Limits Involving Trigonometric Functions

Key Questions

  • 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

  • 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

Questions