Basic Limit Results

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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.
 

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  • #2
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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.
 
  • #3
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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##.
 
  • #4
opus
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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?
 
  • #5
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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.
 
  • #6
opus
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Sometimes these get confusing because it takes me a long time to type out my responses and it throws the whole conversation of balance :DD
I think I got it now. Thank you guys.
 

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