Question

How do you evaluate the limit `(1-cosx)/x` as x approaches `0`?

Answer 1

Because the expression evaluated at the limit is `0/0`, we can use L'Hospital's Rule.

Explanation:

The rule states that, when given the limit of a fraction, `(lim)/(xrarrc) f(x)/g(x)` where `f(c)/g(c)` is, `0/0`, `oo/oo` or `(-oo)/-oo`, then `(lim)/(xrarrc)(f'(x))/(g'(x))` goes to the same limit.

To implement the rule, we take the derivative of the numerator:
`d(1 - cos(x))/dx = sin(x)`

the derivative of the of the denominator:

`dx/dx = 1`

Assemble the new fraction with the same limit:

`(lim)/(xrarr0) sin(x)/1 = 0`

Therefore, `(lim)/(xrarr0) (1 - cos(x))/x = 0`

Answer 2

Usually, this is done after showing that `lim_(xrarr0)sinx/x = 1`. See below.

Explanation:

`(1-cosx)/x = ((1-cosx))/x * ((1+cosx))/((1+cosx))`

`= (1-cos^2x)/(x(1+cosx))`

`= sin^2x/(x(1+cosx))`

`= sinx/x * sinx * 1/(1+cosx)`

Taking the limit as `xrarr0`, we get

`(1)(0)(1/2) = 0`