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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?

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  • $\begingroup$ The more delicate part is justifying why we can bring the square root outside the limit ;-) $\endgroup$
    – Ant
    Nov 19 '15 at 17:15
  • $\begingroup$ I'm going to take it as given for now and worry about it later ;D $\endgroup$
    – user265554
    Nov 19 '15 at 17:23
  • $\begingroup$ 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 :-) $\endgroup$
    – Ant
    Nov 19 '15 at 17:43
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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 $x > N$.

This is rather rigorous, it follows directly from the definition itself
(of what we call limit as x approaches positive infinity).

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  • $\begingroup$ Thanks! And we're interested in what happen for x>N, only when x approaches infinity? I mean, if x approaches N, aren't we interested in what happens when x=N? $\endgroup$
    – user265554
    Nov 19 '15 at 17:22
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    $\begingroup$ @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. $\endgroup$ Nov 19 '15 at 17:25
  • $\begingroup$ 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. $\endgroup$ Nov 19 '15 at 19:30
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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.

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