Question

Evaluate the limit of this:​

Answer 1

Step-by-step explanation:

$\lim _{x -> 0} \frac{ \tan(x) - \sin(x) }{ {x}^{3} }$

Since, it is inte form of 0/0,

So, using l'hispital rule,

$= \lim _{x -> 0} \frac{ \frac{d}{dx}( \tan(x) - \sin(x) ) }{ \frac{d}{dx}( {x}^{3}) }$

$= \lim _{x -> 0} \frac{ \sec^{2} (x) - \cos(x) }{3 {x}^{2} }$

Again using l'hospital rule,

$= \lim _{x -> 0} \frac{2 \sec^{2} (x) \tan(x) + \sin(x) }{6x}$

$= 2\lim _{x -> 0} \frac{ \sec^{2} (x) \tan(x) }{6x} + \lim _{x -> 0} \frac{ \sin(x) }{6x}$

$= \frac{1}{3} \lim _{x -> 0} \sec^{2} (x) \lim _{x -> 0} \frac{ \tan(x) }{x} + \frac{1}{6} \lim _{x -> 0}\frac{ \sin(x) }{x}$

$= \frac{1}{3} \times \sec^{2} (0) \times 1 + \frac{1}{6} \times 1$

$= \frac{1}{3} + \frac{1}{6}$

$= \frac{3}{6} = \frac{1}{2}$