Is it possible to find limx→0tanx−xx2tanx without l'Hospital's rule?
I have limx→0tanxx=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<tanx−xx2tanx≤13⋅tanxx+g(x) with various g and prove that g(x)→0, but with no result.
Answer
L=limx→0x−tanxx2tanx=limx→0cosx−sinxxx2⋅xsinx=limx→01−2sin2x2−cosx2sin(x/2)(x/2)x2
gives L=A+B where:
A=−12+limx→01−cosx2x2=−12+limx→02sin2x4x2=−12+18=−38
and
B=limx→01x2(1−sin(x/2)x/2)=14limx→01x2(1−sinxx)
(assuming such a limit exists) fulfills:
3B=4B−B=limx→0(sin(x/2)x/2−sinxx)=limx→01x2(2sin(x/2)sin(x)−1)
so that:
B=13limx→01x2(1cosx2−1)=13limx→01−cosx2x2=13limx→02sin2x4x2=124
and L=A+B=−38+124=−13.
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