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 $x\to 0$.
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$\begingroup$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 $\frac12<h(x) < 2$ (any upper bound will work). So $$f(x)>\frac12\big(h(x)-1\big)^2$$
Now can you use the squeeze theorem to derive the result?
$\endgroup$ 1 $\begingroup$Suppose there is a sequence $(x_n)$ such that $x_n \to 0$ but $h(x_n) \to 1$ is not true.
Then there is a $\delta > 0$ such that either,
a) $ h(x_n) > 1 + \delta$ for infinitely many values of $n$,or,
b) $0 < h(x_n) < \dfrac{1}{1+\delta}$ for infinitely many values of n.
If b) holds assume (by choosing a subsequence if necessary) that $ 0 < h(x_n) < \dfrac{1}{1+\delta}$ for all $n$.
The boundedness of $h(x_n)$ implies we can find a subsequence $(y_n)$ of $(x_n)$ such that $\lim h(y_n) = L$ exists and $ 0 \leq L \leq \dfrac{1}{1+\delta} < 1 $.
$L \neq 0 $ as a 0 limit will contradict $h(y_n) + \dfrac{1}{h(y_n)} \to 2 $ a consequence of the given condition.
But this implies $\dfrac{1}{h(y_n)}$ also converges to $\dfrac{1}{L}$ with $L + \dfrac{1}{L} = 2$ and $L = 1$, a contradiction to $ L < 1$.
If $h(x_n) > 1 + \delta$ infinitely often, then $ 0 < \dfrac{1}{h(x_n)} < 1/(1+\delta)$ infinitely often, and the argument proceeds similarly to obtain a contradiction.
Hence, for any $x_n \to 0$ $h(x_n) \to 1$, hence $\lim h(x) = 1$ as $x \to 0.$
$\endgroup$ $\begingroup$Let $g(x)=\sqrt{h(x)}$. We have $$h(x)-\frac{1}{h(x)}-2=\left(g(x)-\frac{1}{g(x)}\right)^2.$$
This has limit $0$, so $g(x)-\frac{1}{g(x)}$ has limit $0$. We show that $g(x)$ has limit $1$. Note that $|g(x)|$ is bounded above, so $g^2(x)-1$ has limit $0$. Since $g(x)+1\gt 1$, it follows that $g(x)-1$ has limit $0$.
$\endgroup$ $\begingroup$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!
$\endgroup$ 1 $\begingroup$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 $$\lim_{x\to 0} h(x) = 1$$
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