If $\lim_{x\to 0}\left(f(x)+\frac{1}{f(x)}\right)=2,$ show that $\lim_{x\to 0}f(x)=1$.
1st Solution. Although not the most straightforward one, let me present a quick solution: First, we note that
$$ \lim_{x\to0} \left| f(x) - \frac{1}{f(x)} \right| = \lim_{x\to0} \sqrt{\left(f(x) + \frac{1}{f(x)} \right)^2 - 4} = 0, $$
Then by using $\max\{a,b\} = \frac{a+b}{2} + \frac{|a-b|}{2}$ and $\min\{a,b\} = \frac{a+b}{2} - \frac{|a-b|}{2}$ which hold for any $a, b \in \mathbb{R}$, we get
$$ \lim_{x\to0} \max\biggl\{ f(x), \frac{1}{f(x)} \biggr\} = 1 = \lim_{x\to0} \min\biggl\{ f(x), \frac{1}{f(x)} \biggr\}. $$
Now the desired conclusion follows by the squeezing theorem.
2nd Solution. We have
$$ \left| f(x) - 1 \right| = \frac{f(x)}{f(x)+1} \left|f(x) - \frac{1}{f(x)}\right| \leq \left|f(x) - \frac{1}{f(x)}\right|. $$
Since we know that $\lim_{x\to0} \left| f(x) - \frac{1}{f(x)} \right| = 0$, the desired claim follows by the squeezing theorem.
If the result is false, then there exists $\epsilon>0$ such that no $\delta>0$ works. Thus there exists a sequence $x_n\to 0$ such that $|f(x_n)-1|\ge \epsilon$ for all $n.$ WLOG, $f(x_n)\ge1+\epsilon$ for all $n.$
Let $g(x) = x+1/x$ for $x\in [1,\infty).$ It's easy to see that $g$ is strictly increasing on this interval. Thus we have $(g\circ f)(x_n) \ge g(1+\epsilon) > 1$ for all $n.$ It follows that $\lim_{x\to 0}(f(x)+1/f(x))=1$ is false, contradiction.
By definition of limit we have $\forall \varepsilon>0$
$$\left| f(x) + \frac{1}{f(x)} - 2 \right|=\left| \frac{(f(x)-1)^2}{f(x)} \right| < \varepsilon$$
and since
$$\left| \frac{(x-1)^2}{x} \right| < 1 \implies \left|\frac{x-1}x\right|<\frac{\sqrt 5+1}2<2$$
assuming wlog $\varepsilon <1$ we have
$$\left| \frac{(f(x)-1)^2}{f(x)} \right| =\left|f(x)-1 \right|\left| \frac{f(x)-1}{f(x)} \right|< 2\left|f(x)-1 \right|<\varepsilon \implies \left|f(x)-1 \right|<\frac{\varepsilon}2$$