how do you solve lim x->0 (sin 3x)/7x?

Answer 1

This can be solved without L'Hopital's rule.
You know that
lim x->0 (sin nx)/(nx) = 1
First bring out the 1/7
(1/7) lim x->0 (sin 3x)/x
Now multiply the top and bottom by 3
(1/7) lim x->0 3(sin 3x)/(3x)
= (1/7)(3)(1)
= 3/7

Answer 2

First you have to evaluate the limit as x->0

You would get sin(0)/0 = 0/0, which is undefined. With this condition met you can use L'Hopital's rule to get the limit. This rule states to take the derivative of the numerator and denominator separately, then evaluate the limit.

Derivative of numerator: 3cos(3x)

Derivative of denominator: 7

Now evaluate the limit as x->0

[3 cos(3*0)]/7 --------> limit = 3/7