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Sine, Cosine, Limit summary

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  1. Aug 13, 2015 #1
    Dear PF Forum,
    In previous threads, I have asked about sine and cosine. The answer given by the members/mentors/advisor are very clear. But lengthy. Perhaps these yes/no questions that I can simply remember and not forget it (again).
    So here we are
    1. if h = 0 then sin(h) = 0
    2. if ##\lim_{h \to 0}## then sin(h) 0
    3. If ##\lim_{h \to 0}## then sin(h) has no limit

    4. if h = 0 then sin(h)/h is undefined
    5. if ##\lim_{h \to 0}## then sin(h)/h = 1
    6. If ##\lim_{h \to 0}## then sin(h)/h has limit

    7. if h = 0 then Cin(h) = 1
    8. If ##\lim_{h \to 0}## then cos(h) 1
    9. If ##\lim_{h \to 0}## then cos(h) has limit
    ---------------------------------------------------------------------
    10. if ##\lim_{h \to 0}## does cos(h) - 1 has limit? If yes then what is the value?
    11. if ##\lim_{h \to 0}## does (cos(h) - 1)/h has limit? If yes then what is the value?

    And in the derivative of Sin(x)
    ##\lim_{h \to 0} \sin(x)\frac{\cos(h)-1}{h} + \cos(x)\frac{\sin(h)}{h}##
    ##\lim_{h \to 0} \sin(x)\frac{\cos(h)-1}{h} + \cos(x)##

    So if the derivative of Sin(x) is Cos(x) then...
    ##\lim_{h \to 0} \sin(x)\frac{\cos(h)-1}{h}## should be zero. then...
    12. ##\lim_{h \to 0} \cos(h)-1## should be zero, then...
    13. ##\lim_{h \to 0} \cos(h) ## should be 1, then...
    8. If ##\lim_{h \to 0}## then cos(h) 1
    It seems number 13 contradicts number 8.
    I would be very grateful if someone be sokind to answer me. But before that, could you confirm the yes/no question in number 1 to 9, please.
    Thank you very much.
     
  2. jcsd
  3. Aug 13, 2015 #2

    RUber

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    Number 2 and 3 don't make sense.
    ##\lim_{h \to 0} \sin(h) = 0 ## but if h is approaching 0 then sin(h) 0.
    Similarly, I am not too sure of your notation.
    ##\lim_{h \to 0}sin(h)/h =1## is true.
    No clue what you mean in number 6. Is this supposed to be a counterpoint to number 3 above?
    For 7, do you mean cos(h)? If so, yes.
    Again, for 8 and 9, ##\lim_{h \to 0} \cos(h) = 1## but as h approaches 0, cos(h) approaches but is not equal to 1.
    ---------------------------------------------------------------------
    ##\lim_{h \to 0} (\cos(h) - 1) = \lim_{h \to 0} (\cos(h) )- 1=1 -1 ##
    11 is a good question. Do you know L'Hopital's rule? I would start there.
    That checks out.
    Right. number 8 is wrong.
     
  4. Aug 13, 2015 #3

    DEvens

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    A very long series of "yes-no" questions is not a very efficient way to learn this stuff.

    Why don't you try to learn the concepts and the notations? Memorizing a bunch of "if lim h" kinds of things is really going to do nothing but confuse you. Especially when you have them stated quite clumsily.
     
  5. Aug 13, 2015 #4

    PeroK

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    And, what you are memorising makes no sense. Not least, the derivative of sin is cos, there is no "if" about it. Perhaps more important, your post reveals you have not understood the concept of a limit. You are treating limit h -> 0 as an independent property of h, which is not correct.

    This is not the way to do or learn rigorous mathematics, I'm sorry to say.
     
  6. Aug 13, 2015 #5

    Drakkith

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    Staff: Mentor

    Indeed. The best way to learn about limits is probably to grab a calculus textbook and, starting from the front, work your way through the book, answering every problem as you go.
     
  7. Aug 14, 2015 #6
    Concerning number two and three.
    This is because someone told me that the limit musn't always exist.
    So I am trying to understand this by testing some equations. I choose:
    ##\lim_{h \to 0}## then ##\sin(h)## 0 and
    ##\lim_{h \to 0}## then ##\sin(h)## has no limit.
    ##\lim_{h \to 0}## then ##\cos(h)## has limit.
    Just to make sure if I understand left/right hand correctly.

    What?? That must be the effect of copy-paste.
     
  8. Aug 14, 2015 #7
    So what is the different between
    A: if ##\lim_{h \to 0}## then ##\sin(h) = 0## and
    B: if h = 0 then ##\sin(h) = 0##?
    Isn't the left/right hand of ##\lim_{h \to 0} \sin(h)## is different, but
    The left/right hand of ##\lim_{h \to 0} \frac{6h}{h}## is equal, right.
    Or
    The left/right hand of ##\lim_{h \to 0} \frac{\sin(h)}{h}## is also equal.
    but ##\lim_{h \to 0} \sin(h)## the left/right is different.
     
  9. Aug 14, 2015 #8
    Thanks DEvens for your respons. It's not the equations that I intend to remember. It's the concept of limit that I try to understand. Here, I choose sine and cosine to study limit.
    For example:
    If h = 0 then sin(h) = 0
    if ##\lim_{h \to 0}## is sin(h) = 0? I think sin(h) ≠ 0. But I need confirmation.

    if h = 0 then sin(h)/h is undefined
    if ##\lim_{h \to 0}## then sin(h)/h = 1. This what we were taught in high school. Now, I just want to know 'why'
     
  10. Aug 14, 2015 #9
    Yes, today I learned that I don't know the concept of limit. I tought I knew all the way from highschool. Thanks for pointing it out to me, so I know what to learn.

