Thursday, 21 February 2019

calculus - limit limxto0fractanxxx2tanx without Hospital



Is it possible to find limx0tanxxx2tanx without l'Hospital's rule?



I have limx0tanxx=1 proved without H. but it doesn't help me with this complicated limit (however I'm sure I have to use it somehow).



I know the answer is 13, so I tried to estimate: 0<tanxxx2tanx13tanxx+g(x) with various g and prove that g(x)0, but with no result.


Answer




L=limx0xtanxx2tanx=limx0cosxsinxxx2xsinx=limx012sin2x2cosx2sin(x/2)(x/2)x2
gives L=A+B where:
A=12+limx01cosx2x2=12+limx02sin2x4x2=12+18=38
and
B=limx01x2(1sin(x/2)x/2)=14limx01x2(1sinxx)
(assuming such a limit exists) fulfills:
3B=4BB=limx0(sin(x/2)x/2sinxx)=limx01x2(2sin(x/2)sin(x)1)
so that:
B=13limx01x2(1cosx21)=13limx01cosx2x2=13limx02sin2x4x2=124
and L=A+B=38+124=13.



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