We have two responses for you

Hi Josh,

The theorem that I am sure you know that needs to be used hers is that the limit of sin(t)/t approaches 1 as t approaches 0. Thus if you had

you could make the variable substitution t = 4x. As x approaches 0 certainly 4x = t approaches zero so the limit above is 1. Similarly

Thus

But this isn't your problem, mine has an extra 6x in the numerator and an extra 4x in the denominator, but

Now if you take the limit of the right side as x approach er zero the first fraction approaches 1, the second fraction approaches 1 and the third fraction is (4x)/(6x) = 4/6 = 2/3. Thus the limit is 2/3.

I hope this helps,
Harley

Hi Josh.

Looking at the fraction, I can see that it goes to zero divided by zero, so my first guess would be to see if I can use l'Hôpital's Rule on it: If, as x approaches 0, lim sin(4x) = 0 [and it does] and lim sin(6x) = 0 [it does] and lim (sin 4x)′ / (sin 6x)′ has a finite value is +/- inifinity [we'll check this], then the lim sin(4x)/sin(6x) = lim (sin 4x)′ / (sin 6x)′.

So we need the derivative of sin 4x. Using the chain rule we have (sin 4x)′ = (4x)′ cos(4x) = 4 cos(4x).
And the same with the denominator.

So lim (sin 4x)′ / (sin 6x)′ = lim [4 cos(4x)] / [6 cos 6x]. As x approaches 0, this converges to 2/3. That's a finite value, so that's the same as the original limit.

Hope this helps,
Stephen La Rocque.