# Basic Limit Results

• B
Gold Member

## Main Question or Discussion Point

In my text, it states the Basic Limit Results as follows:

For any real number $a$, and any constant $c$,

(i) $\lim_{x \rightarrow a}{x}=a$
(ii) $\lim_{x \rightarrow a}{c}=c$

Now from the previous chapter, Im used to seeing these as taking the limit of some function as the x values of that function approach some x value (a). This will give some y value if a limit exists.

Now for (i), is this saying that we are taking the limit of some x value as our x values close in on some other x value (a), and the limit is the x value that we're closing in on (a)? I don't know what to make of all the x values and it seems quite confusing.

fresh_42
Mentor
The answer is yes, although I'm not sure whether it should be seen like this. I find it easiest just to consider the function $x \longmapsto f(x)=y=x$ resp. $x \longmapsto f(x)=y=c\,.$ This way you stay in the terms of functions without any "new" interpretation needed. In the first case you're approaching a point on a straight line, namely $(x,y)=(a,a)$. In the second case you're sitting on the line $y=c$ and which ever point you approach, it will remain $(x,y)=(*,c)$, e.g. $(a,c)$. But you don't need this geometric view, you can as well operate with the usual definition $\lim_{x \to a}f(x) = f(a)$ and the functions I mentioned.

• opus
Math_QED
Homework Helper
2019 Award
The answer is yes, although I'm not sure whether it should be seen like this. I find it easiest just to consider the function $x \longmapsto f(x)=y=x$ resp. $x \longmapsto f(x)=y=c\,.$ This way you stay in the terms of functions without any "new" interpretation needed. In the first case you're approaching a point on a straight line, namely $(x,y)=(a,a)$. In the second case you're sitting on the line $y=c$ and which ever point you approach, it will remain $(x,y)=(*,c)$, e.g. $(a,c)$. But you don't need this geometric view, you can as well operate with the usual definition $\lim_{x \to a}f(x) = f(a)$ and the functions I mentioned.
For (i), define $f$ by $f(x) = x$ on the appropriate domain.

Then $\lim_{x \to a} x$ is shorthand for $\lim_{x \to a} f(x)$.

For (ii), similarly, $\lim_{x \to a} c$ is shorthand for $\lim_{x \to a} g(x)$ where $g$ is given by $g(x) = c$.

• opus
Gold Member
x⟼f(x)=y=xx \longmapsto f(x)=y=x resp. x⟼f(x)=y=c
Having a hard time deciphering this. Could you please explain?

In the first case you're approaching a point on a straight line, namely (x,y)=(a,a)(x,y)=(a,a).
Does it have to be a straight line? In drawing it out, although I'm not sure if it's accurate, it looks like this is describing the line y=x. Is this true?

In the second case you're sitting on the line y=cy=c and which ever point you approach, it will remain (x,y)=(∗,c)(x,y)=(*,c), e.g. (a,c)(a,c)
So just a horizontal line as to whatever x value we approach, it will approach the same y value?

fresh_42
Mentor
Having a hard time deciphering this. Could you please explain?
@Math_QED already did. I didn't know how you note a function:
• $x \longmapsto \ldots \text{ expression with x } \ldots$
• $y= \ldots \text{ expression with x } \ldots$
• $f(x) = \ldots \text{ expression with x } \ldots$
so I combined all of them: $x \longmapsto f(x) = y = \ldots \text{ expression with x } \ldots$
Just it have to be a straight line? In drawing it out, although I'm not sure if it's accurate, it looks like this is describing the line y=x. Is this true?
Yes. O.k. straight line is doubled, so either straight or line, but that is only a matter of language, either will do.
So just a horizontal line as to whatever x value we approach, it will approach the same y value?
Yes.

These are the graphs of the functions as stated in @Math_QED 's and my post. It is the geometry behind. If you don't want to rely on graphs, which is often a good idea esp. if the functions are weird, then you can work with the analytic definition of limits as well.

• opus
Gold Member
Sometimes these get confusing because it takes me a long time to type out my responses and it throws the whole conversation of balance I think I got it now. Thank you guys.