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math.stackexchange.com/questions/462199/…

After reading the linked answer, I still don't completely understand what is going on. Using the example $\frac{2xh+3h^2}{h}=2x+3h$ is true for all $h$ except when $h=0$. I understand this. So for all values arbitrarily close to $h=0$ this is true. Hence the functions behaves more like $2x$ as we get closer to $h=0$. However, it never reaches there, so what allows us to substitute $h=0$. We have basically said that this statement is true for all $h$ when not $0$, then have gone ahead and let $h=0$. This seems extremely unintuitive to me.

user21820
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James Chadwick
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  • $h=0$ is substituted only in the right side. What enables that substitution is a different equation, the equation $\lim_{h\to0}(2x+3h)=2x+3\cdot0$. The argument would be: From your equation being true for all $h\neq0$, one applies $\lim_{h\to0}$ on both sides to get $\lim_{h\to0}\frac{2xh+h^2}{h}=\lim_{h\to0}(2x+h)$. Then from the equation I said, you deduce that this is $=2x+3\cdot 0$. The equation that I said needs a proof. – NDB Jul 18 '23 at 19:23
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    @James You said "the linked answer", but you linked the question rather than a particular answer. Which answer were you looking at, exactly? – Ben Grossmann Jul 18 '23 at 19:32
  • $h$ ist not $0$. It goes to $0$. It allows us to say that $\lim\limits_{h\to 0} \frac{c\cdot h}{h}=c$ and $\lim\limits_{h\to 0} c\cdot h=0$ – callculus42 Jul 18 '23 at 19:34
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    If you agree that for $h\ne 0$ we have $\frac{2xh+3h^2}{h}=2x+3h$, you can clear your doubt avoiding substitution altogether by proving that the limit is $2x$ using the definition. Since $|2x+3h-2x|=3|h|$, given an arbitrary $\epsilon>0$ if $0<|h|<\epsilon/3$ we have $|2x+3h-2x|=3|h|<3 \frac{\epsilon}{3}=\epsilon$. That is, $\lim_{h \to 0} (2x+3h)=2x$. No plugging of $h=0$ occured. – Bernkastel Jul 18 '23 at 19:57
  • If no plugging of $h = 0$ occurred, then where did the candidate limit $2x$ come from? – Ben Grossmann Jul 18 '23 at 20:30
  • See related https://math.stackexchange.com/a/1822706/72031 – Paramanand Singh Jul 20 '23 at 08:54

5 Answers5

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On limit arguments you don't set $h =0$. You should read again Blue answer in the linked question. Limits at a point don't care at all about the value of the function at that point. The function doesn't even have to be defined at that point for the limit to exist.

Here is a way you can look at it:

The function $f(h)= \frac{2xh+3h^2}{h}$ has domain $\mathbb{R}\setminus\{0\}$. The function $g(h)=2x+3h$ has domain $\mathbb{R}$. They agree in the domain of $f$, but they are different functions. It doesn't make sense to ask what is $f(0)$ because it is undefined there. However, we can ask what value should $f$ have at $0$ if it was a "nice" function, that is we can ask what is $\lim_{h\to 0} f(h)$. Because limits don't care about the point, you can say this limit is exactly the same as $\lim_{h\to 0} g(h)$. Now $g$ is a continuous function, so $\lim_{h\to 0} g(h) = g(0)$.

