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

• suffian
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.

#### 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)?

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...

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.

thanks! i was getting used to multiplying by constants, but that's pretty ingenious.

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;

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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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...

Cute.

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 :)

do what i said above twice, forgot you asked sin[sin[x]]

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...

when you take the limit it is correct
but don't want to prove that mathematically :)

## 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.