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

Aug 30, 2015

$f \left(x\right)$ is continuous on $\left(- 1 , 1\right)$.

#### 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

$x < 1 \text{ }$ and $\text{ } - x < 1 \implies x > - 1$

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

That means that you need

$\textcolor{b l u e}{{\lim}_{x \to c} f \left(x\right) = f \left(c\right) , \text{ } \left(\forall\right) c \in \left(- 1 , 1\right)}$

So, for any $c \in \left(- 1 , 1\right)$ you know that you have

$f \left(c\right) = \frac{c}{\sqrt{1 - {c}^{2}}}$

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

$c = {\lim}_{x \to c} x \text{ }$ and $\text{ } \sqrt{1 - {c}^{2}} = \sqrt{1 - {\lim}_{x \to c} {x}^{2}}$

Since those limits exist for $c \in \left(- 1 , 1\right)$, you can say that

$f \left(c\right) = \frac{c}{\sqrt{1 - {c}^{2}}} = \frac{{\lim}_{x \to c} x}{\sqrt{1 - {\lim}_{x \to c} {x}^{2}}}$

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

$\frac{{\lim}_{x \to c} x}{\sqrt{{\lim}_{x \to c} 1 - {\lim}_{x \to c} {x}^{2}}} = \frac{{\lim}_{x \to c} x}{\sqrt{{\lim}_{x \to c} \left(1 - {x}^{2}\right)}}$

The denominatoris equivalent to

$\sqrt{{\lim}_{x \to c} \left(1 - {x}^{2}\right)} = {\lim}_{x \to c} \sqrt{1 - {x}^{2}}$

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

$\frac{{\lim}_{x \to c} x}{{\lim}_{x \to c} \sqrt{1 - {x}^{2}}} = {\lim}_{x \to c} \left(\frac{x}{\sqrt{1 - {x}^{2}}}\right) = {\lim}_{x \to c} f \left(x\right)$

Since $f \left(c\right) = {\lim}_{x \to c} f \left(x\right)$ for any $c \in \left(- 1 , 1\right)$, the function will indeed be continuous on its domain, $\left(- 1 , 1\right)$.