Paulo
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  • It is inaccurate (and in my opinion, confusing) to say that "you don't set $h = 0$". We do not plug $h = 0$ into $f$, but we do plug $h = 0$ into $g$. So we do set $h = 0$, but only after "simplifying" the function. – Ben Grossmann Jul 18 '23 at 19:46
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    @BenGrossmann I wouldn't say that we set $h=0$ after simplifying the function. We still have $\lim h \to 0$. But $\lim\limits_{h\to 0} 3h=0$. – callculus42 Jul 18 '23 at 19:54
  • @callculus42 But how do you come to the conclusion that $\lim_{h \to 0} 3h = 0$? Presumably, you reach this conclusion using the fact that $3 \cdot 0 = 0$. That is, you set $h =0$ within the expression $3h$. – Ben Grossmann Jul 18 '23 at 19:55
  • I'll grant you it can be confusing, but to me at least it clarifies things. I dont think it is inaccurate. – Paulo Jul 18 '23 at 19:56
  • @BenGrossmann You cannot find a number which is small enough to express h. h is always smaller than any positive number. Thus $\lim\limits_{h\to 0} c\cdot h=0$ – callculus42 Jul 18 '23 at 20:06
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    @callculus42 In the standard usage of the real numbers, if $h$ is such that "$h$ is always smaller than any positive number", then it follows that $h \leq 0$. In other words, you seem to be implying that $h$ has an infinitesimal value, but infinitesimal values do not exist within the usual framework for calculus. Even if your description of $h$ did make sense, that does not change the fact that the way in which we compute the limit of an expression like $\lim_{h \to 0} 3h$ in a practical setting ultimately uses the fact that $3 \cdot 0 = 0$ – Ben Grossmann Jul 18 '23 at 20:17
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    @callculus42 The fact that substituting $h = 0$ into $3 h$ gives the correct value for $\lim_{h \to 0}3h$ produces the correct answer is something that on some level requires proof via the definition of a limit, but once the continuity of a function like $h \mapsto 3h$ is established, it is absolutely fine to compute $\lim_{h \to 0}3h$ by setting $h = 0$. Moreover, setting $h = 0$ is precisely how both you and Paulo are computing these limits, regardless of whether you explicitly state this in your justification. – Ben Grossmann Jul 18 '23 at 20:20
  • @BenGrossmann "h is always smaller than any positive number", then it follows that h≤0 This would not my conclusion. Tell me the smallest positive number you can think of and I can give you a smaller one. Nevertheless the answers and the comments gives the OP a good idea about limit calculation and how to interpret it. – callculus42 Jul 18 '23 at 20:27
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    @callculus42 You're being unclear about whether you're taking $h$ to be a single value or something changing. The statement "you cannot find a number which is small enough to express h" implies that $h$ has a single value (that is too small to be expressed by a positive real number). On the other hand, "tell me the smallest positive number you can think of and I can give you a smaller one" seems to imply that you're considering progressively smaller values of $h$. The answers and comments are mostly good except that some people are implying that we never plug in $h = 0$. – Ben Grossmann Jul 18 '23 at 20:35
  • +1 ... Because you recommended my answer. ;) – Blue Jul 19 '23 at 05:55
  • @BenGrossmann My idea was a dynamic game. You tell me a (small, positive) value. I tell you a smaller value. You tell me a smaller value ... And we never reach $h$. – callculus42 Jul 19 '23 at 17:14
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    @callculus42 Again, you're implying two contradictory things. The statement "we never reach $h$" implies that $h$ has a single non-negative, non-zero value that it is impossible for us to reach with positive real numbers. As I explain in an earlier comment, this is incompatible with the standard framework for calculus. On the other hand, your dynamic game implies that $h$ is supposed to refer to a succession of progressively smaller real numbers. This is fine, but doesn't answer the original question of what exactly it is that you're doing to determine that $\lim_{h \to 0} 3h$ is equal to $0$. – Ben Grossmann Jul 19 '23 at 18:35
  • I have down voted this post by mistake (probably a touch on phone screen while scrolling) and noticed this now. The vote is locked now and can be undone only when the post is edited. – Paramanand Singh Jul 20 '23 at 22:31
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It is important to understand the question being asked: what we want is the limit of $\frac{2xh + 3h}{h}$ as $h \to 0$, which is different from the value obtained by plugging in $h = 0$ (which is undefined).

The step of ultimately "plugging in" $h$ into the simplified version of the fraction can be justified as follows. We begin with a function of $h$, namely $f(h) = \frac{2xh + 3h^2}{h}$. We note that at all points except $h = 0$, this function behaves exactly like another function, namely $g(h) = 2x + 3h$. Now here's the important point: because $f$ and $g$ agree at all values of $h$ except zero, we can conclude that if $g$ has a limit as $h \to 0$, then $f$ must have the same limit.

With that, we have switched from finding $\lim_{h \to 0}f(h)$ to finding $\lim_{h \to 0}g(h)$. Because $g$ is a "nice" (i.e. linear and hence continuous function) function of $h$, $\lim_{h \to 0}g(h)$ will be the same as $g(0)$. Thus, we have $$ \lim_{h \to 0}f(h) = \lim_{h \to 0}g(h) = g(0) = 2x. $$

Ben Grossmann
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There are two facts that are important to understand.

  1. For any function $f$, its limit as $h\rightarrow 0$ only depends on the values of the function evaluated at points nearby but not equal to $h=0$. In particular, the value of $f(0)$ does not matter for the limit. It does not even matter if $f$ is defined at $0$, only that is defined for all $h$ very close to $0$. As a consequence, if two functions are equal for all points nearby but not equal to $h=0$, then they have the same limit at $h=0$.

The next point may seem to contradict the first at first.

  1. When $f(h)$ is continuous at $h=0$, then $\lim_{h\rightarrow 0}f(h)=f(0)$. So for a continuous function, we can evaluate the function at $0$ to determine a limit.

The first fact says that in general, there is no connection between the value of a function at a point and the limit as you approach the point. The second fact says that for a particular type of function, there is a connection. In case this is not clear: there is generally no relationship between someone's first name and someone's last name, but if you go to a convention for people who have the same first and last name, then there is connection for that particular group of people.

If you understand these two facts, we can solve the problem. Let $f(h) = \frac{2xh + 3h^{2}}{h}$ and let $g(h) = 2x + 3h$. By fact #1, because these functions agree at all points nearby $h=0$, we know that $\lim_{h\rightarrow 0}f(h)=\lim_{h\rightarrow 0}g(h)$. Therefore, we will simply try to determine $\lim_{h\rightarrow 0}g(h)$ now, and we will never need to think about $f$ again or plug anything into it.

