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?

You could not assume the limit of $f$ exists, since the statement doesn’t include it.
By the definition of limit, given $\varepsilon>0,$ there exists $\delta>0$ such that 
$$2-\varepsilon <f(x)+\frac 1 {f(x)}< 2+\varepsilon$$
for $x\in (0,\delta).$ Therefore
$$|\sqrt{f(x)}-\frac 1 {\sqrt{f(x)}}|<\sqrt \varepsilon.$$
Hence
$$|f(x)-1|<\sqrt{\varepsilon f(x)}<\sqrt{\varepsilon(2+\varepsilon)}$$
and the conclusion follows.
