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?
 A: First, note that:
$$a + \frac{1}{a} - 2 = \frac{a^2 - 2a + 1}{a} = \frac{(a - 1)^2}{a}.$$
So, roughly speaking, given we can make $\left|f(x) + \frac{1}{f(x)} - 2\right|$ as small as we like, we should be able to make $\frac{(f(x) - 1)^2}{|f(x)|}$ as small as we like. There are two ways for this to happen: either $f(x)$ is exceptionally close to $1$ or $|f(x)|$ is extremely, unreasonably large. However, in the latter case, $|f(x)|$ being very large means that $(f(x) - 1)^2$ is even larger, which will prevent the whole fraction from being small, so there really was only one option.
Let's formalise this. First, we need to eliminate the possibility of $|f(x)|$ becoming overly large, so let's try to show that $f(x)$ is bounded on some neighbourhood of $x = 0$.
I'm going to pick, fairly arbitrarily, $\varepsilon = 1$, and apply it to the limit definition of $f(x) + 1/f(x) \to 2$. Then, there exists some $\delta_0 > 0$ such that
\begin{align*}
0 < |x| < \delta_0 &\implies \left|f(x) + \frac{1}{f(x)} - 2\right| = \frac{(f(x) - 1)^2}{|f(x)|} < 1 \\
&\implies f(x)^2 - 2f(x) + 1 < |f(x)| \\
&\implies f(x)^2 + (\pm 1 - 2)f(x) + 1 < 0,
\end{align*}
where the $\pm$ depends on the sign of $f(x)$. In either case, we have a convex parabola, and the solutions to the above inequalities will be bounded. That is, there exists some $M$ such that $|f(x)| \le M$ for all $x \in (-\delta_0, \delta_0) \setminus \{0\}$. Since $f(x) \neq 0$ for at least some $x$ near $0$, we know $M > 0$.
Now we prove the limit. Suppose $\varepsilon > 0$. We have, for $0 < |x| < \delta_0$,
\begin{align*}
|f(x) - 1| < \varepsilon &\impliedby (f(x) - 1)^2 < \varepsilon^2 \\
&\impliedby \frac{(f(x) - 1)^2}{M} < \frac{\varepsilon^2}{M} \\
&\impliedby \frac{(f(x) - 1)^2}{|f(x)|} < \frac{\varepsilon^2}{M} \\
&\impliedby \left|f(x) + \frac{1}{f(x)} - 2\right| < \frac{\varepsilon^2}{M}.
\end{align*}
Using the definition of the known limit, there is some $\delta_1$ such that
$$0 < |x| < \delta_1 \implies \left|f(x) + \frac{1}{f(x)} - 2\right| < \frac{\varepsilon^2}{M},$$
and hence
$$0 < |x| < \min\{\delta_0, \delta_1\} \implies |f(x) - 1| < \varepsilon,$$
completing the proof.
A: 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.
A: 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 $x \to 0$.
A: Here is a very tedious answer. User's answer is much more succinct.
Let $\phi(x) = x+ {1\over x}$. Note that $\phi(x) \ge 2 $ for all $x$.
If $y\ge 2$, then $\phi(x)=y$ has at most two solutions given by
$x_1(y) = {1 \over 2} (y-\sqrt{y^2-4})$ and $x_2(y) = {1 \over 2} (y+\sqrt{y^2-4})$.
Suppose $2 \le y \le 3$.
Note that $x_1(y)-1 = {1 \over 2} (y-2-\sqrt{y^2-4}) = {1 \over 2} \sqrt{y-2}(\sqrt{y-2}-\sqrt{y+2})$ and so
$|x_1(y)-1| \le   {1 \over 2} \sqrt{y-2} | \sqrt{y-2}-\sqrt{y+2} | \le {1 +\sqrt{5}\over 2} \sqrt{y-2} \le 2 \sqrt{y-2}$.
The same analysis shows $|x_2(y)-1| \le 2 \sqrt{y-2}$.
In particular, given $\epsilon>0$, if we choose $\delta= \min(1, {\epsilon^2 \over 4})$, then if $|y-2| < \delta$ we have $\max(|x_1(y)-1|, |x_2(y)-1|) < \epsilon$.
We are also given that for any $ \epsilon'>0$ there is some $\delta'>0$ such that if $|x| < \delta'$ then $|\phi(f(x))-2| < \epsilon'$.
Hence we choose $\epsilon>0$, let $\epsilon'= \min(1, {\epsilon^2 \over 4})$ and choose a $\delta' >0$ such that if $|x| < \delta'$ then
$|\phi(f(x))-2| < \epsilon'$. Then since $f(x)\in \{  x_1(\phi(f(x)), x_1(\phi(f(x)) \}$ we have $|f(x)-1| < \epsilon$.
