1 Answer

James Wald · Answer 1 · 2023-10-28T17:07:15+0000

The function itself is not well-defined at
x=2 because both numerator and denominator are equal to zero there, so it's discontinuous there. We attempt to remove this discontinuity.

Rationalize the numerator and denominator.

$\displaystyle (x - √(8 - x^2))/(√(x^2 + 12) - 4) \cdot (x + √(8 - x^2))/(x + √(8 - x^2)) \cdot (√(x^2+12) + 4)/(√(x^2 + 12) + 4)$

$\displaystyle = (x^2 - \left(√(8-x^2)\right)^2)/(\left(√(x^2+12)\right)^2 - 4^2) \cdot (√(x^2 + 12) + 4)/(x + √(8 - x^2))$

$\displaystyle = (x^2 - (8-x^2))/((x^2+12)-16) \cdot (√(x^2 + 12) + 4)/(x + √(8 - x^2))$

$\displaystyle = (2x^2 - 8)/(x^2 - 4) \cdot (√(x^2 + 12) + 4)/(x + √(8 - x^2))$

$\displaystyle = 2 ((x-2)(x+2))/((x-2)(x+4)) \cdot (√(x^2 + 12) + 4)/(x + √(8 - x^2))$

Then in the limit, since
$x\\eq2$ , we can cancel both factors of
x-2 and
x+2 , subsequently removing the discontinuities at
$x=\pm2$ , and we're left with