Let $\lim _{x\to 0}\left(f\left(x\right)+\frac{1}{f\left(x\right)}\right)=2\:$ Prove that $\lim _{x\to 0}\:f\left(x\right)=1\:$ 
Let $f$ be positive function and $\lim _{x\to
0}\left(f\left(x\right)+\frac{1}{f\left(x\right)}\right)=2\:$.  Prove
  that $\lim _{x\to 0}\:f\left(x\right)=1\:$
and prove that $f$  is bounded on a neighbourhood of $0$.

I started to prove it by using $(ε, δ)$-definition 
$$\forall \epsilon >0, \exists \delta>0 :\forall x, |x-c|<\delta\implies|f(x)-L|<\epsilon $$
$\left|f\left(x\right)+\frac{1}{f\left(x\right)}-2\right|< \left|f\left(x\right)\left(f\left(x\right)+\frac{1}{f\left(x\right)}-2\right)\right| = \left|\left(f\left(x\right)-1\right)^2\right|= \left(f\left(x\right)-1\right)^2 <\epsilon$
I get that $|f(x) - 1| < \sqrt{\epsilon}$ and that proves that $f$  is bounded on a neighbourhood of $0$ and for $\delta = \sqrt{\epsilon}$   ,$\lim _{x\to 0}\:f\left(x\right)=1\:$ 
Is that correct ? I do not know how to prove it. 
Thanks .
 A: Since $\lim_{x\rightarrow 0} f(x)+1/f(x)  =2$ then there exists a neighborhood about the origin such that
\begin{align}
\left| f(x)+\frac{1}{f(x)}-2\right|<1  \ \ \implies \ \ f(x)+\frac{1}{f(x)}<3 \ \ \implies \ \ f(x)^2-3f(x)+1<0 
\end{align}
which means $(f(x)-\frac{3}{2})^2<\frac{5}{4}$ or $|f(x)| < \frac{3}{2}+\sqrt{\frac{5}{4}}$. So $f(x)$ is bounded in a neighborhood of the origin. 
Moreover, for every $\varepsilon>0$, there exists $\delta>0$ such that
\begin{align}
\left| f(x)+\frac{1}{f(x)}-2\right|<\varepsilon
\end{align}
when $x \in (-\delta, \delta)$. In particular, we see that
\begin{align}
\left| f(x)+\frac{1}{f(x)}-2\right|=\frac{|f(x)-1|^2}{f(x)}<\varepsilon  \ \implies \ \ |f(x)-1|<\sqrt{\varepsilon f(x)}<M\sqrt{\varepsilon}
\end{align}
since $f(x)$ is bounded. Hence $\lim_{x\rightarrow 0} f(x) = 1$. 
A: $\lim\limits_{x\to0}\left(\sqrt{f(x)}-\frac{1}{\sqrt{f(x)}}\right)^2=0$, so $\lim\limits_{x\to0}\sqrt{f(x)}=1$.
We have used continuity and strict monotonic of function $g(t)=t-\frac{1}{t}$ for $t>0$.
A: Observe that
$$\boxed{\forall f\neq0,\quad \left(f+\dfrac1f -2\right)=\left(f-1 \right)\left(1-\dfrac1f \right)}$$
so
$$(f(x)-1)+\left(\dfrac1{f(x)-1}\right)<\epsilon\tag1$$
because $$\left(f+\dfrac1f -2\right)=\left(f-1 \right)\left(1-\dfrac1f \right)$$
$$(f(x)-1)\left(1-\dfrac1{f(x)}\right)<\epsilon\tag2$$
squaring $(1)$ and using $(2)$
$$0\le(f(x)-1)^2+\left(\dfrac1{f(x)-1}\right)^2\le\epsilon^2+2\epsilon$$
Hence
$$0\le(f(x)-1)^2\le\epsilon^2+2\epsilon$$
$$\Box$$
A: Proof by contradiction would be your best bet. If $\lim_{x\to 0} f(x) = c > 0$ such that $c \neq 1$ then
$$\lim_{x \to 0} f(x) + \frac{1}{f(x)} = c + \frac{1}{c} = \frac{c^2+1}{c} \geq \frac{2c}{c} = 2$$
by AM-GM inequality, with equality only if $c^2 = 1$, a contradiction. 
Then since $\lim_{x\to 0} f(x) = 1$, we have that for a given $\epsilon$, there exists $\delta$ such that
$$|x| < \delta \implies |f(x) - 1| \epsilon$$
thus 
$$|x| < \delta \implies |f(x)| < 1 + \epsilon$$
by triangle inequality.
A: Consider the map $$g(x) = x + \frac{1}{x}-2$$ defined on $I=(0, \infty)$. $g$ is continuous on $I$, non negative and $x=1$ is the only root of $g(x)=0$.
The hypothesis $\lim\limits_{x\to
0}\left(f\left(x\right)+\frac{1}{f\left(x\right)}\right)=2$ implies $\lim\limits_{x \to 0} g(f(x)) = 0$ which with the elements above implies $\lim\limits_{x \to 0} f(x) = 1$.
A: $(\sqrt f -\frac 1 {\sqrt f})^{2}=f+\frac   1 f -2 \to 0$. So there exists $\delta >0$ such that  $|f-1| <\epsilon \sqrt f$ for $|x| <\delta$. This gives $|f| <1+\epsilon \sqrt f$ and hence $(\sqrt f -\frac  {\epsilon } 2)^{2} <1+\frac {\epsilon ^{2}} 4$. Hence $\sqrt f <\frac  {\epsilon } 2+\sqrt (1+\frac {\epsilon ^{2}} 4)$. Now $|f-1| <\epsilon \sqrt f< \epsilon (\frac  {\epsilon } 2+\sqrt (1+\frac {\epsilon ^{2}} 4))<2\epsilon$ for $|x|<\delta$ assuming that  $\epsilon <1$. 
A: rewrite the limit in f inverse t  format 
replace t by 1/t
as no change on rhs u see that f inverse t  = f inverse 1/t
safely we can say t = 1/t replace back in x
hence 2 lim x tend to 0 (f(x))= 2 hence lim x tend to 0 (f(x)) = 1
