Let $f:(-a,a)\setminus\{0\}\to (0,\infty)$ satisfying
$$\lim_{x\to 0} \left(f(x)+ \frac{1}{f(x)}\right) = 2.$$
Show that $\displaystyle\lim_{x\to 0} f(x) = 1$
My attempt:
Given that $\displaystyle\lim_{x\to 0} \left(f(x)+ \frac{1}{f(x)}\right) = 2$, I know $0<f(x)<f(x)+ \dfrac{1}{f(x)}<\epsilon+2$ for $x \in (-\delta, \delta)$, so $f(x)$ is bounded in some neighborhood of $0$.
Now, $\left|f(x)+ \dfrac{1}{f(x)}-2\right| = \left|\dfrac{(f(x)-1)^2}{f(x)}\right| < \epsilon,$ so $|f(x)-1|<\sqrt{f(x)\epsilon} < \sqrt{(\epsilon+2)\epsilon}$ as long as $x \in (-\delta, \delta)$.
Is this proof correct? I am not sure if I proved that $\displaystyle\lim_{x\to 0} f(x) = 1$, because my new "epsilon" for $f(x)$ ended up being $\sqrt{(\epsilon+2)\epsilon}$ .