When can I move the limit operand into a function? When can I move $\lim$ inside an expression? what are the requirements from the function?
For example: $\displaystyle \lim_{x \to \infty}\frac{\sqrt{x^2+2}}{3x-6}$
 A: The function has to be continuous. Since continuity at infinity is a controversial concept, change it to a more comfortable situation by setting $x=1/y$.
A: By definition of the continuous function:

Function is continuous at $x_0$ iff $\lim_{x\to x_0} f(x)$ exists and
$$\lim_{x\to x_0} f(x) = f(x_0)$$


Thus, if function $f$ is continuous at $g$ and $\lim\limits_{x\to x_0} g(x)=g$ then:
$$\lim_{x\to x_0}f(g(x))=f\left(\lim_{x\to x_0} g(x)\right)$$
When you are operating on $\pm\infty$ you can flip the function inside out by substitution $x=\tfrac1u$ so that you are analyzing continuity at $u=0$.
A: Your example:
$$\lim_{x\to\infty}\frac{\sqrt{x^2+1}}{3x-6}=$$
$$\lim_{x\to\infty}\frac{\sqrt{x^2+1}}{3x}=$$
$$\lim_{x\to\infty}\frac{\sqrt{x^2}}{3x}=$$
$$\frac{1}{3}\lim_{x\to\infty}\frac{\sqrt{x^2}}{x}=$$
$$\frac{1}{3}\lim_{x\to\infty}\frac{x}{x}=$$
$$\frac{1}{3}\lim_{x\to\infty}1=\frac{1}{3}$$
A: You can do this, if both the limits exist. In this case both of them are $\infty$, so they don't exist and you can't move the $\lim$ inside.
In case you need help to solve this limit, try dividing numerator and denominator by $x$ or use l'Hopital's rule if you are familiar with it.
