#lim_(x->0) (cos(x)-1)/x = 0#. *We determine this by utilising L'hospital's Rule*.

To paraphrase, L'Hospital's rule states that when given a limit of the form #lim_(x→a)f(x)/g(x)#, where #f(a)# and #g(a)# are values that cause the limit to be indeterminate (most often, if both are 0, or some form of ∞), then as long as both functions are continuous and differentiable at and in the vicinity of #a,# one may state that

#lim_(x→a)f(x)/g(x)=lim_(x→a)(f'(x))/(g'(x))#

Or in words, the limit of the quotient of two functions is equal to the limit of the quotient of their derivatives.

In the example provided, we have #f(x)=cos(x)-1# and #g(x)=x#. These functions are continuous and differentiable near #x=0, cos(0) -1 =0 and (0)=0#. Thus, our initial #f(a)/g(a)=0/0=?.#

Therefore, we should make use of L'Hospital's Rule. #d/dx (cos(x) -1)=-sin(x), d/dx x=1#. Thus...

#lim_(x->0) (cos(x)-1)/x = lim_(x->0)(-sin(x))/1 = -sin(0)/1 = -0/1 = 0#