Squeeze Theorem to evaluate the limit of $\arctan(x)$ as $x$ approaches $0$.

Question

I need to use $$\frac{x}{1+x^2} < \arctan(x) < x$$ to evaluate the limit of $\arctan(x)$ as $x$ approaches $0$. I have attempted to take the limit as $x$ approaches $0$ for each expression in the equality.

So $$\lim_{x\to 0}\frac{x}{1+x^2} < \lim_{x\to 0} \arctan(x)) < \lim_{x\to 0} x$$ as $x$ approaches $0$ but it just leave me with $$0 < \lim_{x\to 0} \arctan(x) < 0$$ which doesn't make sense. What do I do?

Answer 1

Limits do not preserve $<$ (as you have observed: $0<x$, but taking limits as $x\to 0$ clearly doesn't give $0<0$). They do, on the other hand, preserve $\leq$.

So from $$ \frac x{x^2+1}<\arctan(x)<x $$ we can derive $$ \frac x{x^2+1}\leq\arctan(x)\leq x $$ and then, from there, we can just take limits and the inequality still holds.

Theorem: If $f, g:\Bbb R\to \Bbb R$ are functions such that for all $x\in \Bbb R$ we have $f(x)\leq g(x)$, and $\lim_{x\to 0}f(x) = F$ and $\lim_{x\to 0}g(x) = G$ exist, then $F\leq G$.

This theorem is more restrictive than necessary: We don't need $f(x)\leq g(x)$ on all of $\Bbb R$, we don't need the limit to go to $0$, and so on. Yet all these assumptions makes the proof a little bit simpler to read, so I have them in there.

Proof: Assume, for contradiction that $G<F$, and pick $\varepsilon = \frac{F-G}2$. By definition of limit, there is a $\delta>0$ such that for any $x$ with $0<|x|<\delta$ we have $|f(x)-F|<\varepsilon$ and $|g(x)-G|<\varepsilon$.

However, $|f(x)-F|<\frac{F-G}2$ implies $f(x)>\frac{F+G}2$, and similarily we get $g(x)<\frac{F+G}2$. This contradicts $f(x)\leq g(x)$, and finishes the proof.