Question

How do you find the limit of `(1-cosx)/(xsinx)` as `x->0`?

Answer 1

`lim_(x rarr 0) (1- cosx)/(x sinx) = 1/2`

Explanation:

First of all, since as `x rarr 0`, `sinx rarr 0` also, we can rewrite the denominator as `x^2`.

Hence we need to find: `lim_(x rarr 0) (1- cosx)/(x^2)`

Since this still results in an indeterminate `0/0`, we apply L'Hopital's Rule.

`(d/dx(1-cos x)) / (d/dx(x^2)) = sinx/(2x)`

If we substitute 'approaching zero' as a less formal `1/oo`, we arrive at the expression: `(1/oo)/(2/oo)`.

After cancelling out the infinities, this leaves `1/2`.

Or simply let `sinx =x` again, which gives:

`sinx/(2x) = x/(2x) = 1/2`.

Answer 2

Use `lim_(xrarr0)sinx/x=1`

Explanation:

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

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

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

`lim_(x rarr 0) (1- cosx)/(x sinx) = lim_(x rarr 0)(sinx/x * sinx/sinx * 1/(1+cosx))`

`= (1)(1)(1/(1+1))=1/2`