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

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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

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

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