Question

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

Answer 1

`-1/2`

Explanation:

The limit

`L = lim_(x->0)(x-sin(x))/(x-tan(x))`

is of the form `0/0`, Thus, we can use L'hospital's rule, which says

If `f(a)=g(a) = 0`, `lim_{x to a} f(x)/g(x) = lim_{x to a} {f^'(x)}/{g^'(x)}`

Thus,

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

`qquad = lim_(x->0)(1-cos(x))/(1-sec^2(x))`

This, again is of the `0/0` form, so we use L'hospital's rule again

`L = lim_(x->0){d/dx(1-cos(x))}/{d/dx(1-sec^2(x))} = lim_(x->0)(sin(x))/((-2sec^2(x)tan(x)))`

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

`L = lim_(x->0)(sin(x))/((-2sec^2(x)tan(x))) = lim_(x->0)(1)/((-2sec^3(x))`

and thus

`L = -1/2`