<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] $$
a + \frac 1 a = \left( a - 2 + \frac 1 a \right) +2 = \left( \sqrt a - \frac 1 {\sqrt a} \right)^2 + 2 = \text{square} + 2.
$$
So this is $\ge 2$ unless the square is $0$.
$a+ \dfrac 1 a$ cannot get close to $2$ unless $a$ gets close to $1$.
Even if you allow complex numbers (so that the inequality above doesn't hold), we have
$$
a + \frac 1 a = 2 \Longrightarrow a^2 + 1 = 2a
$$
and that is a quadratic equation whose only solution is $a=1$ (a double root).
</s>