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

1 Answer
Aug 30, 2015

Answer:

#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)#.