# Why can we replace a variable with a constant in a limit?

We say $\lim_{x \to c} f(x) = L$ that means $f(x)$ may be as close to $L$ as $x$ tends to $c$. "Tends" here means $x$ approaches $c$ but never actually becomes $c$. If it so, then why do we such easily replace $x$ with value $c$, whenever it is appropriate. E.g. $\lim_{x \to 5} 4 + x = 4 + 5 = 9$, or $\lim_{h \to 0} f(x+h) = f(x)$.

Understand me right. I don't want to discuss cases when we can do the substitution and sometimes not, because we get a division by zero, and we need to do some simplification, etc. I understand all this. I just can't understand if $x$ actually is never $c$, what allows me to write $c$ as a value of $x$? Well, I used to think about it like "ah, as $x \to 0$, x is very small number, let it be zero". But it is not statistics, you know, to close eyes and make approximations.

I met the notion of "infinitesimal" and as I understood it is opposed to "$\delta-\epsilon$" approach. I can't fully understand how they are related to each other and to my question. Maybe I lack of historical context. If it so, please clarify this for me. Thanks.

• As $x$ gets very close to $5$, the function $4+x$ gets very close to $4+5$, so this is the limit. For functions like this where the function is easily defined at the limiting point, we can simply replace the variable with the constant to which we're taking the limit. Limits become more important when dealing with stuff that is less well defined, for example $\lim_{x\to0}\frac{\sin x}{x}$, where substituting in the $0$ doesn't help. Dec 24, 2017 at 13:10
• @drhab, sorry? But the definition states x goes close enough, but never becomes c. Dec 24, 2017 at 13:17
• @JohnDoe, as I mentioned, I understand the cases when we can't just put it. But I am asking completely different. Why am I able actually to think about replacing $x$ with value $c$? Dec 24, 2017 at 13:19
• Well, I understand you, for example polynomials are well defined, so we can put $c$ instead of $x$. But what I ask is not particular cases when we can or not. I ask why is it even eligible to substitute the value for the variable? Why do we have a chance to think about it, why does this choice exist, while $x$ is actually never equal to $c$. Dec 24, 2017 at 13:21
• @drhab What does "well-defined" mean in this context? $\lim_{x\rightarrow c} f(x) = f(c) \Leftrightarrow f$ is continuous at $c$ (this might even be per definition depending on your definitions!). You can also have functions which are defined at $c$, but discontinuous there (i.e. the limit doesn't exist or it only exists at one side or it is a jump on both sides). Dec 24, 2017 at 13:21

In order to compute $\lim_{x \to 5} (4 + x)$, we use the fact that $$\lim_{x \to c} (f(x) + g(x)) = \lim_{x \to c} f(x) + \lim_{x \to c} g(x)$$ whenever both limits on the right exist.

Now, $\lim_{x \to 5} 4 = 4$ holds because as $x$ approaches any value (including $5$), the expression $4$ approaches the value $4$, since it is constant.

Also, $\lim_{x \to 5} x = 5$ because as $x$ approaches $5$, the expression $x$ approaches $5$, trivially.

Of course, all of these results can be rigorously proven with the $(\epsilon, \delta)$-definition of limit.

Thus $\lim_{x \to 5}(4 + x) = 4 + 5 = 9$. It looks as though we just replaced $x$ with $5$, but what is actually going on is quite different.

• Oh, how stupid I am! So we actually don't put a value of c instead of x. But it is the value of $\lim_{x \to c} x = c$, so we put it into expression using the property you mentioned. Dec 24, 2017 at 13:27
• The argument why $\lim_{x\rightarrow 5}x = 5$ is probably the same as why $\lim_{x\rightarrow 5}(4+x)=4+5|_{x=5}$: The functions $f,g: \mathbb{R}\rightarrow\mathbb{R}$ with $f(x)=4+x$ and $g(x)=5$ are both continuous at 5. So if the function being regarded is continuous, you can simply plug in (as in symbolic replacement) your c. Dec 24, 2017 at 13:30
• @ComFreek, yeah, technically we just put the value of c. But actually in the expression it is the value of the limit of the term $x$. As it is a polynomial, its limit $L$ is equal to $f(c)$, i.e. $f(x) = x \implies f(c)= c \implies L = c$. Dec 24, 2017 at 13:35

First of all a limit operation $\lim_{x\to a}$ does not involve assigning values to $x$ which are successively close to $a$ although the phrase tends to suggests a similar meaning. That phrase tends to has a meaning but only in full context like $f(x)$ tends to $L$ as $x$ tends to $a$. Moreover the last sentence has a precise meaning which is often diluted heavily in various textbook expositions of the concept of limit. This dilution has a big cost in terms of creating confusion in minds of students.

The concise equation $\lim_{x\to a} f(x) =L$ means that a very specific logical statement is true. It is the complicated form of that logical statement which makes the overall concept of limit slightly difficult to understand. Also contrary to the popular belief (shared by many educators) the difficulty in this logical statement is not related to structure of the statement but it is rather due to the inequalities involved and a lack of appreciation of order relations in the overall high school math education.

The logical statement here is that it can be ensured that all the values of $f(x)$ are as close to $L$ as we want by constraining the values of $x$ close to $a$. It does not mean that the value of $x$ is close to $a$, but rather that we can guarantee some kind of pattern in the values of $f(x)$ (the pattern being that these values are near $L$) by restricting values of $x$ near $a$. To put the matter in crude terms, we are guaranteeing that something can be done, not that we actually do it. That is, in this entire discussion we are not assigning any values to $x$.

The above discussion in no way helps us to figure out (or evaluate) $L$ given the function $f$ and point $a$. That job is done via the help of theorems which are popularly known as algebra of limits. And further this is greatly simplified by noting that for a large class of functions commonly seen in calculus the limit of a function at a point is same as the value of the function at that point (this fact is again established using algebra of limits). This point has been already discussed in great detail in this answer.

The presentation of these ideas using the notion of infitesimals is somewhat different. And the infinitesimals are mainly used to remove the inequalities which are inherent in the logical statement related to limits. But proponents of infinitesimal approach have very smartly avoided mentioning the above fact and instead stress that their presentation is aimed at simplifying the structure of the logical statements. And a huge amount of foundational work has been done in the theory of infinitesimals to avoid these inequalities which are involved in the usual $\epsilon, \delta$ approach. Thus the complete understanding of infinitesimal approach is is no way simpler than the usual / standard approach which involves just an appreciation of order relations.

• Thank you for the great answer! Dec 25, 2017 at 10:57