If $\lim (f(x) + 1/f(x)) = 2 $ prove that $\lim_{x \to 0} f(x) =1 $ 
Let $f:(-a,a) \setminus \{ 0 \} \to (0 , \infty) $ and assume $\lim_{x
 \to 0} \left( f(x) + \dfrac{1}{f(x) } \right) = 2$. Prove using the
definition of limit that $\lim_{x \to 0} f(x) = 1$

Attempt:
Let $L = \lim_{x \to 0} f(x) $. Let $\epsilon > 0$ be given. If we can find some $\delta > 0$ with $|x| < \delta $ such that $|f(x) - 1 | < \epsilon $ then we will be done.
We know that since $f(x) > 0$, then $\lim 1/f(x) $ is defined. In fact, applying the limit to hypothesis, we end up with
$$ L+ \dfrac{1}{L} = 2 $$
and certainly $L=1$ as desired. I am having difficulties making this proof formal in $\delta-\epsilon$ language. Can someone assist me?

Since $f(x) + 1/f(x) \geqslant 2$, for any $\epsilon > 0$  there exists $\delta > 0$ such that if $|x| < \delta$ we have
$$0 \leqslant f(x)  +\frac{1}{f(x)} - 2 =\underbrace{f(x) -1 + \frac{1}{f(x)} - 1}_{A} = \underbrace{(f(x) -1)\left(1 - \frac{1}{f(x)}\right)}_{B} < \epsilon,$$
Hence, 
$$(f(x) -1)^2 \leqslant (f(x) -1)^2 + \left(\frac{1}{f(x)} - 1\right)^2 = A^2 + 2B < \epsilon^2 + 2\epsilon,$$
and for all $|x| < \delta$ we have $|f(x) - 1| < \sqrt{\epsilon^2 + 2\epsilon}$. Therefore,  $f(x) \to 1$ as $x \to 0$.
