Question

x-tant
is equal to :
x-sinx
7 )limx-0
A)0
B)3​

Answer 1

Step-by-step explanation:

We have,

$\lim_{x \rarr0 } ( \frac{x - \tan(x) }{x - \sin(x) } )$

Since, it is 0/0 from, so using l'hospital rule

$\lim_{x \rarr0 } \frac{ \frac{d}{dx}(x - \tan(x) )}{ \frac{d}{dx} (x - \sin(x)) }$

$= \lim_{x \rarr0 } \frac{1 - \sec ^{2} (x) }{1 - \cos(x) }$

$= - \lim_{x \rarr0 } \frac{( 1 - \cos ^{2} (x)) }{ \cos^{2} (x)( 1 - \cos(x)) }$

$= - \lim_{x \rarr0 } \frac{(1 + \cos(x))(1 - \cos(x)) }{ \cos^{2} (x)(1 - \cos(x)) }$

$= - \lim_{x \rarr0 } \frac{1 + \cos(x) }{ \cos^{2} (x) }$

$= - \frac{1 + 1}{1} = - 2$