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 $x\to 0$.

Let $g(x) = x + x^{-1}$. Then your limit is
$$ \lim_{x \to 0} g(h(x)) = 2 $$
$g$ isn't quite an invertible function: for every point $a \in (2, \infty)$ there are two solutions to $g(x) = a$. But in some sense it's a continuous two-valued 'function': if I let $f^+$ and $f_-$ be the larger and smaller of the two solutions, then they are both continuous functions:
$$ \lim_{x \to 0} f_+(g(h(x))) = f_+(2) = 1 $$
$$ \lim_{x \to 0} f_-(g(h(x))) = f_-(2) = 1 $$
So in some sense, we whould be able to say
$$ \lim_{x \to 0} g^{-1}(g(h(x)) = g^{-1}(2) = \{1\} $$
$$ \lim_{x \to 0} \{ h(x), h(x)^{-1} \} = \{1\} $$
and since we just get one limit point, the individual limits should exist (i.e. $h(x) \to 1$). All that's left is finding a way to make a precisely statement that captures this idea, and then prove the theorem saying we can find limits like this. This is a fair bit of work, but a good idea in the long run since it gives us a new tool useful for dealing with more problems!
