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