Prove limit of function

Question

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 $\,\lim_{x\to 0} f(x) = 1$.

Answer 1

Hint Prove first that there exists some $\delta$ such that $f$ is positive and bounded on $(x- \delta, x+ \delta)$.

Hint 2: $$\lim \limits_{x \to a}f(x)+\frac{1}{f(x)}=2 \Leftrightarrow \\ \lim \limits_{x \to a}(f(x)+\frac{1}{f(x)}-2)=0 \Rightarrow \\ \lim \limits_{x \to a}(f(x)^2-2 f(x)+1)=0 $$ with the last step following from the fact that $f$ is positive and bounded.

Thus $$\lim \limits_{x \to a}(f(x)-1)^2=0 $$ Tke the square root now.

Answer 2

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.

Answer 3

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 $$\lim_{x\to1}f(x)=1.$$

Answer 4

Let \begin{align} &a=\liminf_{x\to 0}f(x)&&b=\limsup_{x\to 0}f(x) \end{align} Since $$\liminf_{x\to 0}\left(f(x)+\frac 1{f(x)}\right)\leq\liminf_{x\to 0}f(x)+\limsup_{x\to 0}\frac 1{f(x)}\leq\limsup_{x\to 0}\left(f(x)+\frac 1{f(x)}\right)$$ we get $$2\leq a+\frac 1a\leq 2$$ from which $a=1$. Similarly, $b=1$, hence $\lim_{x\to 0}f(x)=1$.

Answer 5

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 $\lim_{x\to 0} f(x)=1$.

Answer 6

Write $c(x) = f(x) + \frac{1}{f(x)}$. Solve a quadratic equation to see that $f(x)$ is either $(c(x)+\sqrt{c(x)^2-4}\;)/2$ or $(c(x)-\sqrt{c(x)^2-4}\;)/2$ . So, for all $x$, we have $$ \frac{c(x)-\sqrt{c(x)^2-4}}{2} \le f(x) \le \frac{c(x)+\sqrt{c(x)^2-4}}{2} $$ But we are told that $\lim c(x) = 2$, so that $$ \lim \frac{c(x)-\sqrt{c(x)^2-4}}{2} = 1\quad\text{and}\quad \frac{c(x)+\sqrt{c(x)^2-4}}{2} = 1. $$ Our function $f(x)$ is between these, so $\lim f(x) = 1$.