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Question: Suppose $f:(-\delta,\delta)\to (0,\infty)$ has the property that $$\lim_{x\to 0}\left(f(x)+\frac{1}{f(x)}\right)=2.$$ Show that $\lim_{x\to 0}f(x)=1$.

My approach: Let $h:(-\delta,\delta)\to(-1,\infty)$ be such that $h(x)=f(x)-1, \forall x\in(-\delta,\delta).$ Note that if we can show that $\lim_{x\to 0}h(x)=0$, then we will be done. Now since we have $$\lim_{x\to 0}\left(f(x)+\frac{1}{f(x)}\right)=2\implies \lim_{x\to 0}\frac{(f(x)-1)^2}{f(x)}=0\implies \lim_{x\to 0}\frac{h^2(x)}{h(x)+1}=0.$$ Next I tried to come up with some bounds in order to use Sandwich theorem to show that $\lim_{x\to 0} h(x)=0,$ but the bounds didn't quite work out. The bounds were the following: $$\begin{cases}h(x)\ge \frac{h^2(x)}{h(x)+1},\text{when }h(x)\ge 0,\\h(x)<\frac{h^2(x)}{h(x)+1},\text{when }h(x)<0.\end{cases}$$

How to proceed after this?

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  • $\begingroup$ 𝑓:(−𝛿,𝛿)→(0,∞). I would have thought 𝑓:ℝ→ℝ, no? $\endgroup$
    – mjw
    Aug 28 '20 at 20:26
  • $\begingroup$ @mjw I don't think you need $f$ to be defined on the entire real line. It is enough for $f$ to be defined on some $\delta-$neighborhood of $0$. $\endgroup$ Aug 28 '20 at 20:30
  • $\begingroup$ Okay, now it is clear. $f$ maps an open interval to the positive $x-$axis. For some reason, read it as an ordered pair. Notation is a bit overloaded $\cdots.$ Thanks for clarifying. $\endgroup$
    – mjw
    Aug 28 '20 at 20:32
  • $\begingroup$ Please see this thread. There are many other threads dealing with same question which you can find via approach0. I am not closing this as a dupe because you want help with your specific approach. $\endgroup$ Aug 29 '20 at 5:25
  • $\begingroup$ Why not just use AM-GM inequality? $\endgroup$
    – user600016
    Sep 10 '20 at 5:58

11 Answers 11

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1st Solution. Although not the most straightforward one, let me present a quick solution: First, we note that

$$ \lim_{x\to0} \left| f(x) - \frac{1}{f(x)} \right| = \lim_{x\to0} \sqrt{\left(f(x) + \frac{1}{f(x)} \right)^2 - 4} = 0, $$

Then by using $\max\{a,b\} = \frac{a+b}{2} + \frac{|a-b|}{2}$ and $\min\{a,b\} = \frac{a+b}{2} - \frac{|a-b|}{2}$ which hold for any $a, b \in \mathbb{R}$, we get

$$ \lim_{x\to0} \max\biggl\{ f(x), \frac{1}{f(x)} \biggr\} = 1 = \lim_{x\to0} \min\biggl\{ f(x), \frac{1}{f(x)} \biggr\}. $$

Now the desired conclusion follows by the squeezing theorem.


2nd Solution. We have

$$ \left| f(x) - 1 \right| = \frac{f(x)}{f(x)+1} \left|f(x) - \frac{1}{f(x)}\right| \leq \left|f(x) - \frac{1}{f(x)}\right|. $$

Since we know that $\lim_{x\to0} \left| f(x) - \frac{1}{f(x)} \right| = 0$, the desired claim follows by the squeezing theorem.

