nona.m.nona wrote:36: Find the limit: lim, x -> 0, (sin 4x)/(sin 6x)
I'm pretty sure I need to use the fact that lim, x->0, (sin x)/(x) = 1....
You're exactly right about the tool you need to use. There's a bit of a trick to setting it up, though. For the moment, let's leave off the "limit" bit, and just deal with the fraction. You've got "sin(stuff)" and need "[sin(stuff)] / (stuff)". So multiply, top and bottom, by whatever you need (just like when you were converting fractions to common denominators).
. . . . .
. . . . .}}\, =\, \left( \frac{1}{\sin{(6x)}}\right) \left(\frac{6x}{6x}\right)\, =\, \left(\frac{1}{6x}\right)\left(\frac{6x}{\sin{(6x)}}\right))
Put them together, and you get:
. . . . . \left(\frac{\sin{(4x)}}{4x}\right) \left(\frac{1}{\left(\frac{\sin{(6x)}}{6x}\right)}\right))
The first fraction simplifies nicely, and the other two are evaluated easily by using the limit they gave you, and noting that the particular form of the "(stuff)" doesn't matter, as long as the limit, as the variable goes to zero, of "(stuff)" is zero.

Eliz.
P.S. This trick doesn't come up often, but you should expect to see it on your next test, and maybe on this semester's final.