    Perhaps the word "IF" is the wrong word for mathematic.
    I should have written.
    Because the derivative of Sin(x) is Cos(x) so the proof is:
    ##\lim_{h \to 0} \sin(x)\frac{\cos(h)-1}{h}## should be zero. then, etc...

    But, I'm a computer programmer, I use if ... then ... instinctively.
     
  11. Aug 14, 2015 #10

    PeroK

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    What you have written still makes no mathematical sense. For example, as sin and cos are continuous functions we have:

    ##\forall x## ##\lim_{h \rightarrow x}sin(h) = sin(x)## and ##\lim_{h \rightarrow x}cos(h) = cos(x)##

    There is no issue or mystery about these limits. The more interesting ones are:

    ##\lim_{h \rightarrow 0} \frac{sin(h)}{h} = 1## and ##\lim_{h \rightarrow 0} \frac{cos(h)-1}{h} = 0##

    Which can be proved in any number of ways. For example, the second of these is actually the derivative of cos at 0:

    ##\lim_{h \rightarrow 0} \frac{cos(h)-1}{h} = \lim_{h \rightarrow 0} \frac{cos(h)-cos(0)}{h-0} = cos'(0) = -sin(0) = 0##

    That is mathematics.

    (And being a computer programmer is no excuse!)
     
  12. Aug 14, 2015 #11

    RUber

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    As PeroK wrote, the limit of a continuous function is always defined and equal to the value of the function at the limit point.
    The limit should be written ##\lim_{h\to 0} \sin h ## no if, no then. And it is said, the limit as h goes to zero of sine of h.
    The left/right limits have to be evaluated for discontinuous functions like 1/x.

    Can you plot the sin and coz functions? Visually, the derivatives and limits are rather easy to see.
     
  13. Aug 14, 2015 #12

    RUber

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    Anything divided by zero is undefined--always. However, if the function is continuous on either side of the singularity (division by zero), then you can still do most things you would like to do with that function by just defining a value for the function at zero.
    Example ##f(x) = sin x / x## is undefined at zero, but
    ##g(x)=\left\{ \begin{array}{l l} 1& if\, x=0\\ (\sin x)/x&if\, x\neq 0 \end{array}\right.##
    Is equal to f(x) almost everywhere and is also continuous for all x.
     
  14. Aug 14, 2015 #13

    jbriggs444

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    I would like to draw attention to what PeroK pointed out in response #4 and which does not seem to have sunk in.

    The notation ##\lim_{h \to 0} \sin(h) = 0## is not a statement about h. It is not a statement about the value of sin(h) for any particular h. It is not a statement about the value of sin(0). Instead, it is a statement about the values that the ##\sin## function takes on in the neighborhood of zero. It is (roughly) a statement that as the neighborhoods around 0 get very very small, the values of the sin function within such neighborhoods are all very very close to zero.

    Tearing the notation in two and treating it as if it were a statement about the value of the sin function at a single value is not correct.
     
  15. Aug 14, 2015 #14
    Thanks a lot. I'll remember all those equations thet you wrote. It's not memorizing. It's the 'misleading' that I'm afraid of.. Supposed, if I consider h = 0 and 6h/h = 6., then, I'll be mislead forever.
    Of course being a computer programmer is no excuse, everybody has to grasp and grab basic math. What excuse is that I wrote
    "if the derivative of sin(x) is cos(x) then.." While what I meant is "Because the derivative of sin(x) is cos(x) then the proof is ..."
    "IF" implies that the derivative of sin(x) is might/might not be cos(x).
    Believe me, it was typo. We were taught since high school that sin(x) is cos(x), but here in PF Forum for the first time the mentors/advisors enlight me the proof!. I must have missed class and fooled around that day :smile:
     
  16. Aug 14, 2015 #15
    Thanks a lot.
    Plot? What do you mean by plot? Drawing it? Well, it will be difficult with the tools in my computer.
    But sin and cos curve are like wave, right?
    Sine, with peak at pi/2 and trough at 0, Pi, 2Pi, 3Pi, etc...
    And if we shift the cosine curve pi/2 to the left/right, it will match sine curve. I hope that is right.
     
  17. Aug 14, 2015 #16
    Thanks. I'll remember the concept.
     
  18. Aug 14, 2015 #17
    Ok, I'll contemplate your answer.
     
  19. Aug 14, 2015 #18

    RUber

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    What tools do you have on your computer? sin and cos are very common functions that most calculators can graph for you. There are also many online options, wolframalpha.com is useful.

    Cosine shifted to the right, not left, by pi/2 will match sine.
     
  20. Aug 14, 2015 #19
    Tools? Only Microsoft Paint and Microsoft Excel. Of course I could make a small computer program to PutPixel(X,ScreenHeight-Sin(X/100*Pi)*100), but.. that will be tedious.
    "Cosine shifted to the left not match sine"? I'll try to imagine that. But I want to answer your post first.
    Correction:PutPixel(X,Round(ScreenHeight-Sin(X/100*Pi)*100))
     
    Last edited: Aug 14, 2015
  21. Aug 14, 2015 #20
    That's right cosine shift to the left by 1.5Pi.
    Peat at 1 and trough at -1, right. I imagined Peat at 1 and trough only at zero. Okay the period of the graph for cosine to sine is 0.5pi to the right.
    [Add: Why didn't you correct my post? :smile:
    ]
     
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