How to find continuities with square root? I don't understand how to find $$\frac{4-x^2}{3-\sqrt{x^2+5}}$$
The book says to multiply the equation by $\frac{3 + \sqrt {x^2+5}}{3 + \sqrt {x^2+5}}$. I don't understand where that comes from. It says the multiplication simplifies to "$3 + \sqrt {x^2+5}$" - I don't see how that's possible. Is the book wrong?

Just use the binomial identity
$$(a-b)(a+b)=a^2-b^2,$$
which for $a=3$ and $b=\sqrt{x^2+5}$ yields
$$\frac{(4-x^2)(3+\sqrt{x^2+5})}{(3-\sqrt{x^2+5})(3+\sqrt{x^2+5})}=\frac{(4-x^2)(3+\sqrt{x^2+5})}{9-(x^2+5)}=\frac{(4-x^2)(3+\sqrt{x^2+5})}{(4-x^2)}$$
which is your desired result $(3+\sqrt{x^2+5}).$
