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?

For this simplification, you need to know that $(a+b)(a-b)=a^2-b^2$. It is important to recognize this, as this is something you need very often. You also need this when simplifying expressions with square roots in the denominator. In this case, this gives us the following simplification: $(3+\sqrt{x^2+5})(3-\sqrt{x^2+5})=3^2-\left(\sqrt{x^2+5}\right)^2=9-(x^2+5)=4-x^2$.
For this, we subsitute $a=3$ and $b=\sqrt{x^2+5}$ in the identity $(a+b)(a-b)=a^2-b^2$. 
This means that \begin{align*} \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)\left(3+\sqrt{x^2+5}\right)}{\left(3-\sqrt{x^2+5}\right)\left(3+\sqrt{x^2+5}\right)} \\ &= \frac{(4-x^2)\left(3+\sqrt{x^2+5}\right)}{4-x^2} \\ &= 3+\sqrt{x^2+5}\end{align*}

However, this is not valid for all $x$, since the former expression is not defined for all $x$, while the latter expression is. This is where the question is actually about: Finding the discontinuities of the function. For this, the following is useful:

Fact. If $f(x)$ and $g(x)$ are continuous, then $\frac{f(x)}{g(x)}$ is continuous where $g(x) \neq 0$. 

So we need to find where $3-\sqrt{x^2+5}=0$. So $\sqrt{x^2+5}=3$, so $x^2+5=9$ and $x^2=4$ which means that $x=2$ or $x=-2$. These both satisfy the given equation. 
