Question

Ltx>0 x-tanx/x-sinx​

Answer 1

Question :

Evaluate:

$\sf \lim _{x \to0} \: \dfrac{x - \tan \: x}{x - \: sin \: x}$

Solution :

$\sf \lim _{x \to0} \: \dfrac{x - \tan \: x}{x - \: sin \: x}$

Since 0/0 is of indeterminate form.

Then ,apply L'Hospital's Rule.

$\implies \: \sf \lim _{x \to0} \: \dfrac{x - \tan \: x}{x - \: sin \: x} = \sf \lim _{x \to0} \: \dfrac{ \frac{d(x - \tan \: x)}{dx} }{ \frac{d(x - \sin \: x)}{dx} }$

$\sf = \lim _{x \to0} \: \dfrac{1 - \sec {}^{2}x }{1 - \cos \: x}$

This is again is of the 0/0 form , Use L'Hospital's rule again.

$\sf = \lim _{x \to0} \: \dfrac{ - 2\sec {}^{2}x \tan \: x }{ \sin \: x}$

$\sf = \lim _{x \to0} \dfrac{ - 2}{ \cos {}^{2} x} \times \dfrac{ \frac{ \cancel{ \sin \: x}}{ \cos \: x} }{ \cancel{\sin \: x}}$

$\sf = \lim _{x \to0} \: \dfrac{ - 2}{\cos {}^{3} \: x}$

$\sf = \lim _{x \to0} - 2 \sec {}^{3} x$

$= - 2 \sec {}^{3} (0)$

$= - 2$

Therefore ,

$\bf \lim _{x \to0} \: \dfrac{x - \tan \: x}{x - \: sin \: x} = - 2$

_____________________

About L'HOSPITAL'S Rule :

If f(a)=g(a)= 0 , then

$\sf \lim _{x \to \: a} \: \frac{f(x)}{g(x)} = \sf \lim _{x \to \: a} \frac{f'(x)}{g'(x)}$

Answer 2

Answer :-

$\sf{\implies -2}$

To solve :-

$\displaystyle \lim_{x \to 0} \: \: \: \: \: \: \: \: \frac{x - \tan(x) }{x - \sin(x) }$

Applying L - hospital rule.

$\implies \displaystyle \lim_{x \to a } \:\:\:\: \frac{f(x)}{g(x)} = \displaystyle \lim_{x \to a } \:\:\:\: \frac{f'(x)}{g'(x)}$

$\implies \displaystyle \lim_{x \to 0 } \:\:\:\: \frac{\frac{d}{dx} (x - \tan x ) }{\frac{d}{dx}( x - \sin x)}$

$\implies \frac{d}{dx} (x) - \frac{d}{dx} ( \tan x ) = 1{x}^{0} - {\sec}^{2} x$

$\implies \frac{d}{dx} (x) - \frac{d}{dx} ( \sin x ) = 1{x}^{0} - \cos x$

$\implies \displaystyle \lim_{x \to 0} \: \: \: \: \frac{1 - { \sec(x) }^{2} }{1 - \cos(x) }$

Again applying L - hospital rule

$\implies \frac{d}{dx} (1) - \frac{d}{dx} ( {\sec}^{2} x ) = 0 - 2{\sec}^{2} x \tan x$

$\implies \frac{d}{dx} (1) - \frac{d}{dx} ( \cos x ) = 0 - \sin x$

$\implies \displaystyle \lim_{x \to a } \:\:\:\: \frac{-2{\sec}^{2} x \tan x }{\sin x}$

$\implies \displaystyle \lim_{x \to a } \:\:\:\: \frac{-2 {\sec}^{2} x ( \tan x)}{\sin x}$

$\implies \displaystyle \lim_{x \to a } \: \: \: \frac{ - 2 { \sec(x)^{2} \frac{ \sin(x) }{ \cos(x) }} }{ \sin(x) }$

$\implies \displaystyle \lim_{x \to a } \:\:\:\:- 2 {\sec}^{2} x \frac{1}{\cos x}$

$\implies \displaystyle \lim_{x \to a } \:\:\:\: -2 {\sec}^{3} x$

$\implies -2 \displaystyle \lim_{x \to a } {\sec}^{3} x$

$\large{\sf{\implies -2}}$