Question

Question #59af9

Answer 1

`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]}