# How do you show that the function f(x)=1-sqrt(1-x^2) is continuous on the interval [-1,1]?

May 1, 2015

$f \left(x\right) = 1 - \sqrt{1 - {x}^{2}}$ has domain $\left[- 1 , 1\right]$

For $a$ in $\left(- 1 , 1\right)$, ${\lim}_{x \rightarrow a} f \left(x\right) = f \left(a\right)$ because

${\lim}_{x \rightarrow a} f \left(x\right) = {\lim}_{x \rightarrow a} \left(1 - \sqrt{1 - {x}^{2}}\right) = 1 - {\lim}_{x \rightarrow a} \sqrt{1 - {x}^{2}}$

= 1-sqrt(lim_(xrarra) (1-x^2)) = 1-sqrt(1-lim_(xrarra) x^2))

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

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

Similar reasoning will show that

${\lim}_{x \rightarrow - {1}^{+}} f \left(x\right) = f \left(- 1\right)$ and

${\lim}_{x \rightarrow {1}^{-}} f \left(x\right) = f \left(1\right)$

So $f$ is continuous on $\left[- 1 , 1\right]$.