Question

hlo guys
plzzzz give correct answer​

Answer 1

$\huge{\mathcal{\underline{\green{ANSWER}}}}$

lim x->0 (tan x-sin x)/x^3 this is 0/0 form,

Apply hospital’s rule

lim x->0 {( sec x)^2 -cos x}/3x^2 Again 0/0 form,

lim x->0{(2sec^2x.tan x)+sin x}/6x , 0/0 form

lim x->0{2tan x(1+tan^2x)+sin x}/6x

lim x->0{2 tan x+2( tan x)^3+sin x}/6x

Apply hospital’s rule

lim x->0{2(sec x)^2+6(tan x.sec x)^2+cos x}/6

=3/6

=1/2 answer.