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

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  1. Jan 30, 2015 #1
    As I read in the James Stewart's Calculus 7th edition, he said:
    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!!!
     
    Last edited: Jan 30, 2015
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  3. Jan 30, 2015 #2

    Orodruin

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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?
     
  4. Jan 30, 2015 #3

    ShayanJ

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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.
     
  5. Jan 30, 2015 #4
    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.
     
  6. Jan 30, 2015 #5

    Mark44

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    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.
     
  7. Jan 30, 2015 #6

    Orodruin

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    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.
     
  8. Jan 30, 2015 #7

    Mark44

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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.
     
  9. Jan 30, 2015 #8
    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:)
     
  10. Jan 30, 2015 #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.
     
  11. Jan 30, 2015 #10
    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].
     
    Last edited: Jan 30, 2015
  12. Jan 30, 2015 #11

    Orodruin

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    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.
     
  13. Jan 30, 2015 #12
    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].
     
  14. Jan 30, 2015 #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##.
     
  15. Jan 30, 2015 #14
    Wow~ Thank you so much:)!!!!!
     
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