Finally, because $g(h)$ is a continuous function, we can evaluate $\lim_{h\rightarrow 0}g(h)=g(0)=2x$ by plugging in $h=0$ to it. Note that we don't need to plug $h=0$ into anyone else, just $g$.

Writing this in one line: $\lim_{h\rightarrow 0}f(h) = \lim_{h\rightarrow 0}g(h) = g(0) = 2x$. So the limit we wanted is $2x$.

Mikhail Katz
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f3qgrgdf
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The shortest explanation is that $x$ is a standard number, $h$ is an infinitesimal and therefore nonstandard, and to pass from $2x+3h$ to $2x$, one takes the standard part (the standard part of every infinitesimal is zero). This is equivalent to the limit definition, and in fact the concept of limit can be defined in terms of the standard part. For example, the limit of $f(x)$ as $x$ tends to $0$ is the standard part of $f(h)$ where $h$ is infinitesimal.

This is pretty much how Leibniz himself thought of the process of calculating $\frac{dy}{dx}$ when he spoke about deleting negligible infinitesimal terms, and since he is the one who invented the whole thing in the 17th century, we should certainly be paying attention to his intuitions about this process. There is of course also the slick trick of doing this without infinitesimals that was invented at the end of the 19th century, namely the epsilon-delta paraphrase of the original infinitesimal definitions. However, if such slick tricks make the subject of derivatives and integrals impenetrable, then Leibniz's intuitions could and should come to the rescue. For further details see this answer.

Mikhail Katz
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    It seems unlikely that OP meant to use hyperreal numbers, or even has the prerequisite knowledge to understand your answer. Obtaining quick answers using slick tricks that can be justified, but whose justification is beyond the grasp of the student, is an excellent way to make a subject seem impenetrable and magic. My own understanding of combinatorics was held up quite a bit by showing results obtained by generating function manipulations without explaining formal power series and how they work; rather regrettable. – Marc van Leeuwen Jul 19 '23 at 21:28
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    @MarcvanLeeuwen, yes, a rigorous version of non-standard analysis is probably not appropriate for the OP, but, in fact, the currently-standard "definition" of limit is arguably not the most intuitive thing in the world, either. In my own experience, the game of having a thing not be zero, so we can divide, and then set it to 0, was weird. (Bishop Berkeley's essay...) By now, for elementary calculus, I think the implicit method of "take the standard part" (if not said so explicitly) is the most succinct description of "take the limit..." Sure, tastes vary. :) – paul garrett Jul 19 '23 at 22:28
  • @MarcvanLeeuwen, thanks for your comments, which I appreciate. Sometimes folks downvote without leaving comments, which can be frustrating. I updated my answer to clarify the point you raised. – Mikhail Katz Jul 20 '23 at 06:44
  • @paulgarrett, thanks for your input. – Mikhail Katz Jul 20 '23 at 06:44
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You have some good answers already but here is a less formal and maybe more intuitive style.

As everyone says: an important point about limits is that we do not need to evaluate the expression at the limit ($h = 0$ in your case). It is not necessary that the expression is even defined at the limit point. If it is then we still don't use the value at that point.

You can think of the usual $\epsilon - \delta$ definition of limits as a game.

I claim that a function $f(x)$ has the limit $y$ at $x = a$. You can now challenge me by picking positive values of $\epsilon$ and I need to answer with a positive value of $\delta$ which gets the function within $\epsilon$ of my claimed limit. If I can always manage this then I have won the game: the function has the claimed limit. If you defeat me, you find an $\epsilon$ and I cannot answer with a suitable $\delta$ then you have won: the function does not have the claimed limit or maybe has no limit.

Your example is simple: I will pick my answer for $\delta$ as $\frac{1}{3}$ of your challenge value of $\epsilon$.

The value of $\delta$ does not need to be the best in any sense (e.g. only just good enough), it just needs to be good enough. So, I might answer $1$ anytime that your challenge value is $3$ or more. If you say $1000000$ then I do not need to bother calculating $1000000 / 3$.

In the case when the function is defined at $x = a$ you can compare the limit to $f(a)$. If the limit exists and equals $f(a)$ then we say that the function is continuous at $a$. Note that this is an optional extra step in the case that the function is defined at $a$, you can calculate the limit even if it is not.

You will also see limits which appear to involve $\infty$. These are just suggestive notations and the formal definitions do not involve $\infty$ in the same way that a limit at a (normal finite) value does not need you to use the value of the function at that point.

Limits are most often studied with the real and complex numbers but it is possible to study them elsewhere if you have a way to define closeness. You can use the same definition with the rational numbers or even the integers. With the rationals, the limit may fail to exist even it looks as if it should. With the integers, limits are even more strange. Ask additional questions if you want to know more.

badjohn
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