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Homework Help: The limit Sin[ Sin[x] ] / x as x->0

  1. Mar 19, 2003 #1
    I encountered this problem in a set of limit problems:

    Limit[ Sin[ Sin[x] ] / x , x-> 0 ]

    According to what my book says, if the interior function in the sine approaches zero and the denominator also approaches zero, then the limit is 1; which, as I verified, is the answer. But is there a way to solve this limit by analytic means by using the simple limit rules combined with the basic trig limits (i.e. not L'Hopital's Rule)?
     
  2. jcsd
  3. Mar 19, 2003 #2
    lim sin(sinx)/x=(lim sin(sinx)/sinx)*(lim sinx/x) (1)
    lim sin(sinx)/sinx=lim sint/t, where t=sinx; x->0;t->0;
    lim sint/t=1;lim sinx/x=1;
    From (1) we obtain that the value of the limit is 1;
    No L'Hospital...
     
  4. Mar 19, 2003 #3

    Tom Mattson

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    Very clever, bogdan!

    Of course, you wouldn't know that lim sin(x)/x=1 without implicitly using L'Hopital, but you don't lose any points for that. :wink:
     
  5. Mar 19, 2003 #4
    thanks! i was getting used to multiplying by constants, but that's pretty ingenious.
     
  6. Mar 20, 2003 #5
    Look, no L'Hospital :
    lim (sin(0+x)-sin0)/x=cos 0;x->0;(the derivitive..."derivate" in 0...bad english...);
    lim (sin(0+x)-sin0)/x=lim sinx/x=cos 0=1;


    :smile:

    unfortunately calculating the "derivate" of sin we have to use
    lim sinx/x=1...but...using the fact that sin is derivable on R (it's "derivate" it's finite everywere...and it exists) and the fact that sin(-x)=-sinx (it is "antisimmetric") then we obtrain that the left derivate of sin in 0 is equal to the right derivate...and because the function is "antisimetric" the derivate in 0 must be 1...isn't it right ?
     
    Last edited: Mar 20, 2003
  7. Mar 20, 2003 #6
    or...
    in a vecinity of 0 we have :
    sinx<x<tgx;(using some drawing...triangle areas...)
    1<x/sinx<1/cosx;
    jumping to limit we obtain:
    lim 1<=lim x/sinx<=lim 1/cosx;x->0
    1<=lim x/sinx<=1;=>lim x/sinx=1, from where lim sinx/x=1...
    Again no L'Hospital...
     
  8. Mar 21, 2003 #7
  9. Mar 23, 2003 #8
    for very small values of x sin[x] becomes almost x
    so lim x->o sin[x]/x becomes x/x = 1

    find this an easier way :)
     
  10. Mar 23, 2003 #9
    do what i said above twice, forgot you asked sin[sin[x]]
     
  11. Mar 24, 2003 #10
    almost ?
    okay...but that has to be proven mathematically...precisely...
    you can say that sinx=sqrt(1-cosx) for small values...but it's not correct...
     
  12. Mar 24, 2003 #11
    when you take the limit it is correct
    but don't want to prove that mathematically :)
     
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