lim theta --> 0 [sin4(theta) / sin 6 (theta)?

Answer 1

For simplicity, I'll replace your "theta" with x.

Your first step would be to determine whether you get the indeterminate form of zero over zero [0/0]. The reason why we do this is because this is one of the conditions required to use L'Hospital's rule. What you would do is plug in 0 for your function. So for

lim (x => 0, (sin4x)/(sin6x))

we get sin4(0)/sin6(0) = sin(0)/sin(0) = 0/0

So we do indeed get that form. Now, we apply L'Hospital's rule; we take the derivative of the top and the derivative of the bottom to get a new limit.

lim (x => 0, [cos(4x)*4]/[cos(6x0*6])

Note; when taking the derivative, we apply the chain rule, hence why you see a 4 and a 6 there.

Now, we pull out all constants and take them outside of the limit. Constants usually get in the way when solving derivatives, integrals, and limits, so it's always a good idea to pull them out as soon as you can.

4/6 * lim (x => 0, cos(4x)/cos(6x))

Now, we plug in 0 for the function. cos(4*0)/cos(6*0) = cos(0)/cos(0) = 1/1 = 1.

Since the limit worked out to an actual number, we can actually plug in for what we solved for.

4/6 * (1) = 4/6

So your answer to your limit is 4/6.

Answer 2

purely manage each summand one after the other. The reduce of cosine is purely cos(0) (the function is continuous at 0. As for a million / sin(theta), the reduce does no longer exist. So, the reduce contained in the challenge does no longer exist.