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

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Answer 1 · 2015-05-01T20:31:56

$f(x) = 1 - sqrt(1-x^2)$ has domain $[-1,1]$

For $a$ in $(-1,1)$ , $lim_(xrarra) f(x) = f(a)$ because

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

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

$=1-sqrt(1-a^2) = f(a)$

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

Similar reasoning will show that

$lim_(xrarr-1^+) f(x) = f(-1)$ and

$lim_(xrarr1^-) f(x) = f(1)$

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

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