Question

How do you find the interval notation to prove `f(x)= x/(sqrt(1-x^2))` is continuous?

Answer 1

`f(x)` is continuous on `(-1, 1)`.

Explanation:

First, identify the domain of the function.

Since you're dealing with the square root of an expression, you need that expression to be positive for any value of `x` in the function's domain.

Moreover, you need that expression to be positive and different from zero, since taking the square root of zero would produce a division by zero.

So, you need

`1 - x^2 > 0`

`x^2 < 1`

`sqrt(x^2) < sqrt(1) implies |x| < 1`

This means that you have

`x < 1" "` and `" "-x < 1 implies x > -1`

The domain of the function will be `(-1, 1)`. In order for the function to be continuous on its domain, you need it to be continuous at each point `c in (-1,1)`.

That means that you need

`color(blue)(lim_(x -> c) f(x) = f(c)," "(AA) c in (-1,1))`

So, for any `c in (-1,1)` you know that you have

`f(c) = c/sqrt(1-c^2)`

Now focus on the numerator and denominator if this fraction. You can write

`c = lim_(x ->c) x" "` and `" "sqrt(1-c^2) = sqrt(1 - lim_(x ->c)x^2)`

Since those limits exist for `c in (-1,1)`, you can say that

`f(c) = c/sqrt(1-c^2) = (lim_(x ->c)x)/sqrt(1 - lim_(x->c)x^2)`

You know that the limit of a constant is equal to that constant, so you can write

`(lim_(x->c)x)/sqrt(lim_(x->c)1 - lim_(x->c)x^2) = (lim_(x->c)x)/sqrt(lim_(x->c)(1-x^2))`

The denominatoris equivalent to

`sqrt(lim_(x->c)(1-x^2)) = lim_(x->c)sqrt(1-x^2)`

which means that you get, using the fact that the quotient of the limits is equal to the limit of the quotient

`(lim_(x->c)x)/(lim_(x->c)sqrt(1-x^2)) = lim_(x->c)(x/sqrt(1-x^2)) = lim_(x->c)f(x)`

Since `f(c) = lim_(x->c)f(x)` for any `c in (-1,1)`, the function will indeed be continuous on its domain, `(-1,1)`.