Question #59af9

1 Answer
Feb 15, 2018

#lim_(x->0)tan(x)/sin(3x)=1/3#

Explanation:

Recall first the fundamental trigonometric limit #lim_(x->0)sin(x)/x=1# and that #tan(x)=sin(x)/cos(x)#.

Then,
#\ \ \ \ \ \ lim_(x->0)tan(x)/sin(3x)#

#=lim_(x->0)1/cos(x)*sin(x)*1/sin(3x)#

#=lim_(x->0)1/3*1/cos(x)*sin(x)/x * (3x)/sin(3x)#

#=lim_(x->0)(1/3*1/cos(x))*lim_(x->0)(sin(x)/x) * lim_(x->0)((3x)/sin(3x))#

For the first limit, just substitute the values in to get #1/3#. For the second limit, use the fundamental trigonometric limit to get #1#. For the third limit, as #x->0#, #3x->0#. Thus, the fundamental trigonometric limit applies, and the answer is #1#.

Thus, the answer is
#=1/3*1*1#

#=1/3#

We can verify this with a graph:
graph{y=tan(x)/sin(3x) [-2.5 2.5 -1.25 1.25]}