Question

How do you find the limit of `sin^2x/x` as `x->0`?

Answer 1

The limit is `0`.

Explanation:

Consider the well-known limit `lim_(x->0) sinx/x = 1`. We will need to rewrite this expression in order to use this property.

`=lim_(x-> 0)( sinx/x xx sinx)`

`=lim_(x->0) sinx/x xx lim_(x->0)sinx`

`=1 xx 0`

`=0`

Hopefully this helps!