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    $\begingroup$ Very similar to what I was thinking: if $g(x) := f(x) + \frac{1}{f(x)}$, then $f(x) = \frac{1}{2} (g(x) \pm \sqrt{(g(x))^2 - 4})$ so in particular $\frac{1}{2} (g(x) - \sqrt{(g(x))^2 - 4}) \le f(x) \le \frac{1}{2} (g(x) + \sqrt{(g(x))^2 - 4})$. $\endgroup$ Aug 28 '20 at 19:56
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    $\begingroup$ Hi Sangchul. I hope that you are doing well and continuing to stay safe and healthy. For your "2nd Solution," How do you justify the equality $|f-1|=\sqrt{f\left(f-\frac1f\right)^2}$? Unless I am miscalculating in my head, which is always possible, I don't believe that can be correct. $\endgroup$
    – Mark Viola
    Aug 28 '20 at 21:05
  • $\begingroup$ @MarkViola, I am doing well, and hop you do so! It seems that I made some mistakes there, missing some factors. I will try to fix it. $\endgroup$ Aug 28 '20 at 21:08
  • $\begingroup$ Thank you my friend! My family and I are doing well. Here, $$f^2-1=f^2\left(f-\frac1f\right)^2$$ $\endgroup$
    – Mark Viola
    Aug 28 '20 at 21:10
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    $\begingroup$ @SangchulLee Now that works and you used $f>0$, which was a given piece of information. (+1) $\endgroup$
    – Mark Viola
    Aug 28 '20 at 21:20
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If the result is false, then there exists $\epsilon>0$ such that no $\delta>0$ works. Thus there exists a sequence $x_n\to 0$ such that $|f(x_n)-1|\ge \epsilon$ for all $n.$ WLOG, $f(x_n)\ge1+\epsilon$ for all $n.$

Let $g(x) = x+1/x$ for $x\in [1,\infty).$ It's easy to see that $g$ is strictly increasing on this interval. Thus we have $(g\circ f)(x_n) \ge g(1+\epsilon) > 1$ for all $n.$ It follows that $\lim_{x\to 0}(f(x)+1/f(x))=1$ is false, contradiction.

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By definition of limit we have $\forall \varepsilon>0$

$$\left| f(x) + \frac{1}{f(x)} - 2 \right|=\left| \frac{(f(x)-1)^2}{f(x)} \right| < \varepsilon$$

and since

$$\left| \frac{(x-1)^2}{x} \right| < 1 \implies \left|\frac{x-1}x\right|<\frac{\sqrt 5+1}2<2$$

assuming wlog $\varepsilon <1$ we have

$$\left| \frac{(f(x)-1)^2}{f(x)} \right| =\left|f(x)-1 \right|\left| \frac{f(x)-1}{f(x)} \right|< 2\left|f(x)-1 \right|<\varepsilon \implies \left|f(x)-1 \right|<\frac{\varepsilon}2$$

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Perhaps a somewhat amusing solution, this is a special case of a question I asked a few years ago. If $a_n,b_n$ are two sequences (real or complex), such that $a_n+b_n\to 2$, and $a_nb_n\to 1$, then $a_n$ and $b_n$ both converge to $1$. There are a few different proofs of that on the page I linked to.

In this case, we take $a_n = f(x_n)$ and $b_n = 1/f(x_n)$ for any sequence $x_n\to 0$. Then by assumption $a_n + b_n \to 2$ and $a_nb_n$ is identically equal to $1$, so the hypotheses are satisfied. Note that the hypothesis that $f$ be a strictly positive function is not necessary.

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Just as @ Sangchul Lee, we can get $$\lim_{x\to0} \left| f(x) - \frac{1}{f(x)} \right| = \lim_{x\to0} \sqrt{\left(f(x) + \frac{1}{f(x)} \right)^2 - 4} = 0.$$ It is easy to see that $$\lim_{x\to0} \left| f(x) - \frac{1}{f(x)} \right|=0\iff \lim_{x\to0} \left( f(x) - \frac{1}{f(x)} \right)=0.$$ By $$f(x)=\frac{1}{2}\left[\left( f(x)+\frac{1}{f(x)} \right)+\left( f(x) - \frac{1}{f(x)} \right)\right],$$ we know $$\lim_{x\to 0}f(x)=1.$$

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  • $\begingroup$ (+1) I am surprised that I missed this easy solution. Nice! $\endgroup$ Aug 30 '20 at 15:52
  • $\begingroup$ No best but only better! $\endgroup$
    – Riemann
    Aug 31 '20 at 4:46
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Alternative quick method:

We can easily deduce that, $$0<\liminf_{x\to 0}f(x)≤\limsup_{x\to 0}f(x)<+ \infty$$

