Evaluate the limit of the function sin(3x)/tan(5x), if x goes to 0?

If we'll replace x by 0, we'll get the indetermination
"0/0" type. Therefore, we can use L'Hospital's rule, to determine the limit of the
quotient of derivatives.

lim f(x)/g(x) = lim
f'(x)/g'(x)

lim sin(3x)/tan(5x) = lim
[sin(3x)]'/[tan(5x)]'

We'll apply chain rule to
differentiate with respect to x:

lim [sin(3x)]'/[tan(5x)]'
= lim 3*cos (3x)/5/[cos (5x)]^2

We'll replace again x by 0
and we'll get:

lim 3*cos (3x)/5/[cos (5x)]^2 = 3*cos
(0)/5/[cos (0)]^2

Since cos 0 = 1, we'll
have:

lim 3*cos (3x)/5/[cos (5x)]^2 =
3/5

The requested limit of the function
sin(3x)/tan(5x), when x approaches to 0, is: lim sin(3x)/tan(5x) =
3/5.