Is the limit of functions necessarily equal to "itself"?

  • #1
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As I read in the James Stewart's Calculus 7th edition, he said:
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An alternative notation for

[itex]\displaystyle\lim_{x\rightarrow a}f(x) = L[/itex]

is

[itex]f(x) \rightarrow L[/itex] as [itex]x \rightarrow a[/itex]

which is usually read "f(x) approaches L as x approaches a"
My question is: Is [itex]f(x)\rightarrow 0[/itex] the same as [itex]f(x) = L[/itex]?

For example,

[itex]f(x) = x^2[/itex]

[itex]\displaystyle\lim_{x\rightarrow 5}f(x) = 25[/itex]

I can say that [itex]f(x) = x^2[/itex] approaches 25 as [itex]x[/itex] approaches 5.

Therefore, can I say that the f(x) is not equal to 25 unless x is not approaching 5, but x, itself, is 5?

Thanks for reading!!!
 
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Answers and Replies

  • #2
Orodruin
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My question is: Is "f(x) approaches L" the same as "f(x) equals L"?
No. You may have a function which has a discontinuity at x=a. Consider the function f(0) = 1 and f(x) = 0 if x is different from 0. It has 0 as both left and right limits and yet the function value at x=0 is one.
so, can I say that the f(x) is not equal to 25 unless x is not approaching 5, and x, itself, is 5?
This is a different question and formulated with so many negations that I am not completely sure what question you actually intend to ask. Can you reformulate it?
 
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  • #3
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Here you're asking about the notion of continuity of functions. A continuous function, by definition, has the property that its limit at a point is equal to its value at that point. But there are functions that are not continuous and so don't have this property. See the section Non-examples in the given link.
 
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  • #4
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No. You may have a function which has a discontinuity at x=a. Consider the function f(0) = 1 and f(x) = 0 if x is different from 0. It has 0 as both left and right limits and yet the function value at x=0 is one.

This is a different question and formulated with so many negations that I am not completely sure what question you actually intend to ask. Can you reformulate it?
Sorry about that. It's a little difficult for me to express this abstract question in English.

The proposition is : [itex]x + 2 \neq 7[/itex] unless [itex] x = 5[/itex]
Is that proposition correct? I guess someone's answer is no because when x approaches 5, [itex]x + 2[/itex] would also be 7.
 
  • #5
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Sorry about that. It's a little difficult for me to express this abstract question in English.

The proposition is : [itex]x + 2 \neq 7[/itex] unless [itex] x = 5[/itex]
Is that proposition correct? I guess someone's answer is no because when x approaches 5, [itex]x + 2[/itex] would also be 7.
When x approaches 5, x + 2 approaches 7, but won't necessarily be equal to 7. When we talk about x approaching 5, what we're saying is that x is in some neighborhood around 5 (i.e., |x - 5| is "small"), so x + 2 will be in some neighborhood around 7 (i.e., |x + 2 - 7| is also "small"). x + 2 won't be equal to 7 if x ≠ 5.
 
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  • #6
Orodruin
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Sorry about that. It's a little difficult for me to express this abstract question in English.

The proposition is : [itex]x + 2 \neq 7[/itex] unless [itex] x = 5[/itex]
Is that proposition correct? I guess someone's answer is no because when x approaches 5, [itex]x + 2[/itex] would also be 7.
No, for the function x+2, the statement is true. Even if it approaches 7 as x approaches 5, it is not 7 except for when x=5.

However, this is not always the case. Take the function f(x)=0 for all x. The limit when x goes to 0 is 0 and also f(0)=0. There are also examples where you will find that f takes the same values in very different points. That f(x) = a for only one x is a property called injective (also formulated as f(x) = f(y) implies x = y) and a linear function with non-zero slope has this property.
 
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  • #7
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A better example would be the function f given by ##f(x) = \frac{x^2 + 3x + 2}{x + 1}##. We can talk about this limit:
##\lim_{x \to -1} f(x)##, which turns out to be 1, even though f is not defined at x = -1.
 
  • #8
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Mark44, thank you very much! You help me a lot. Now here's my next relevant question.
Actually, my question has something to do the following physics question:

In the uniform circular motion, is the instantaneous centripetal acceleration perpendicular to instantaneous velocity?

[itex]\vec{a}_{c}=\displaystyle \lim_{\Delta t\rightarrow 0}\frac{\Delta \vec{v}}{\Delta t}[/itex]

If we differentiate the velocity and solve the inner product [itex]\vec{a}\cdot \vec{v}[/itex], then we'll get zero.
However, if we imagine the [itex]\Delta \vec{v}[/itex] then we would think [itex]\vec{v}[/itex] will never be perpendicular to [itex]\Delta \vec{v}[/itex].
Someone would say we have to reconcile these two facts.

And I think the reason why they think those two argument are contradictory is that
they mistaken the direction of [itex]\vec{a}_{c}=\displaystyle \lim_{\Delta t\rightarrow 0}\frac{\Delta \vec{v}}{\Delta t}[/itex] as the direction of [itex]\Delta \vec{v}[/itex].

Just like what you told me, the direction of [itex]\vec{a}_{c}[/itex] is what [itex]\frac{\Delta \vec{v}}{\Delta t}[/itex] approaches as [itex]\Delta t[/itex] approaches [itex]0[/itex].
Therefore, the direction of [itex]\vec{a}_{c}[/itex] is not the same as the direction of [itex]\frac{\Delta \vec{v}}{\Delta t}[/itex].
Is that correct?

Thanks so much:)
 
  • #9
Orodruin
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Well, the direction is the limit of the direction of ##\Delta \vec v## as ##\Delta t \to 0##. This is just by definition of the derivative.
 
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  • #10
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Yes, Orodruin. So, there's no contradiction, right?
That means, the direction of [itex]\Delta \vec{v}[/itex] is not the direction of [itex]a_{c}[/itex].

And the we should say the direction of [itex]a_{c}[/itex] is what [itex]\Delta \vec{v}[/itex] approaches as [itex]\Delta t[/itex] approaches zero, rather than the direction of [itex]\Delta \vec{v}[/itex].
 
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  • #11
Orodruin
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Yes, Orodruin. So, there's no contradiction, right?
That means, the direction of [itex]\Delta \vec{v}[/itex] is not the direction of [itex]a_{c}[/itex].
As long as ##\Delta t## remains finite, the direction of ##\Delta \vec v## is not orthogonal to ##\vec v##. Acceleration is a derivative of the velocity and thus defined in the limit.
 
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  • #12
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Sorry, Orodruin. I want to know more about "thus defined in the limit". Does that mean what I said?

The direction of [itex]a_{c}[/itex] is defined as the direction of what [itex]\Delta \vec{v}[/itex] approaches as [itex]\Delta t[/itex] approaches [itex]0[/itex], rather than the direction of [itex]\Delta \vec{v}[/itex].
 
  • #13
Orodruin
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Yes. Or rather if you just talk about direction, the direction of ##\vec a## is the what the direction of ##\Delta \vec v## approaches as ##\Delta t \to 0##.
 
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  • #14
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Wow~ Thank you so much:)!!!!!
 

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