TITLE: If $\lim_{x\to 0}\left(f(x)+\frac{1}{f(x)}\right)=2,$ show that $\lim_{x\to 0}f(x)=1$.
QUESTION [16 upvotes]: Question: Suppose $f:(-\delta,\delta)\to (0,\infty)$ has the property that $$\lim_{x\to 0}\left(f(x)+\frac{1}{f(x)}\right)=2.$$ Show that $\lim_{x\to 0}f(x)=1$.

My approach: Let $h:(-\delta,\delta)\to(-1,\infty)$ be such that $h(x)=f(x)-1, \forall x\in(-\delta,\delta).$ Note that if we can show that $\lim_{x\to 0}h(x)=0$, then we will be done. Now since we have $$\lim_{x\to 0}\left(f(x)+\frac{1}{f(x)}\right)=2\implies \lim_{x\to 0}\frac{(f(x)-1)^2}{f(x)}=0\implies \lim_{x\to 0}\frac{h^2(x)}{h(x)+1}=0.$$ Next I tried to come up with some bounds in order to use Sandwich theorem to show that $\lim_{x\to 0} h(x)=0,$ but the bounds didn't quite work out. The bounds were the following: $$\begin{cases}h(x)\ge \frac{h^2(x)}{h(x)+1},\text{when }h(x)\ge 0,\\h(x)<\frac{h^2(x)}{h(x)+1},\text{when }h(x)<0.\end{cases}$$
How to proceed after this?

REPLY [2 votes]: Setting $y = f(x) + \dfrac{1}{f(x)} \to 2$ we have
$$
f(x) = \frac{y\pm\sqrt{y^2-4}}{2} \to \frac{2\pm\sqrt{2^2-4}}{2} = 1.
$$
The limit can be justified using the squeeze theorem, since $f(x) = \frac{y\pm\sqrt{y^2-4}}{2},$ i.e. $f(x)$ equals either $\frac{y-\sqrt{y^2-4}}{2}$ or $\frac{y+\sqrt{y^2-4}}{2},$ implies $\frac{y-\sqrt{y^2-4}}{2} \leq f(x) \leq \frac{y+\sqrt{y^2-4}}{2}$.