Computing limx→0+tan(x)−xx3 without L'Hopital
Say limx→0+tan(x)−xx3=L
For L:
L=limx→0tanx−xx3L=limx→0tan2x−2x8x34L=limx→012tan2x−xx33L=limx→012tan2x−tanxx3=limx→0tanxx11−tan2x−1x2=limx→0(tanx)3x3=1L=13
I found that in another Q, can someone tell me why
L=limx→0tanx−xx3=limx→0tan2x−2x8x3
Answer
If x=2y then y→0 when x→0, so limx→0f(x)=limy→0f(2y).
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