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?

$$\frac{4-x^2}{3-\sqrt{x^2+5}}=\frac{4-x^2}{3-\sqrt{x^2+5}}\cdot\frac{3+\sqrt{x^2+5}}{3+\sqrt{x^2+5}}=\frac{(4-x^2)(3+\sqrt{x^2+5})}{3^2-(x^2+5)}$$ $$=\frac{(4-x^2)(3+\sqrt{x^2+5})}{4-x^2}=3+\sqrt{x^2+5}$$
