#lim_(x->0) sin(x)/x = 1#. *We determine this by the use of 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 #oo#), 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) = sin(x)# and #g(x) = x#. These functions are continuous and differentiable near #x=0#, #sin(0) = 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 sin(x) = cos(x), d/dx x = 1#. Thus...

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