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Thread: Limits?

  1. #1

    Default Limits?

    Ok, I took a calc test the other day, and there was one question that has been driving me mad. If anyone knows the answer to this, then please tell me!! I will give you a cookie if you do! (Although I might not be able to get you the cookie unless I meet you in person... O.o)

    Here it is, in all its annoying glory:

    Limit Sin ( 1/x ) as x goes to 0.

    I talked with my friends, and they think the answer is 0, but I think the limit doesn't exist. This is driving me crazy, I already took the class in highschool, so I shouldn't be getting simple stuff, like limits, wrong. >.< Help me...

  2. #2

  3. #3

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    I took calc 8 years ago, so I may not be perfect. I believe it's essentially the same as sin(y) where y approaches infinite. I don't believe you'll find convergence.

  4. #4

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    Quote Originally Posted by NightFox View Post
    I took calc 8 years ago, so I may not be perfect. I believe it's essentially the same as sin(y) where y approaches infinite. I don't believe you'll find convergence.
    Exactly! That's what I was trying to tell them! I knew I wasn't crazy...

    My one friend is like, a math genious, he is almost never wrong about anything math related, won't it be funny if I can prove him wrong? XD

  5. #5

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    Technically, that limit does not exist (you cannot approach infinite); in practice, setting that limit expression equal to infinity is understood and accepted.

  6. #6

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    Quote Originally Posted by Tyki View Post
    Exactly! That's what I was trying to tell them! I knew I wasn't crazy...
    Don't think of infinite. Think of actual numbers. What's happening to 1/x as X gets smaller and smaller (closer and closer to zero)? Take 1/1 as a starting point. 1/0.001 is bigger. 1/0.00000001 is even bigger. 1/0.0000000000000000000000000000000001 is H-U-G-E. The closer x gets to zero, the closer 1/x gets to infinite (I'm parsing what infinite is, but you get the point).

    As I remember, the point of a limit is that it subs for an equals sign when you can't have one- if you're looking for end behavior approaching infinite or negative infinite, or infinitesimality, or an asymtope.

    ---------- Post added at 02:06 ---------- Previous post was at 02:04 ----------



    Quote Originally Posted by Traemo View Post
    Technically, that limit does not exist (you cannot approach infinite); in practice, setting that limit expression equal to infinity is understood and accepted.
    Is "limits approaching infinite" more accurately sequences and series? That's what I seem to remember. The notation is extremely similar but the application isn't.

  7. #7

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    The way I tried to explain it to them was like this:

    The limit of the sin (x) as x goes to infinity doesn't exist. (we agree on this)

    And the limit of 1/x as x goes to 0 also, does not exist (positive infinity). (we agree on this.)

    Then I explained that the limit of the sin ( 1/x ) as x goes to 0 can't exist, because we stated that as x goes to 0, 1/x goes to positive infinity, and from the first statement, we already know that the sin (positive infinity) doesn't exist.

    That, and I also tried explaining it to them by saying that while the x-value of the function is approaching 0, the y-value is not necessarily doing the same, the range of the function is still -1 to 1, no matter how close to 0 x gets. Therefore, since the function continues to oscillate around the x-axis, and is not approaching any specific y-value, there is no limit (does not exist).

    They didn't really have any counterexamples... they just said that 0 works and makes sense... but it doesn't :P

    I like how much more sophisticated my language sounds when I talk about math... :3

  8. #8

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    Quote Originally Posted by Tyki View Post
    That, and I also tried explaining it to them by saying that while the x-value of the function is approaching 0, the y-value is not necessarily doing the same, the range of the function is still -1 to 1, no matter how close to 0 x gets. Therefore, since the function continues to oscillate around the x-axis, and is not approaching any specific y-value, there is no limit (does not exist).
    I was always bad with explaining the theory. I like making tables and drawing graphs. Make a table with X and Y values for y = sin(1/x). Use x values of 1, 0.1, 0.0001, 0.000001, 0.00000001, etc. Then fill in the Y side. Show essentially, if we rename 1/x as Z, that Z is just getting larger and larger. Sin(Z), where Z is a set of all positive numbers, is just an oscillating line. Challenge him to show you where the convergence is.

  9. #9

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    I'm taking Calc 1 for the second time (my major requires I take their version...)
    Wolfram Alpha is really helpful for figuring problems out.

    Check it out.

    I have been getting a few proofs like this here and there.
    For those, you have to use the formula sin(theta)=1 or something. Then do some plugging in. I forget how to do it, but it's usually this equation.

  10. #10

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    Quote Originally Posted by Tyki View Post
    They didn't really have any counterexamples... they just said that 0 works and makes sense... but it doesn't :P
    You explained well that the limit doesn't exist and they seem to agree with you. Maybe they have difficulty grasping the concept that some maths formula doesn't have a defined value?

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