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Proof of lim x->0 x/sinx

  1. Sep 22, 2014 #1

    mrg

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    1. The problem statement, all variables and given/known data

    I'd like to see a proof that lim x->0 x/sinx = 1.

    2. Relevant equations

    It is given that lim x->0 sinx/x = 1

    3. The attempt at a solution

    I can confirm this limit quite easily by graphing - the curve obviously approaches one, and the left- and right-handed limits are equal. However, assuming that either I can't graph or don't have the time to plot points, (on a timed standardized test that doesn't allow graphing calculators, for example) how would I know this to be true?

    I thought about the fact that perhaps some property out there exists saying that if lim x->a f(x)/g(x) = L, then its reciprocal, lim x->a g(x)/f(x) = 1/L. However, I don't believe this to be true, since lim x->-2 (x^2+4x+4)/(x+2) = 0, and the reciprocal lim x->-2 (x+2)/(x^2+4x+4) = DNE

    Perhaps this is because the reciprocal of 0 is undefined, and because of this, this is one instance where my property fails. In which case, I can make a simple restriction which allows me to use this property for all other instances. Of course, this isn't very convincing, and leaves me quite skeptical.

    Also - I know of L'Hopital's rule, and could easily use that in this instance, but we just learned limits, so that's off the table for now.

    If you must use some special techniques to prove this, please do. But I believe if I can show that my property works, then I will have this proof completed. Furthermore, I could use that line of thinking for other limits I run into to save me some time.

    Thanks for your help in this matter!
     
  2. jcsd
  3. Sep 22, 2014 #2

    LCKurtz

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    What if you assume ##L\ne 0##? Then does the reciprocal give ##1/L##?
     
  4. Sep 23, 2014 #3

    pasmith

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    Do some geometry: Draw a unit circle. Let the centre be O. Draw a radius and let it intersect the circle at A. Draw another radius intersecting the circle at B. Draw a line throguh B perpendicular to OA. Let this intersect OA at a point C. Now let [itex]x[/itex] be the angle AOB. Since [itex]x[/itex] is in radians, the length of the arc of the circle from A to B is [itex]x[/itex]. The length of the line AC is [itex]\sin(x)[/itex]. Now imagine moving the point B along the circle closer and closer to A, whilst keeping CB perpendicular to OA. What happens to the difference between the lengths of the arc and the line?
     
  5. Sep 23, 2014 #4

    HallsofIvy

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    What definition of "sin(x)" are you using? How you prove something depends heavily on what definitions you are using.
     
  6. Sep 23, 2014 #5

    olivermsun

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    mrg,
    pasmith alludes to the geometric proof, which is very elegant. I guess HallsofIvy is asking whether you have a series definition of sin(x), in which case there is also an algebra proof which anticipates L'Hopital's Rule (but which does not rely upon it).
     
  7. Sep 23, 2014 #6

    LCKurtz

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    And I am assuming the OP has seen the limit for ##\frac {\sin x} x## and the question is really about limits of reciprocals.
     
  8. Sep 23, 2014 #7

    olivermsun

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    Good point. If that's it, then I'm hoping the OPs notes already provide the reciprocal rule and the conditions for its use. ;)
     
  9. Sep 24, 2014 #8

    D H

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    With regard to the inverse of a limit:
    Limits aren't about the value of some function at some point because the function may have no value at that point in question. For example, f(x)=sin(x)/x is undefined at x=0. Limits instead address the behavior of the function in an ever-shrinking neighborhood of the point in question, but always excluding the point itself.

    Consider some function g(x) whose limit at some point x0 is L≠0 and whose value is strictly non-zero for some neighborhood of x0. The the limit of h(x) = 1/g(x) as x approaches x0 is 1/L. In other words, it's perfectly valid to use 1/L as the limit so long as L is non-zero and so long as you can find that neighborhood of x0 where g(x) is always non-zero.

    With regard to sin(x)/x:
    A simple way to show that this approaches one as x approaches zero is to use the geometric concepts pasmith alluded to earlier. You can show that 1<x/sin(x)<sec(x) for all x between -pi/2 and pi/2 (exclusive), except of course for x=0. Since sec(0)=1, the squeeze theorem says that x/sin(x) approaches one as x approaches zero. Now invert that and you get sin(x)/x approaches one. Alternatively, the geometric argument can also be used to show that cos(x)<sin(x)/x<1 in this same region, and once again the squeeze theorem says the limit is one (but now directly).
     
  10. Sep 24, 2014 #9

    Mark44

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    That's my guess as well.
     
  11. Sep 24, 2014 #10

    LCKurtz

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    And, in any case, you might notice the OP never returned to this thread to clarify it.
     
  12. Sep 24, 2014 #11
    I was just searching for scripts and looking your post to learn
    how you did ##1/L##?(1 over L) ... Can someone show me where they are? I can see only the two sub and süper scripts which are between plus symbol and drafts symbols.

    Thank you.
     
    Last edited by a moderator: Sep 24, 2014
  13. Sep 24, 2014 #12

    Mark44

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    I removed your erroneous quote in your previous post. Here's a link to a brief how-to on using LaTeX: https://www.physicsforums.com/threads/physics-forums-faq-and-howto.617567/#post-3977517.
     
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