## Calculus (3rd Edition)

Let $f(x)=g(x)=\sqrt{x}$. One can see that $\lim\limits_{x \to -1}f(x)$ and $\lim\limits_{x \to -1}g(x)$ do not exist (since we can not take the square root of a negative). But the limit $\lim\limits_{x \to -1}f(x)g(x)$ does exist: $$\lim\limits_{x \to -1}f(x)g(x)=\lim\limits_{x \to -1}(\sqrt{x})^2=\lim\limits_{x \to -1}x=-1.$$