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Prove a limit

  1. Oct 25, 2006 #1
    prove that limit of (sin (x^0)/x) as x tends to zero is π/180(ie pi by 180)
     
  2. jcsd
  3. Oct 25, 2006 #2

    arildno

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    DO pay attention to what you are writing! Or are you comatose, perhaps??

    What you've written is utterly meaningless.

    If you want to say to a friend: "I really like you", do you say to him "bungafloop-floop"??

    That's basically what you've written above in the mathematical language.
    Is it really that difficult for you to form a proper sequence of mathematical symbols?


    I assume you are talking about the sine function, where the argument is given in degrees, rather than radians, but that does not excuse your cavalier attitude with respect to notation.
     
  4. Oct 25, 2006 #3
    by sin x^0/x ,i mean sine x to the power 0 divided by x .i have been using this notation for visual basic.sorry!
     
    Last edited: Oct 25, 2006
  5. Oct 26, 2006 #4

    HallsofIvy

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    Now it's even more confusing than before!
    I wasn't aware that visual basic allowed such sloppy, anti-mathematical notation. "sine x to the power 0" is (sin x)^0, not sin x^0 which means "sin of (x^0)= sin(1)". In any case, (sin x)^0= 1 so you are really asking about 1/x as x goes to 0. Of course, that does not converge at all, certainly not to [itex]\pi/180[/itex]

    If you meant, by x^0, "x written in degrees", as arildno suggested, then, since the value "in radians" would be [itex]\pi x/180[/itex] your sin(x)/x becomes [itex]sin(\pi x/180)/x[/itex] You could multiply both numerator and denominator of that by [itex]\pi /180[/itex] and let [itex]y= \pi x/ 180[/itex] to get
    [tex]\frac{\pi}{180}\frac{sin y}{y}[/tex]
    which does have the limit you ask for.
     
  6. Oct 26, 2006 #5

    arildno

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    Remember that the degree-circle is NOT a power!
    The degree-circle works in a similar manner as "m" for "meters":

    2m means "two meters".
     
  7. Oct 26, 2006 #6
    I agree with arild.

    Now, there are several ways to "prove" limits, but in this case it is simply easier to use the definition of limits: "As the function of X approaches a singular point from the left and the right, the limit is the point that they are coming close to."

    Therefore, we can plot out the point and come to the point it is coming close to. In this case, here's our information:

    f(x) = sin(x)/x

    We're looking for:

    Lim(x->0)[sin(x)/x] = ??

    Well, create a chart proof:

    From right to left:
    x: y:
    .1 .998334
    .01 .999983
    .001 .999999
    0 undef
    -.001 .999999
    -.01 .999983
    -.1 .998334

    The function as X is approaching 0 is 1. Therefere by the definition of limits:

    Lim(x->0)[sin(x)/x] = 1

    Q.E.D.
     
  8. Oct 27, 2006 #7

    arildno

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    No, it isn't GoldPhoenix.
    Most probably, OP is given x as measured in DEGREES, rather than in radians.
    Check your calculator again.
     
  9. Oct 27, 2006 #8

    HallsofIvy

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    I would also point out that the fact that
    1 .998334
    .01 .999983
    .001 .999999
    0 undef
    -.001 .999999
    -.01 .999983
    -.1 .998334

    looks like it is getting close to 1 proves absolutely nothing about the limit. It would be very easy to make up functions that have exactly those values but are, say, -1000000 for x <b>close</b> to 0 ("close" here meaning "less than 0.00000000001 from 0".
     
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