# Is my understanding of limits correct?

I want to explain the basic concept of limits to see if my understanding is correct or not.

If we have a function in the form of a fraction and for a value of $$x$$ the numerator and denominator $$=0$$ , the graph of the function will have a perforation. In order to find the coördinates of the perforation we can use limits to rewrite the function in a form so that we can put in the value of $$x$$ that would have resulted in $$0$$.

Because of the limits we can rewrite, for example, $$f(x)= \frac {x^2 - 5x + 6}{x-2}$$ = $$\frac {(x-3)(x-2)}{x-2}$$ to $$\displaystyle \lim_{x \to 2} x-3 = -1$$

We normally are not allowed to divide by $$x-2$$ because we do not know if $$x-2$$ could equal $$0$$. In this case we know $$x=2$$ will result in $$0$$, so the limit basically says that we take a $$x$$ that is very close to $$2$$, like $$1.9999999999999$$ but does not equal $$2$$ and therefore we can divide by $$x-2$$ and simplify the function to a form where we can input $$x=2$$

Please correct me if I am wrong and apologies for some of the formatting, I do not know how to format the limit correctly.

• Roughly, the notation $\lim_{x \to 2}$ assumes that $x \neq 2$, so division by $x-2$ is okay. Aug 22, 2022 at 16:18
• For formatting, $\lim_{x \to a} \frac{f}{g}$ makes $\lim_{x \to a} \frac{f}{g}$. Aug 22, 2022 at 16:19
• This is an utterly valid argumentation. For every $x\ne 2$ , $f(x)$ and $x-3$ have the same value hence the simplification is allowed although $f(x)$ is not even defined at $x=2$. The simplified expression $x-3$ is innocent and the limit easily follows. Aug 22, 2022 at 16:25
• @Randall But $\lim_{x \to 2}$ also assumes a number that is extremely close to $2$ ,right? on top of $x$ not equal to $2$? Aug 22, 2022 at 16:33
• Yes (but not really), but "extremely close" does not permit equal to 2. Aug 22, 2022 at 16:35

Too long for a comment.

The meaning of a limit is literally in the name: the limit. It's the value a function approaches as the input approaches a certain value $$a$$. Notice the word "approaches." Meaning neither does the input, nor the function actually reach the point of interest at $$x=a$$. The limit is the value that's strictly surrounded by the function values as you approach $$a$$; you never get there, that's why it's called the limit. Keep in mind that the limit doesn't have to be a possible output of the function.

Limits are not only interesting when the function is not defined at $$a$$; we can also use them to check a function is continuous (i.e is connected with no cuts) by checking the limit is the same as the value of the function at $$a$$. Continuous functions admit very nice theorems and are easier to work with in general.

That's why you can cancel the factors that cause problems when taking the limit; you are essentially removing the discontinuity, i.e replacing the function with a continuous one, both should approach the same value as you approach $$a$$ because they only differ in the definition at $$a$$ which we don't use anyway to evaluate the limit.

Yeah, you're just about right. It's a shame that limits are introduced in high school without ever truly defining what they are. Maybe this helps a bit: One usually starts with sequences, not functions: Let $$(x_n)_{n\in\mathbb{N}}\subset\mathbb{R}$$ be real sequence, i.e. $$x_1,x_2,x_3,...\in\mathbb{R}$$.

Then, one says that this sequence has $$x'\in\mathbb{R}$$ as its limit, and writes $$\lim\limits_{n\rightarrow\infty} x_n = x',$$ if the quantity $$|x'-x_n|$$ becomes arbitrarily small for big $$n\in\mathbb{N}$$, which is usually defined as follows: $$\lim\limits_{n\rightarrow\infty} x_n = x'\Longleftrightarrow\text{ for all } \varepsilon>0 \text{ there exists a N\in\mathbb{N} with } |x'-x_n|<\varepsilon \text{ for all n\geq N.}$$

Now, what about functions $$f:\mathbb{R}\rightarrow\mathbb{R}$$? Take any $$x'\in\mathbb{R}$$, then there are a lot of ways to reach this point as the limit of a sequence. E.g. $$x'=0$$ is the limit of the sequences $$(\frac{1}{n})_{n\in\mathbb{N}}$$, $$\left(\frac{(-1)^n}{n^2}\right)_{n\in\mathbb{N}}$$, etc.

If there exists a $$y\in\mathbb{R}$$, so that $$\lim\limits_{n\rightarrow\infty} f(x_n)=y$$ for all possible sequences $$(x_n)_{n\in\mathbb{N}}$$ with $$\lim\limits_{n\rightarrow\infty} x_n = x'$$, then one writes $$\lim\limits_{x\rightarrow x'} f(x) = y.$$

Concretely, the expression $$\lim\limits_{x\rightarrow 2} f(x) = -1$$ from your original post means that the limit of $$(f(x_n))_{n\in\mathbb{N}}$$ is $$-1$$ for every possible sequence $$(x_n)_{n\in\mathbb{N}}$$ that converges towards $$2$$.

In theory, this would also allow the sequence $$x_n = 2$$ for all $$n\in\mathbb{N}$$, which would yield a division by zero in your function. That's why a priori (before computing the limit), your function is only defined on $$I=\mathbb{R}\setminus\{2\}$$. Thus, every possible sequence in $$I$$ that converges towards $$2$$ never takes $$2$$ as its value, and so you can divide by $$(x-2)$$ without running into problems.

• I do not know which high school you mean. The $\epsilon$-$\delta$ formalism is standard and I am not aware of a high school not dealing with it. Aug 23, 2022 at 7:35