Question

How do you find the limit of `(sin4x)/(sin6x)` as `x->0`?

Answer 1

`2/3`

Explanation:

Multiply and divide by `4x` and by `6x`, then rearrange as follows:

`sin(4x)/sin(6x) * (4x)/(4x) * (6x)/(6x) = sin(4x)/(4x) * (6x)/sin(6x) * (4x)/(6x)`

As you might have noticed, now we have in the first two terms the known limit

`lim_{y\to 0} sin(y)/y`

which is always `1`, no matter what expression we insert as `y`, as long as it tends to zero (and of course, both `4x` and `6x` tend to zero as `x\to 0`).

The third term is simply `4/6=2/3`, since we can cancel the `x` factor.

So, the product of the limits is `1*1*2/3=2/3`