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
