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]}