If $\lim (f(x) + 1/f(x)) = 2 $ prove that $\lim_{x \to 0} f(x) =1 $

First, note that: $$a + \frac{1}{a} - 2 = \frac{a^2 - 2a + 1}{a} = \frac{(a - 1)^2}{a}.$$ So, roughly speaking, given we can make $\left|f(x) + \frac{1}{f(x)} - 2\right|$ as small as we like, we should be able to make $\frac{(f(x) - 1)^2}{|f(x)|}$ as small as we like. There are two ways for this to happen: either $f(x)$ is exceptionally close to $1$ or $|f(x)|$ is extremely, unreasonably large. However, in the latter case, $|f(x)|$ being very large means that $(f(x) - 1)^2$ is even larger, which will prevent the whole fraction from being small, so there really was only one option.

Let's formalise this. First, we need to eliminate the possibility of $|f(x)|$ becoming overly large, so let's try to show that $f(x)$ is bounded on some neighbourhood of $x = 0$.

I'm going to pick, fairly arbitrarily, $\varepsilon = 1$, and apply it to the limit definition of $f(x) + 1/f(x) \to 2$. Then, there exists some $\delta_0 > 0$ such that \begin{align*} 0 < |x| < \delta_0 &\implies \left|f(x) + \frac{1}{f(x)} - 2\right| = \frac{(f(x) - 1)^2}{|f(x)|} < 1 \\ &\implies f(x)^2 - 2f(x) + 1 < |f(x)| \\ &\implies f(x)^2 + (\pm 1 - 2)f(x) + 1 < 0, \end{align*} where the $\pm$ depends on the sign of $f(x)$. In either case, we have a convex parabola, and the solutions to the above inequalities will be bounded. That is, there exists some $M$ such that $|f(x)| \le M$ for all $x \in (-\delta_0, \delta_0) \setminus \{0\}$. Since $f(x) \neq 0$ for at least some $x$ near $0$, we know $M > 0$.

Now we prove the limit. Suppose $\varepsilon > 0$. We have, for $0 < |x| < \delta_0$, \begin{align*} |f(x) - 1| < \varepsilon &\impliedby (f(x) - 1)^2 < \varepsilon^2 \\ &\impliedby \frac{(f(x) - 1)^2}{M} < \frac{\varepsilon^2}{M} \\ &\impliedby \frac{(f(x) - 1)^2}{|f(x)|} < \frac{\varepsilon^2}{M} \\ &\impliedby \left|f(x) + \frac{1}{f(x)} - 2\right| < \frac{\varepsilon^2}{M}. \end{align*} Using the definition of the known limit, there is some $\delta_1$ such that $$0 < |x| < \delta_1 \implies \left|f(x) + \frac{1}{f(x)} - 2\right| < \frac{\varepsilon^2}{M},$$ and hence $$0 < |x| < \min\{\delta_0, \delta_1\} \implies |f(x) - 1| < \varepsilon,$$ completing the proof.

You could not assume the limit of $f$ exists, since the statement doesn’t include it.

By the definition of limit, given $\varepsilon>0,$ there exists $\delta>0$ such that $$2-\varepsilon <f(x)+\frac 1 {f(x)}< 2+\varepsilon$$ for $x\in (0,\delta).$ Therefore $$|\sqrt{f(x)}-\frac 1 {\sqrt{f(x)}}|<\sqrt \varepsilon.$$ Hence $$|f(x)-1|<\sqrt{\varepsilon f(x)}<\sqrt{\varepsilon(2+\varepsilon)}$$ and the conclusion follows.

Since $f(x) + 1/f(x) \geqslant 2$, for any $\epsilon > 0$ there exists $\delta > 0$ such that if $|x| < \delta$ we have

$$0 \leqslant f(x) +\frac{1}{f(x)} - 2 =\underbrace{f(x) -1 + \frac{1}{f(x)} - 1}_{A} = \underbrace{(f(x) -1)\left(1 - \frac{1}{f(x)}\right)}_{B} < \epsilon,$$ Hence,

$$(f(x) -1)^2 \leqslant (f(x) -1)^2 + \left(\frac{1}{f(x)} - 1\right)^2 = A^2 + 2B < \epsilon^2 + 2\epsilon,$$

and for all $|x| < \delta$ we have $|f(x) - 1| < \sqrt{\epsilon^2 + 2\epsilon}$. Therefore, $f(x) \to 1$ as $x \to 0$.