Question

What is the limit as x approaches 0 of `sin(3x)/sin(4x)`?

Answer 1

The answer is `3/4`.

You both `sin(3*0) = 0` and `sin(4*0)=0`, so you can use l'hopitals rule. This is:
`lim_(x->0) sin(3x)/sin(4x) = lim_(x->0) ([sin(3x)]')/([ sin(4x)]') = lim_(x->0) (3cos(3x))/(4cos(4x)) =(3cos(0))/(4cos(0)) = 3/4`

The first step is taking the derivative of both the nominator and the denominator; the last step is just filling in zero, you're allowed to do this because `cos(0) = 1`, so you don't risk dividing by zero.