How do you use L'hospital's rule to find the limit #lim_(x->0)(x-sin(x))/(x-tan(x))# ?

To use L'Hôpital's Rule to find the limit (\lim_{x \to 0} \frac{x - \sin(x)}{x - \tan(x)}), we can apply the rule to the fraction (\frac{0}{0}) form.

First, we differentiate the numerator and the denominator separately:

1. Differentiate the numerator: [ \frac{d}{dx}(x - \sin(x)) = 1 - \cos(x) ]

2. Differentiate the denominator: [ \frac{d}{dx}(x - \tan(x)) = 1 - \sec^2(x) ]

Next, we substitute these derivatives back into the original limit expression:

[ \lim_{x \to 0} \frac{1 - \cos(x)}{1 - \sec^2(x)} ]

Now, evaluate this new limit expression:

1. Substitute (x = 0) into the expression: [ \frac{1 - \cos(0)}{1 - \sec^2(0)} = \frac{1 - 1}{1 - 1} ]

2. Simplify the expression: [ \frac{0}{0} ]

Since we still have an indeterminate form, we can apply L'Hôpital's Rule again. Differentiate the numerator and denominator:

1. Differentiate the numerator: [ \frac{d}{dx}(1 - \cos(x)) = \sin(x) ]

2. Differentiate the denominator: [ \frac{d}{dx}(1 - \sec^2(x)) = -2\sec(x)\tan(x) ]

Substitute these derivatives back into the limit expression:

[ \lim_{x \to 0} \frac{\sin(x)}{-2\sec(x)\tan(x)} ]

Now, evaluate the limit as (x) approaches 0:

1. Substitute (x = 0) into the expression: [ \frac{\sin(0)}{-2\sec(0)\tan(0)} = \frac{0}{0} ]

Since we still have an indeterminate form, we can apply L'Hôpital's Rule one more time. Differentiate the numerator and denominator:

1. Differentiate the numerator: [ \frac{d}{dx}(\sin(x)) = \cos(x) ]

2. Differentiate the denominator: [ \frac{d}{dx}(-2\sec(x)\tan(x)) = -2\sec(x)\tan(x) - 2\sec^3(x) ]

Substitute these derivatives back into the limit expression:

[ \lim_{x \to 0} \frac{\cos(x)}{-2\sec(x)\tan(x) - 2\sec^3(x)} ]

Finally, evaluate the limit as (x) approaches 0:

1. Substitute (x = 0) into the expression: [ \frac{\cos(0)}{-2\sec(0)\tan(0) - 2\sec^3(0)} = \frac{1}{0 - 2} = -\frac{1}{2} ]

So, (\lim_{x \to 0} \frac{x - \sin(x)}{x - \tan(x)} = -\frac{1}{2}).

The limit

Thus,

#lim_(x->0)(x-sin(x))/(x-tan(x)) = lim_(x->0){d/dx(x-sin(x))}/ {d/dx(x-tan(x))}#

We could use L'hospital's rule yet again, but it is much simpler to use simple trigonometry to go ahead :

and thus