Special Limits Involving sin(x), x, and tan(x)

Key Questions

  • We will use l'Hôpital's Rule.

    l'Hôpital's rule states:

    #lim_(x->a) f(x)/g(x) = lim_(x->a) (f'(x))/(g'(x))#

    In this example, #f(x)# would be #sinx#, and #g(x)# would be #x#.

    Thus,

    #lim_(x->0) (sinx)/x = lim_(x->0) (cosx)/(1)#

    Quite clearly, this limit evaluates to #1#, since #cos 0# is equal to #1#.

  • Answer
    By using: #lim_{x to 0}{sinx}/{x}=1#, #lim_{x to 0}{tanx}/x=1#.

    Explanation
    Let us look at some details.
    Since #tanx={sinx}/{cosx}#,

    #lim_{x to 0}{tanx}/x = lim_{x to 0}{sinx}/{x}cdot1/{cosx}#

    by the Product Rule,

    #=(lim_{x to 0}{sinx}/x)cdot(lim_{x to 0}1/{cosx})#

    by #lim_{x to 0}{sinx}/{x}=1#,

    #=1cdot1/{cos(0)}=1#

  • One of some useful limits involving #y=sin x# is
    #lim_{x to 0}{sin x}/{x}=1#.
    (Note: This limit indicates that two functions #y=sin x# and #y=x# are very similar when #x# is near #0#.)

Questions