Question

Use Taylor series to evaluate
limx→0(tan x − x)/x^3

Answer 1

Recall that

tan(x) = sin(x)/cos(x)

and

sin(x) = x - x ³/6 + x ⁵/120 - x ⁷/5040 + …

cos(x) = 1 - x ²/2 + x ⁴/24 - x ⁶/720 + …

Truncate the series to three terms. Then

`\displaystyle \lim_(x\to0)(\tan(x)-x)/(x^3) = \lim_(x\to0)((x-x^3/6+x^5/120)/(1-x^2/2+x^4/24)-x)/(x^3) \\\\ = \lim_(x\to0)\left((x-x^3/6+x^5/120)/(x^3-x^5/2+x^7/24)-\frac1{x^2}\right) \\\\ = \lim_(x\to0)\left((1-x^2/6+x^4/120)/(x^2-x^4/2+x^6/24)-\frac1{x^2}\right) \\\\ = \lim_(x\to0)\left((1-x^2/6+x^4/120)/(x^2\left(1-x^2/2+x^4/24\right))-\frac1{x^2}\right) \\\\ = \lim_(x\to0)\left((1-x^2/6+x^4/120)/(x^2\left(1-x^2/2+x^4/24\right))-(1-x^2/2+x^4/24)/(x^2\left(1-x^2/2+x^4/24\right))\right) \\\\ = \lim_(x\to0)(x^2/3-x^4/30)/(x^2\left(1-x^2/2+x^4/24\right)) \\\\ = \lim_(x\to0)(1/3-x^2/30)/(1-x^2/2+x^4/24) = \boxed{\frac13}`