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I need to prove this limit:

Given $f:(-1,1) \to \mathbb{R}\,$ and $\,f(x)>0,\,$ if $\,\lim_{x\to 0} \left(f(x) + \dfrac{1}{f(x)}\right) = 2,\,$ then $\,\lim_{x\to 0} f(x) = 1$.

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Is it not rigorous enough to do this by direct substitution? – Ovi Jul 16 '13 at 15:54
@Ovi $f$ has to be continuous for direct substitution to be possible. – sigmatau Jul 16 '13 at 15:56

Assume $f(x)$ does not converge towards $1$ when $x\to 0$. That means there exists $\epsilon>0$ and a sequence $(x_n)_{n\in\mathbb{N}}$ such that $x_n\to 0$, and $|f(x_n)-1|>\epsilon$ for all $n$, so either $f(x_n)>1+\epsilon$ or $f(x_n)<1-\epsilon$.

Now, notice that the function $g:x\mapsto x+1/x$ admits a strict minimum for $x=1$. That means there exists $\sigma\in\mathbb{R}$ such that in both of the above cases, $g(f(x_n))>2+\sigma$, which is impossible, hence the result.

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Let $z=\lim_{x\to 0} f(x)$, then the first limit can be written as, $z+\dfrac{1}{z}=2$. Equivalently, we can write, $z^2 -2z +1=0$ or $(z-1)^2=0$.

Thus $z=1$ and so $\lim_{x\to 0} f(x)=1$.

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This assumes that $\lim \limits_{x\to 0}(f(x))$ exists. That's the hard part. – Git Gud Jul 16 '13 at 16:01
the function is bounded below. a theorem in analysis affirm that is this situation the limit exists. – math student Jul 16 '13 at 16:04
@LeandroTavares Why is it bounded below? And why isn't the limit $0$? – Git Gud Jul 16 '13 at 16:05
@ Git Gud because $0 < f(x)$ – math student Jul 16 '13 at 16:06
Note that $\lim_{x\to 0}f(x)+\frac1{f(x)}$ might exist even if $\lim_{x\to 0}f(x)$ does not exist. – Hagen von Eitzen Jul 16 '13 at 16:10

Let $g(x)=\max\{f(x),\frac1{f(x)}\}$ and note that $$\begin{align}h\colon [1,\infty)&\to[2,\infty)\\x&\mapsto x+\frac1x\end{align}$$ is a continous bijection with continuous inverse $$\begin{align}h^{-1}\colon [2,\infty)&\to[1,\infty)\\y&\mapsto \left(\frac{\sqrt{y-2}+\sqrt{y+2}}2\right)^2\end{align}$$ (the square roots ocurring here are $\sqrt x\pm\frac1{\sqrt x}$ because $y\pm 2=x\pm2+\frac1x$).

Since $h(g(x))=f(x)+\frac1{f(x)}\to 2$ as $x\to 0$, we conclude $g(x)\to h^{-1}(2)=1$. Hence if $\epsilon>0$. Then for $x$ sufficiently close to $1$ we have $g(x)<1+\epsilon$, i.e. $$1-\epsilon=\frac{1-\epsilon^2}{1+\epsilon}<\frac1{1+\epsilon}<f(x)<1+\epsilon.$$ This precisely says that $$\lim_{x\to1}f(x)=1.$$

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