Question

How do you find the limit of `(1-cosx)/(2x^2 - x)` as x approaches 0?

Answer 1

`lim_(x->0) (1-cosx)/(2x^2-x) = 0`

Explanation:

We can use the Maclaurin series for `cos`...

`cos x = 1/(0!)-x^2/(2!)+x^4/(4!)-x^6/(6!)+...`

So:

`1 - cos x = 1/2x^2-O(x^4)`

and:

`lim_(x->0) (1-cosx)/(2x^2-x) = lim_(x->0) (1/2x^2-O(x^4))/((2x-1)x)`

`color(white)(lim_(x->0) (1-cosx)/(2x^2-x)) = lim_(x->0) (1/2x-O(x^3))/(2x-1)`

`color(white)(lim_(x->0) (1-cosx)/(2x^2-x)) = 0/(-1)`

`color(white)(lim_(x->0) (1-cosx)/(2x^2-x)) = 0`

Answer 2

If you've learned them, use the fundamental trigonometric limits.

Explanation:

`lim_(xrarr0)sinx/x = 1` (By an argument based on some geometry and the squeeze theorem.)

`lim_(xrarr0)(1-cosx)/x = 0` (Use the limit above and trig identities.)

`lim_(xrarr0)(1-cosx)/(2x^2-x) = lim_(xrarr0)(1-cosx)/x *1/(2x-1)`

`= lim_(xrarr0)(1-cosx)/x * lim_(xrarr0) 1/(2x-1)`

`= 0 * 1/-1 = 0`