Prove that the limit exists and then find its limit We are given a function $h(x)$ which strictly positive. The function $h$ is defined on $\mathbb{R}$ and that satisfies the following:\

$$\lim_{x\to 0}(h(x)+\frac{1}{h(x)})=2$$

The question is to prove that the limit of $h$ exists at$ 0$ and then find its limit as $x\to 0$

If we assume that the limit exists (say it's equal to $\lambda$, then we have $\lambda+\frac{1}{\lambda}=2$ and by solving this equation, we get $\lambda=1$. However, I am completely clueless on how to prove the existence of the limit of $h(x)$ when $x\to 0$.
We have $h(x) +\dfrac 1{h(x)} \to 2$, so $$f(x) :=h(x) +\dfrac 1{h(x)}-2=\frac{h(x)^2 -2h(x) + 1}{h(x)}=\frac{(h(x) - 1)^2}{h(x)}\to 0$$as $x \to 0$. 

Show that for $x$ sufficiently small $h(x)$, we have $\frac12&lt;h(x) &lt; 2$ (any upper bound will work). So $$f(x)&gt;\frac12\big(h(x)-1\big)^2$$

Now can you use the squeeze theorem to derive the result?
Note that if either $h(x)$ or $\frac{1}{h(x)}$ is not bounded in $(-\varepsilon, \varepsilon)$ then the original limit cannot exist. We'll use that property further, refering it as $(*)$

$$\lim_{x\to 0}\left(h(x) + \frac{1}{h(x)}\right) = 2 \Leftrightarrow \lim_{x\to 0}\left(h(x) -2 + \frac{1}{h(x)}\right) = 0 \Leftrightarrow \lim_{x\to 0}\left(\sqrt{h(x)} - \frac{1}{\sqrt{h(x)}}\right)^2 = 0 \Leftrightarrow \lim_{x\to 0}\left(\frac{h(x) - 1}{\sqrt{h(x)}}\right)^2 = 0 \Leftrightarrow \lim_{x\to 0}\frac{(h(x) - 1)^2}{h(x)} = 0 \stackrel{(*)}{\Rightarrow} \lim_{x\to 0}(h(x)-1)^2 = 0$$

Wich means that $$\lim_{x\to 0} h(x) = 1$$