If $\lim_{x\to 0}\left(f(x)+\frac{1}{f(x)}\right)=2,$ show that $\lim_{x\to 0}f(x)=1$.

1^st 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.

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2^nd 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

$$\left| \frac{(f(x)-1)^2}{f(x)} \right| =\left|f(x)-1 \right|\left| \frac{f(x)-1}{f(x)} \right|< 2\left|f(x)-1 \right|<\varepsilon \implies \left|f(x)-1 \right|<\frac{\varepsilon}2$$