Let $f:(-a,a)\setminus\{0\}\to (0,\infty)$ satisfying $\lim_{x\to 0} (f(x)+ \frac{1}{f(x)}) = 2.$ Show that $\lim_{x\to 0} f(x) = 1$ 
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 $\sqrt{(\epsilon+2)\epsilon}$ .
 A: $$\begin{align}
  & Set\ g\left( x \right)=f\left( x \right)+\frac{1}{f\left( x \right)} \\ 
 & So \\ 
 & \quad \ g\left( x \right)f\left( x \right)={{f}^{2}}\left( x \right)+1 \\ 
 & Or \\ 
 & {{f}^{2}}\left( x \right)-g\left( x \right)f\left( x \right)+1=0 \\ 
 & f\left( x \right)=\frac{g\left( x \right)\pm \sqrt{g{{\left( x \right)}^{2}}-4}}{2} \\ 
 & \underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=\frac{\underset{x\to 0}{\mathop{\lim }}\,g\left( x \right)\pm \sqrt{\underset{x\to 0}{\mathop{\lim }}\,g{{\left( x \right)}^{2}}-4}}{2}=\frac{2\pm \sqrt{{{2}^{2}}-4}}{2}=1 \\ 
\end{align}$$
A: May I guess that you are looking for something like this:
Given $\epsilon$, pick the $\delta$ for which 
$$|x| < \delta \implies\left|\dfrac{(f(x)-1)^2}{f(x)}\right| < \min\{\frac{\epsilon^2}{\epsilon +2} \ , \ \epsilon \} $$
holds. 
Then $|x| < \delta $ implies 
$$\left|\dfrac{(f(x)-1)^2}{f(x)}\right| < \frac{\epsilon^2}{\epsilon +2} $$
implies
$$|f(x)-1|^2 < \frac{\epsilon^2|f(x)|}{\epsilon +2} < \frac{\epsilon^2(\epsilon+2)}{\epsilon +2} = \epsilon^2$$
since $0<f(x) < \epsilon + 2$. And therefore
$$|f(x)-1| < \epsilon$$
And we are done.