Let, $\liminf_{x\to 0}f(x)=M, M>0$ and $\limsup_{x\to 0}f(x)=N, N>0$

Then, we have:

$$\begin{align}2=\limsup_{x\to 0}\left(f(x)+\frac{1}{f(x)}\right)≥\liminf_{x\to 0} f(x)+\limsup\dfrac {1}{ f(x)} \Longrightarrow \liminf_{x\to 0} f(x)+\dfrac {1}{\liminf_{x\to 0} f(x)} ≤2 \Longrightarrow M+\dfrac 1M ≤2 \Longrightarrow M+\dfrac 1M -2≤0 \Longrightarrow \dfrac{ \left(M-1\right)^2}{M}≤0\Longrightarrow \left(M-1\right )^2≤0 \Longrightarrow M=1\end{align}$$

$$\begin{align}2=\limsup_{x\to 0}\left(f(x)+\frac{1}{f(x)}\right)≥\limsup_{x\to 0} f(x)+\liminf\dfrac {1}{ f(x)} \Longrightarrow \limsup_{x\to 0} f(x)+\dfrac {1}{\limsup_{x\to 0} f(x)} ≤2 \Longrightarrow N+\dfrac 1N ≤2 \Longrightarrow N+\dfrac 1N -2≤0 \Longrightarrow \dfrac{ \left(N-1\right)^2}{N}≤0 \Longrightarrow \left(N-1\right )^2≤0 \Longrightarrow N=1\end{align}$$

Finally, we get $$\begin{align} \liminf _{x\to 0}f(x)=\limsup_{x\to 0}f(x)=\lim_{x\to 0}f(x)=1.\end{align}$$

I used :

  • $$\limsup\limits_{n \rightarrow \infty} a_n + \liminf\limits_{n \rightarrow \infty} b_n \leq \limsup\limits_{n \rightarrow \infty} (a_n + b_n).$$
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    $\begingroup$ First sentence: why is that obvious? $\endgroup$
    – zhw.
    Aug 28 '20 at 19:47
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    $\begingroup$ You have shown that if the limit of $f$ exists, it must be nonzero finite. But it is not obvious (to me) that the limit must exist $\endgroup$
    – user658409
    Aug 28 '20 at 19:48
  • $\begingroup$ This would be a lot easier to read if you let $\lim \sup_{x \to 0}f(x)=M$ and $\lim \inf_{x \to 0}=m.$ $\endgroup$ Aug 29 '20 at 3:56
  • $\begingroup$ @DanielWainfleet is my solution correct? $\endgroup$ Aug 29 '20 at 8:11
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    $\begingroup$ yes it's correct. $\endgroup$ Aug 29 '20 at 9:02
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Setting $y = f(x) + \dfrac{1}{f(x)} \to 2$ we have $$ f(x) = \frac{y\pm\sqrt{y^2-4}}{2} \to \frac{2\pm\sqrt{2^2-4}}{2} = 1. $$

The limit can be justified using the squeeze theorem, since $f(x) = \frac{y\pm\sqrt{y^2-4}}{2},$ i.e. $f(x)$ equals either $\frac{y-\sqrt{y^2-4}}{2}$ or $\frac{y+\sqrt{y^2-4}}{2},$ implies $\frac{y-\sqrt{y^2-4}}{2} \leq f(x) \leq \frac{y+\sqrt{y^2-4}}{2}$.

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I will start off with a fairly general lemma about limiting behavior of roots of a parameterized polynomial equation:

Lemma: Consider the polynomial equation $x^n + t_{n-1} x^{n-1} + t_{n-2} x^{n-2} + \cdots + t_0 = 0$. Then as $t_{n-1}, \ldots, t_0 \to 0$, all $n$ complex roots of this equation also approach 0. To be precise: for every $\epsilon > 0$, there exists $\delta > 0$ such that whenever $|t_i| < \delta$ for $i = 0, \ldots, n-1$ and $x^n + t_{n-1} x^{n-1} + \cdots + t_0 = 0$, it follows that $|x| < \epsilon$.

