Verified answer

Not exactly. No need of De L'Hospital, but that's not the relation you have to apply.

Multiply by (cosx+1)/(cosx+1) so that the top turns into a square difference:

lim (x->0) (cos²x-1)/x(cosx+1) = lim (x->0) -(1-cos²x)/x(cosx+1)

Apply this fundamental relation: cos²x + sin²x = 1 ==> sin²x = 1 - cos²x

lim (x->0) -sin²x/x(cosx+1)

Now you apply the limit above and you're left with:

lim (x->0) -sinx/(cosx+1) = 0

lim x->0 (cos(x) - 1) / x

Plug in 0:

(cos(0) - 1) / 0 = (1 - 1) / 0 = 0 / 0

Since this is indeterminate, we can use L'Hopital's rule and find the derivative of the top and bottom of the expression, getting:

-sin(x) / 1 or just -sin(x)

Take the limit of this:

lim x-> 0 -sin(x)

Plug in 0:

-sin(0) = -(0) = 0

Thus, the limit is 0.

lim x-->0 (cosx-1)/x= undifined