The limit Sin[ Sin[x] ] / x as x->0

  • Thread starter suffian
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  • #1
suffian
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)?
 

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  • #2
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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...
 
  • #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:
 
  • #4
suffian
thanks! i was getting used to multiplying by constants, but that's pretty ingenious.
 
  • #5
191
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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:
  • #6
191
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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...
 
  • #7
petitbear
Cute.:smile:
 
  • #8
24
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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 :)
 
  • #9
24
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do what i said above twice, forgot you asked sin[sin[x]]
 
  • #10
191
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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...
 
  • #11
24
0
when you take the limit it is correct
but don't want to prove that mathematically :)
 

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