Limits Involving Trigonometric Functions

Key Questions

  • What is the limit #lim_(x->0)(cos(x)-1)/x#?

    #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#

    Psykolord1989 . · 2 · Oct 11 2014
  • What is the limit #lim_(x->0)sin(x)/x#?

    #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#

    Psykolord1989 . · · Oct 11 2014

Questions

Differentiating Trigonometric Functions