feb 25, 2018 at 9:29
  • $\begingroup$ @Surb he actually answered this question before it was edited, that's why he couldn't change it.I have also done same with my answer.I have explained about t tends to 0 not dt. $\endgroup$
    – user517784
    Commented Feb 25, 2018 at 9:33

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