Sunday, 21 August 2016

limits - Find limxto0fracsinleft(1fracsin(x)xright)x2. Is my approach correct?



Find:
L=limx0sin(1sin(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=limx0sin(1sin(x)x)1sin(x)x1x2(1sin(x)x)


Then, if the limits
L1=limx0sin(1sin(x)x)1sin(x)x,



L2=limx01x2(1sin(x)x)


exist, then L=L1L2.



For the first one, by making the substitution u=1sin(x)x, we have
L1=limuu0sin(u)u,


where
u0=limx0(1sin(x)x)=0.

Consequently,
L1=limu0sin(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 x0,
sinx=xx36+O(x5)

gives

1sinxx=x26+O(x4)
then
sin(1sinxx)=x26+O(x4)
and




sin(1sinxx)x2=16+O(x2)




from which one may conclude easily.


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