Proof: If $x$ is a root of the polynomial equation, then it follows that $|t_{n-i} x^{n-i}| \ge \frac{1}{n} |x|^n$ for some $i \in \{ 1, \ldots, n \}$ -- since otherwise, we would have $|x^n + t_{n-1} x^{n-1} + \cdots + t_0| \ge |x|^n - |t_{n-1} x^{n-1}| - \cdots - |t_0| > 0$, giving a contradiction. Therefore, for this value of $i$, we have $|x| \le |t_{n-i} n|^{1/i}$ ("even if $x=0$"). Since $|t_{n-i} n|^{1/i} \to 0$ for each $i$ as $t_0, \ldots, t_{n-1} \to 0$, the desired result follows. $\square$


Now, to apply this lemma to the original problem, let us set $g(x) := f(x) + \frac{1}{f(x)} - 2$. Then $f(x) - 1$ satisfies the equation $(f(x) - 1)^2 - g(x) (f(x) - 1) - g(x) = 0$; and by assumption, we have $g(x) \to 0$ as $x \to 1$. Therefore, by a typical "composition of limits" type argument combined with the lemma above, we can conclude that $f(x) - 1 \to 0$ as $x \to 1$.

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  • $\begingroup$ Huh, it just occurred to me that I'd never really considered the possibility of defining the "limit of a relation" before, treating the relation as representing a multi-valued function. $\endgroup$ Aug 28 '20 at 23:14
  • $\begingroup$ I do have to admit, though, that this solution was mostly inspired by the thoughts of.. well, we can solve for $f$ in terms of $g$ using the quadratic equation. But what about similar problems involving higher order polynomials which can't be solved by radicals anymore? e.g. show that if $(f(t))^5 - t f(t) - t = 0$ then $f(t) \to 0$ as $t \to 0$. $\endgroup$ Aug 28 '20 at 23:30
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There must exist $d>0$ such that $$ 0<|x|<d\implies 0<f(x)\le 3$$ because $f(x)> 3 \implies f(x)+1/f(x)>3$... (and because $f(x)<0\implies f(x)+1/f(x)<0,$ while $f(x)+1/f(x)$ does not exist if $f(x)=0).$

For such $d$ we have $$0<|x|<d\implies \frac {(f(x)-1)^2}{3}\le \frac {(f(x)-1)^2}{f(x)}.$$ Hence $(f(x)-1)^2\to 0,$ so $f(x)-1\to 0.$

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Assuming that $t>0$, let $\phi(t) = t+ {1 \over t}$ and note that $\phi(t) = y $ iff $t = {1 \over 2} (y \pm\sqrt{y^2-4})$.

Suppose $x_n \to 0$ and let $t_n = f(x_n)$. We are given that $y_n =\phi(t_n) \to 2$ (note that we must have $y_n \ge 2$).

We have $t_n \in \{ {1 \over 2} (y_n - \sqrt{y_n^2-4}), {1 \over 2} (y_n + \sqrt{y_n^2-4}) \}$ from which it follows that $t_n \to 1$.

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  • $\begingroup$ I obviously missed your solution when I posted practically the same solution. $\endgroup$
    – md2perpe
    Aug 30 '20 at 10:50
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The condition $\lim_{x\rightarrow0}\Big(f(x)+\frac{1}{f(x)}\big)=2$ along with the assumption $f(x)>0$ implies that $f(x)$ is bounded in some subinterval $(-a,a)\setminus\{0\}$ with $0<a<\delta$, for there is $0<a<\delta$ such that $$ \Big|f(x)+\frac{1}{f(x)} -2\Big|<1 $$ which is equivalent to $|f(x)-1|^2<f(x)$ and so $\alpha=1+\frac{1-\sqrt{5}}{2}<f(x)<1+\frac{1+\sqrt{5}}{2}=\beta$.

This any sequence $\{x_n\}\subset(-a,a)$ that converges to $0$ has a subsequence $x_{n'}$ such that $f(x_{n'})$ converges to some number $p$ between $\alpha$ and $\beta$. Hence

$$p+\frac{1}{p}=2$$ which means that $p=1$. This is independent of the sequence $x_n\rightarrow0$; consequently,

  • $f(x)$ converges as $x\rightarrow0$
  • $\lim_{x\rightarrow0}f(x)=1$
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  • $\begingroup$ My edit was to change "ichi" to "which". We should not assume $f(x)>0$ for all $x$. But we must have $f>0$ on a nbhd of $0$, else $f(x)+1/f(x)$ could not converge to a positive value as $x\to 0$. $\endgroup$ Sep 1 '20 at 3:46

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