#lim_(x->0)(tan x - x)/x^3 = # ?

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Answer 1 · 2017-03-01T18:12:32

#-1/6#

We know that #sin x = x-x^3/(3!)+x^5/(5!)- cdots = sum_(k=0)^oo(-1)^k x^(2k+1)/((2k+1)!)#

and also

#sin x -x = -x^3/(3!)+x^5/(5!)- cdots#

Now #(tan x - x)/x^3 = (sinx-x cos x)/(x^3 cos x)# and for small values on #x# we have

#(tan x - x)/x^3 approx (sin x- x)/x^3# because then #cos x approx 1#

Putting all together

#lim_(x->0)(tan x - x)/x^3 = lim_(x->0)(sin x- x)/x^3 = lim_(x->0)(-x^3/(3!)+x^5/(5!)- cdots)/x^3 = -1/(3!) = -1/6#