Question

What is the limit as x approaches 0 of `tan(6x)/sin(2x)`?

Answer 1

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:
`lim_{x \to 0}tan(6x)/sin(2x) = tan(6*0)/sin(2*0) = tan(0)/sin(0) = (0/0)`

This is an impossible answer, but whenever we find that we have `(0/0)`, there's a trick we can use. It's called L'Hôpital's Rule. It states formally that:
`lim_{x \to a}f(x)/g(x) =lim_{x \to a} (f'(x))/(g'(x))`
In this formula, `f'(x)` means the derivative of f(x) and `g'(x)` means the derivative of g(x). If you haven't learned about them yet, you can learn them here.

Let's apply this rule to our problem.

`f(x) = tan(6x)`
`f'(x) = d/dx tan(6x) = sec^2(6x)*d/dx(6x)` (chain rule)
`= 6*sec^2(6x) = 6/(cos^2(6x))` (since `sec = 1/cos`)
So that's our first derivative. Now for our second:
`g(x) = sin(2x)`
`g'(x) = d/dx sin(2x) = cos(2x)*d/dx(2x)` (chain rule)
`= 2*cos(2x)`

Now, all we need to do is combine both of them.

`lim_{x \to 0}tan(6x)/sin(2x) = lim_{x \to 0}((6*sec^2(6x))/(2*cos(2x)))`
You can bring the number in front of the limit:
`=6/2lim_{x \to 0}((sec^2(6x))/(cos(2x)))`
Now we can replace `sec` by `1/cos`
`= 3*lim_{x \to 0}(1/(cos^2(6x)*cos(2x)))`

Now let's try entering `x` into this formula:
`= 3*(1/(cos^2(6*0)*cos(2*0)))`
`= 3*(1/(cos^2(0)*cos(0)))`
`= 3*1` (since `cos(0)=1`)
`= 3`

And there you have your final answer. Let me know if anything is not clear.
I hope this helped.

Answer 2

`lim_(x->0) (tan 6x) / (sin 2x) = 3`

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 `theta = 2x` we find:

`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`