<s>[INST] Show that $\lim\limits_{x\rightarrow 0}f(x)=1$ 
Suppose a function $f:(-a,a)-\{0\}\rightarrow(0,\infty)$ satisfies  $\lim\limits_{x\rightarrow 0}\left(f(x)+\frac{1}{f(x)}\right)=2$. Show that $$\lim\limits_{x\rightarrow 0}f(x)=1$$


Let $\epsilon>0$ , then there exists a $\delta>0$ such that $$\left(f(x)+\frac{1}{f(x)}\right)-2<\epsilon\;\;\;\text{and}\;\;\;|x|<\delta$$
Then 
\begin{align}
&\left(f(x)+\frac{1}{f(x)}\right)-2\\
=&\left(f(x)-1\right) +\left(\frac{1}{f(x)}-1\right)<\epsilon\tag{1}
\end{align}
Squaring $(1)$ both sides gives $$\left(\left(f(x)-1\right) +\left(\frac{1}{f(x)}-1\right)\right)^2<\epsilon^2\tag{2}$$
Since $(f(x)-1)^2\leq\left(\left(f(x)-1\right) +\left(\frac{1}{f(x)}-1\right)\right)^2$, by using $(2)$, $(f(x)-1)^2<\epsilon^2\Rightarrow f(x)-1<\epsilon$; therefore, as $\epsilon$ is arbitrary, $\lim\limits_{x\rightarrow 0}f(x)=1$

Can someone give me a hint to do this question without using epsilon-delta definition? Thanks
 [/INST] Define $h(u) = \frac{u + \sqrt{u^2 - 4}}{2}$ (this is the inverse function of $x \mapsto x + \frac{1}{x}$). Note that $h(2) = 1$ and that $h$ is continuous at $u = 2$. Thus,
$$ 1 = \lim_{u \to 2} h(u) = \lim_{x \to 0} h \left( f(x) + \frac{1}{f(x)} \right) = \lim_{x \to 0} \frac{f(x) + \frac{1}{f(x)} + \sqrt{\left( f(x) + \frac{1}{f(x)} \right)^2 - 4}}{2} = \lim_{x \to 0} \frac{f(x) + \frac{1}{f(x)} + \sqrt{ \left( f(x) - \frac{1}{f(x)} \right)^2}}{2} = 1 + \lim_{x \to 0} \left| f(x) - \frac{1}{f(x)} \right| $$
which implies that 
$$\lim_{x \to 0} \left( f(x) - \frac{1}{f(x)} \right) = 0. $$
Combining both results together we have
$$ 2 = 2 + 0 = \lim_{x \to 0} \left( f(x) + \frac{1}{f(x)} \right) + \lim_{x \to 0} \left( f(x) - \frac{1}{f(x)} \right) = \lim_{x \to 0} 2f(x) $$
which implies that $\lim_{x \to 0} f(x) = 1$.
</s>