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Limits & Continuity

  1. Dec 22, 2004 #1
    I'm having trouble with limits that involve 0+ and 0-. Can someone show me how the answers to the following limits are obtained?

    [tex]f(x) = \frac{1}{1+e^{\frac{1}{x}}}[/tex]

    [tex]\lim_{x\rightarrow0^{+}} = 0[/tex]

    [tex]\lim_{x\rightarrow0^{-}} = 1[/tex]

    Now, my second query involves continuity. I understand that:

    [tex]f(x) \in C \Leftrightarrow \lim_{x \rightarrow a} f(x) = f(a)[/tex]

    Say we have:

    [tex]f(x) = \frac{\sin x}{x}[/tex]

    Is f(x) continuous at x=0? My book says it is if f(0) is defined as 1. What am I missing?

    Finally, what exactly is Cn?
     
  2. jcsd
  3. Dec 22, 2004 #2

    matt grime

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    as x tends to zero and is positive, 1/x tends to infinity so e^{1/x} tends to infinity, hence f tends to zero as x tends to 0 from above

    as x tends to zero from below, 1/x goes it minus infinity, and e^[1/x} goes to zero hence f tends to 1 as x goes to 0 from below.

    f is continuous at 0 with the assignment of f(0)=1 this can be proven by, say, l'hopital's rule.

    C^n is the space of functions that are differentiable n times and where the n'th derivative is continuous. C is the continuous fucntions.

    eg as functions on R |x| is C but not C^1, x|x| is C^1 but not C^2
     
  4. Dec 22, 2004 #3
    Thanks for the reply matt. I just have two questions:

    e^(1/x) goes to zero if 1/x goes to minus infinity?

    How is f(0)=1?
     
  5. Dec 22, 2004 #4

    matt grime

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    1+e^{1/x} tends to 1, doesn't it? so 1/(1+e^{1/x}) tends to 1 as wel. all as x tends to 0 from below
     
  6. Dec 22, 2004 #5
    Well yes, but I'm not understanding how e^(1/x) tends to zero if (1/x) tends to minus infinity. Shouldn't it also tend to minus infinity?
     
  7. Dec 22, 2004 #6

    matt grime

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    you do know what the graph of e^x looks like?

    e^-x = 1/e^x, so if e^x goes to INFINITY (CORRECTED TYPO) as x goes to infinty then e^x must tend to zero as x goes to minus infinity.
     
    Last edited: Dec 22, 2004
  8. Dec 22, 2004 #7
    But you said in your first post:
    "1/x tends to infinity so e^{1/x} tends to infinity"
     
  9. Dec 22, 2004 #8

    matt grime

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    and that is correct whilst x is always positive. so?


    are you refering to my typo that i'll correct
     
  10. Dec 22, 2004 #9
    If you look at the graph, when x tends to -oo then e^x tends to zero

    set x=1/u and you have

    (1/u)->-oo => e^(1/u)->0
     
  11. Dec 22, 2004 #10
    Oh! So if x tends to minus infinity, e^x is actually e^(-x) where x tends to infinity and thus e^x tends to 1/infinity=0? I think I got it now. Thanks matt. :smile:

    I know I'm being a bother but... My second question was how can f(0)=1 (where f(x)=sinx/x)?
     
  12. Dec 22, 2004 #11

    matt grime

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    like i said, l'hopital's rule, though this is sort of a cheat.

    there is a geometric proof somewhere, but i don't konw where to find a copy of it.
     
  13. Dec 22, 2004 #12
    So f(0)=1 is actually lim[f(x)] as x tends to 0?
     
  14. Dec 22, 2004 #13
    the limit of f(x) is 1 when (x->0+) or (x->0-)

    By the way f(x) cant have a value when x=0

    But if we define - as you said - that f(0)=1,
    then we made f(x) continuous
    (left limit = right limit = f(0))
     
  15. Dec 22, 2004 #14
    So f(0)=1 is just a definition? But does that mean that it's not necessarily true?
     
  16. Dec 22, 2004 #15
    I'll recall another section

    what is a power of a^0 ?

    By definition, a^n=a*a*...a (n factors)
    we can't find out what a^0 means

    but a^0=1 works (if, for example, we think of (a^7)/(a^7)=1)

    So we DEFINE a^0=1

    The same as 0! (factorial) - we define it although there is not a factorial

    If you read at your book that we define f(0)=1, that is,
    we do that just because it works!

    Actually there is not f(0) (sin0/0)
    It's just a definition , but this still works
     
  17. Dec 22, 2004 #16

    matt grime

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    if we chose any other number (and we must choose some number to make it a function from R to R) then it wouldn't be continuous there, so it is a reasonalbe choice - remember you must define the function in some way for all points in the domain or it isn't a function.
     
  18. Dec 22, 2004 #17
    Okay. So f(0)=1 is simply a defined value.

    Thanks for your help guys.
     
  19. Dec 23, 2004 #18

    matt grime

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    yes, it's called a removably singularity: although the function (sinx)/x isn't "techincally" defined at 0 it is clear how he ought to define it
     
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