What is the limit as x approaches 0 of #tan(6x)/sin(2x)#?
2 Answers
The answer is 3:
How did I get there?
The first thing you should always try with limits is just to enter the x value in the function:
This is an impossible answer, but whenever we find that we have
In this formula,
Let's apply this rule to our problem.
So that's our first derivative. Now for our second:
Now, all we need to do is combine both of them.
You can bring the number in front of the limit:
Now we can replace
Now let's try entering
And there you have your final answer. Let me know if anything is not clear.
I hope this helped.
Explanation:
By de Moivre's theorem:
#cos 3 theta + i sin 3 theta = (cos theta + i sin theta)^3#
#color(white)(cos 3 theta + i sin 3 theta) = cos^3 theta + 3i cos^2 theta sin theta - 3 cos theta sin theta - i sin^3 theta#
Equating imaginary parts:
#sin 3 theta = 3 cos^2 theta sin theta - sin^3 theta#
#color(white)(sin 3 theta) = 3 (1 - sin^2 theta) sin theta - sin^3 theta#
#color(white)(sin 3 theta) = 3 sin theta - 4sin^3 theta#
Putting
#lim_(x->0) (tan 6x) / (sin 2x) = lim_(theta->0) (sin 3 theta) / (sin theta cos 3 theta)#
#color(white)(lim_(x->0) (tan 6x) / (sin 2x)) = lim_(theta->0) (3 sin theta - 4 sin^3 theta) / (sin theta cos 3 theta)#
#color(white)(lim_(x->0) (tan 6x) / (sin 2x)) = lim_(theta->0) (3 - 4 sin^2 theta) / (cos 3 theta)#
#color(white)(lim_(x->0) (tan 6x) / (sin 2x)) = (3-0)/1#
#color(white)(lim_(x->0) (tan 6x) / (sin 2x)) = 3#