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

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In summary, the given limit is equal to 1 by using basic limit rules and trigonometric limits. However, it can also be solved without using L'Hopital's Rule by using the property that sin is derivable on all of R and its derivative is equal to cos. This allows us to show that the left and right derivatives in 0 are equal, resulting in a limit of 1. Another approach is to use the fact that for small values of x, sin[x] is almost equal to x, resulting in a limit of 1 as well.
  • #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
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
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
thanks! i was getting used to multiplying by constants, but that's pretty ingenious.
 
  • #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 ?
 
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  • #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...
 
  • #7
Cute.:smile:
 
  • #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 :)
 
  • #9
do what i said above twice, forgot you asked sin[sin[x]]
 
  • #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...
 
  • #11
when you take the limit it is correct
but don't want to prove that mathematically :)
 

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

What is the limit of Sin[ Sin[x] ] / x as x approaches 0?

The limit of \(\frac{\sin(\sin(x))}{x}\) as \(x\) approaches 0 is 1. This limit can be evaluated using standard limit laws and the small-angle approximation for sine.

How can this limit be calculated?

This limit can be calculated by recognizing that as \(x\) approaches 0, \(\sin(x)\) also approaches 0. Therefore, the expression becomes similar to \(\frac{\sin(y)}{y}\) as \(y\) approaches 0, which is a well-known limit in calculus equal to 1.

Why is the limit not 0, considering sin(x) approaches 0?

While \(\sin(x)\) approaches 0, the rate at which \(\sin(\sin(x))\) approaches 0 is proportional to the rate at which \(x\) approaches 0. This proportionality results in the limit being 1 rather than 0.

Can this limit be solved using L'Hôpital's Rule?

L'Hôpital's Rule could be applied after reformulating the limit into a form that results in an indeterminate 0/0 as \(x\) approaches 0. However, in this case, understanding the behavior of the sine function at small angles is more straightforward.

Is this limit a special case in trigonometry or calculus?

This limit is a particular example of how trigonometric functions behave near 0 and illustrates an important concept in calculus, which is the behavior of functions as they approach a point.

Does the small-angle approximation apply to this limit?

Yes, the small-angle approximation (where \(\sin(x) \approx x\) when \(x\) is near 0) is relevant here. It simplifies the calculation by allowing us to approximate \(\sin(\sin(x))\) as \(\sin(x)\) when \(x\) is close to 0.

How does this limit relate to the concept of continuity?

This limit demonstrates the continuity of the sine function at 0. It shows that as \(x\) gets very close to 0, the value of \(\frac{\sin(\sin(x))}{x}\) approaches a definite number (1), indicating the function is continuous at that point.

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