Question

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

Answer 1

Use the fundamental trigonometric limits and algebra.

Explanation:

The fundamental trigonometric limits are:

`lim_(xrarr0)(cosx-1)/x = 0` and `lim_(xrarr0)sinx/x=1`

If the problem is
`(cosx-1)/sin(x^2)`, then use

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

So the limit we seek is equal to `lim_(xrarr0)(cosx-1)/x^2`

Looking for a trick that might help, we it may eventually occur to us to try mulltiplying by `(cosx+1)/(cosx+1)` to replace the subtraction that goes to `0` with an addition that goes to `2`.

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

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

The limit as `xrarr0` is `-(1)*(1)*1/(1+1) = -1/2`

Bonus

If the problem is intended to be `(cosx-1)/(sinx)^2`, then use the same trick to get:

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

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

and, again, the limit is `-1/2` (Without using fundamental trigonometric limits).

Answer 2

Alternatively, you may also use L'Hospital's Rule to still get the same answer to the limit as `-1/2`.

Explanation:

Since the limit is an indeterminant type of form `0/0`, we may apply L'Hospital and differentiate the top and bottom twice to eventually obtain :

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

`=lim_(x->0)(-cosx)/(-4x^2sinx+2cosx^2)`

`=-1/2`