Prove limit of function I need to prove this limit:

Given $f:(-1,1) \to \mathbb{R}\,$ and $\,f(x)>0,\,$ if $\,\lim_{x\to 0} \left(f(x) + \dfrac{1}{f(x)}\right) = 2,\,$ then $\,\lim_{x\to 0} f(x) = 1$.

 A: Assume $f(x)$ does not converge towards $1$ when $x\to 0$. That means there exists $\epsilon>0$ and a sequence $(x_n)_{n\in\mathbb{N}}$ such that $x_n\to 0$, and $|f(x_n)-1|>\epsilon$ for all $n$, so either $f(x_n)>1+\epsilon$ or $f(x_n)<1-\epsilon$.
Now, notice that the function $g:x\mapsto x+1/x$ admits a strict minimum for $x=1$. That means there exists $\sigma\in\mathbb{R}$ such that in both of the above cases, $g(f(x_n))>2+\sigma$, which is impossible, hence the result.
A: Hint Prove first that there exists some $\delta$ such that $f$ is positive and bounded on $(x- \delta, x+ \delta)$.
Hint 2:
$$\lim \limits_{x \to a}f(x)+\frac{1}{f(x)}=2 \Leftrightarrow \\
\lim \limits_{x \to a}(f(x)+\frac{1}{f(x)}-2)=0 \Rightarrow \\
\lim \limits_{x \to a}(f(x)^2-2 f(x)+1)=0 
$$ 
with the last step following from the fact that $f$ is positive and bounded.
Thus
$$\lim \limits_{x \to a}(f(x)-1)^2=0 
$$ 
Tke the square root now.
A: Let $g(x)=\max\{f(x),\frac1{f(x)}\}$ and note that $$\begin{align}h\colon [1,\infty)&\to[2,\infty)\\x&\mapsto x+\frac1x\end{align}$$ is a continous bijection with continuous inverse $$\begin{align}h^{-1}\colon [2,\infty)&\to[1,\infty)\\y&\mapsto \left(\frac{\sqrt{y-2}+\sqrt{y+2}}2\right)^2\end{align}$$ (the square roots ocurring here are $\sqrt x\pm\frac1{\sqrt x}$ because $y\pm 2=x\pm2+\frac1x$).
Since $h(g(x))=f(x)+\frac1{f(x)}\to 2$ as $x\to 0$, we conclude $g(x)\to h^{-1}(2)=1$.
Hence if $\epsilon>0$. Then for $x$ sufficiently close to $1$ we have $g(x)<1+\epsilon$, i.e. $$1-\epsilon=\frac{1-\epsilon^2}{1+\epsilon}<\frac1{1+\epsilon}<f(x)<1+\epsilon.$$
This precisely says that $$\lim_{x\to1}f(x)=1.$$
A: Let $z=\lim_{x\to 0} f(x)$, then the first limit can be written as, $z+\dfrac{1}{z}=2$. Equivalently, we can write, $z^2 -2z +1=0$ or $(z-1)^2=0$.
Thus $z=1$ and so $\lim_{x\to 0} f(x)=1$.
A: Let
\begin{align}
&a=\liminf_{x\to 0}f(x)&&b=\limsup_{x\to 0}f(x)
\end{align}
Since
$$\liminf_{x\to 0}\left(f(x)+\frac 1{f(x)}\right)\leq\liminf_{x\to 0}f(x)+\limsup_{x\to 0}\frac 1{f(x)}\leq\limsup_{x\to 0}\left(f(x)+\frac 1{f(x)}\right)$$
we get
$$2\leq a+\frac 1a\leq 2$$
from which $a=1$.
Similarly, $b=1$, hence $\lim_{x\to 0}f(x)=1$.
A: Write $c(x) = f(x) + \frac{1}{f(x)}$.  Solve a quadratic equation to see that
$f(x)$ is either $(c(x)+\sqrt{c(x)^2-4}\;)/2$ or $(c(x)-\sqrt{c(x)^2-4}\;)/2$ .  So, for all $x$, we have
$$
\frac{c(x)-\sqrt{c(x)^2-4}}{2} \le f(x) \le \frac{c(x)+\sqrt{c(x)^2-4}}{2}
$$
But we are told that $\lim c(x) = 2$, so that
$$
\lim \frac{c(x)-\sqrt{c(x)^2-4}}{2} = 1\quad\text{and}\quad \frac{c(x)+\sqrt{c(x)^2-4}}{2} = 1.
$$
Our function $f(x)$ is between these, so $\lim f(x) = 1$.
