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