Find:
L=limx→0sin(1−sin(x)x)x2
My approach:
Because of the fact that the above limit is evaluated as 00, we might want to try the De L' Hospital rule, but that would lead to a more complex limit which is also of the form 00.
What I tried is:
L=limx→0sin(1−sin(x)x)1−sin(x)x1x2(1−sin(x)x)
Then, if the limits
L1=limx→0sin(1−sin(x)x)1−sin(x)x,
L2=limx→01x2(1−sin(x)x)
exist, then L=L1L2.
For the first one, by making the substitution u=1−sin(x)x, we have
L1=limu→u0sin(u)u,
where
u0=limx→0(1−sin(x)x)=0.
Consequently,
L1=limu→0sin(u)u=1.
Moreover, for the second limit, we apply the De L' Hospital rule twice and we find L2=16.
Finally, L=116=16.
Is this correct?
Answer
In a slightly different way, using the Taylor expansion, as x→0,
sinx=x−x36+O(x5)
1−sinxx=x26+O(x4)
sin(1−sinxx)=x26+O(x4)
sin(1−sinxx)x2=16+O(x2)
from which one may conclude easily.
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