Question

How do you find the limit of `(sin3x)/x` as `x->0`?

Answer 1

The limit is `3`.

Explanation:

`lim_(xrarr0)(sin3x)/x = lim_(xrarr0)(3sin(3x))/((3x))`

`= 3 lim_(xrarr0)(sin(3x))/((3x))`

Now use `lim_(theta rarr0) sintheta / theta = 1` with `theta = 3x` to get

`= 3(1)=3`