Approach warp speed

Repeated use of L'Hopital's Rule.

$f(x)=\frac{tan{x} -x}{x^3}$ f^'(x)= \frac{sec^{2}x -1}{3x^2} f^''(x)=\frac{sec^{2}x tanx}{3x} f^'''(x)=\frac{sec^{2}x tan^{2}x+sec^{4}x}{3}

At last you have a form which is defined for $x=0$

Plugging in $x=0$ you get $lim_{x\rightarrow 0}= \frac{1}{3}$

The series of $tan x= x+\frac{x^3}{3}+..$ We need not consider the next terms as x is too small.Putting it in our expression we get, $\frac{x+\frac{x^3}{3}-x}{x^3}=1/3=\boxed{0.333}$

this question solve by L'Hopital's rule, in L'Hopital's rule take derivative 3 times, and by apply limit we get 1/3 or 0.333........

use L'Hopital's rule and take derivative three time and apply the limit you will get 1/3=.33333333