Question

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

Answer 1

`-1/6`

Explanation:

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`