How do you find the limit of #x/sin(x)# as x approaches 0?

1 Answer
May 25, 2016

Manipulate the fundamental trigonometric limit to get #lim_(x->0)x/sinx=1#.

Explanation:

Evaluating the limit directly produces the indeterminate form #0/0# so we'll have to use an alternate method.

We know that #lim_(x->0)sinx/x=1#, so we'll work from there.

Note that #x/sinx=1/(sinx/x)#, which means #lim_(x->0)x/sinx# can be equivalently expressed as #lim_(x->0)1/(sinx/x)#. Using the properties of limits, this becomes:
#(lim_(x->0)1)/(lim_(x->0)sinx/x#

Well, #lim_(x->0)1=1# for all #x#, and #lim_(x->0)sinx/x=1# (fundamental trigonometric limit). That means we have:
#1/1#

Which is simply #1#.