Question

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

Answer 1

`0`

Explanation:

Know the following limit identity:

`lim_(xrarr0)sinx/x=1`

We can rewrite the given function so that we can make use of the fact that `lim_(xrarr0)sinx/x=1`.

The question rewritten is

`lim_(xrarr0)sin^2x/x`

Notice that we can isolate `sinx/x` from this.

`=lim_(xrarr0)sinx/x(sinx)`

Limits can be multiplied, as follows:

`=lim_(xrarr0)sinx/x*lim_(xrarr0)sinx`

Since the first part equals just `1`, this simplifies to be

`=lim_(xrarr0)sinx`

We can now evaluate the limit by plugging in `0` for `x`.

`=sin(0)=0`

The function should approach `0` at `x=0:`

graph{(sinx)^2/x [-6.243, 6.243, -1, 1]}