# Non-rigorous limits to infinity trouble

I had to solve this problem: $$\lim \limits_{x \to ∞} {x\over \sqrt {3x^2+2}}$$ and I had no idea how to get rid of the square root from the denominator. I googled for some time and found out that you can do this: $$\lim \limits_{x \to ∞} {x\over \sqrt {3x^2+2}} = \lim \limits_{x \to ∞} {\sqrt {x^2}\over \sqrt {3x^2+2}} = \sqrt {\lim \limits_{x \to ∞} {x^2\over {3x^2+2}}} = ...={1\over \sqrt3}$$ The part that bothers me is $$x=\sqrt {x^2}$$ because it works like this only if x is not negative. If, for example, x=3, it works, but if x=-3, then it says that x=-3=3. Can we still do this because technically x is non-negative since it's approaching infinity? It bothers me because x really isn't non-negative, we don't know what it is, it just approaches infinity. This is high school calculus so we treat limits very non-rigorously; could someone shed some light on this with something with a bit more rigor?

• The more delicate part is justifying why we can bring the square root outside the limit ;-)
– Ant
Nov 19 '15 at 17:15
• I'm going to take it as given for now and worry about it later ;D Nov 19 '15 at 17:23
• If it interests you, since $f(x) = \sqrt x$ is a continuos function, this means that $$\lim_{x \to x_0} f(x) = f(x_0) \implies \lim_{x \to x_0} \sqrt x = \sqrt{\lim_{x \to x_0} x}$$ so you can bring the limit outside :-)
– Ant
Nov 19 '15 at 17:43

This is fine as you're looking for the limit as x approaches positive infinity. So you're not interested in any negative values for x. In fact for any given real number N, you don't care what happens when $x \le N$, you're only interested in what happens for $x > N$.
• @user265554: If $x\to N$, then you are emphatically not interested at what happens at $x=N$; this is the point of the "$0<$" in the $\delta$-half of the limit definition. Nov 19 '15 at 17:25
• If x approaches N, you're interested only in an "epsilon-area" (I am not sure of the English term) around N. So you're not interested in any values $x \ge N+\epsilon$ or any values $x \le N-\epsilon$. And you can pick epsilon to be as small as you want. Nov 19 '15 at 19:30
The part that bothers me is $$x=\sqrt {x^2}$$ because it works like this only if x is not negative.
A substitution that works for all $\mathbb{R}$ is $$x = \begin{cases} \sqrt{x^2} & \text{for } x \ge 0 \\ -\sqrt{(-x)^2} & \text{for } x \le 0 \end{cases}$$ Albeit for your specifc limit you will use only the non-negative case at some point, so you do not have to worry about the other case.