# If $\lim_{x\to 0}\left(f(x)+\frac{1}{f(x)}\right)=2,$ show that $\lim_{x\to 0}f(x)=1$.

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?

• 𝑓:(−𝛿,𝛿)→(0,∞). I would have thought 𝑓:ℝ→ℝ, no? – mjw Aug 28 '20 at 20:26
• @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 $0$. – Sanket Biswas Aug 28 '20 at 20:30
• 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. – mjw Aug 28 '20 at 20:32
• 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. – Paramanand Singh Aug 29 '20 at 5:25
• Why not just use AM-GM inequality? – user600016 Sep 10 '20 at 5:58

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.

• 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 $\frac{1}{2} (g(x) - \sqrt{(g(x))^2 - 4}) \le f(x) \le \frac{1}{2} (g(x) + \sqrt{(g(x))^2 - 4})$. – Daniel Schepler Aug 28 '20 at 19:56
• 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. – Mark Viola Aug 28 '20 at 21:05
• @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. – Sangchul Lee Aug 28 '20 at 21:08
• Thank you my friend! My family and I are doing well. Here, $$f^2-1=f^2\left(f-\frac1f\right)^2$$ – Mark Viola Aug 28 '20 at 21:10
• @SangchulLee Now that works and you used $f>0$, which was a given piece of information. (+1) – Mark Viola Aug 28 '20 at 21:20

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.

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

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

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.

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

• (+1) I am surprised that I missed this easy solution. Nice! – Sangchul Lee Aug 30 '20 at 15:52
• No best but only better！ – Riemann Aug 31 '20 at 4:46

# 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 :

• $$\limsup\limits_{n \rightarrow \infty} a_n + \liminf\limits_{n \rightarrow \infty} b_n \leq \limsup\limits_{n \rightarrow \infty} (a_n + b_n).$$
• First sentence: why is that obvious? – zhw. Aug 28 '20 at 19:47
• 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 – user658409 Aug 28 '20 at 19:48
• This would be a lot easier to read if you let $\lim \sup_{x \to 0}f(x)=M$ and $\lim \inf_{x \to 0}=m.$ – DanielWainfleet Aug 29 '20 at 3:56
• @DanielWainfleet is my solution correct? – lone student Aug 29 '20 at 8:11
• yes it's correct. – DanielWainfleet Aug 29 '20 at 9:02

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 $$\frac{y-\sqrt{y^2-4}}{2} \leq f(x) \leq \frac{y+\sqrt{y^2-4}}{2}$$.

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 $$x \to 1$$.

• 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. – Daniel Schepler Aug 28 '20 at 23:14
• 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 $t \to 0$. – Daniel Schepler Aug 28 '20 at 23:30

There must exist $$d>0$$ such that $$0<|x| because $$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 $$0<|x| Hence $$(f(x)-1)^2\to 0,$$ so $$f(x)-1\to 0.$$

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 $$t_n \to 1$$.

• I obviously missed your solution when I posted practically the same solution. – md2perpe Aug 30 '20 at 10:50

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 $$0, for there is $$0 such that $$\Big|f(x)+\frac{1}{f(x)} -2\Big|<1$$ which is equivalent to $$|f(x)-1|^2 and so $$\alpha=1+\frac{1-\sqrt{5}}{2}.

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 $$x\rightarrow0$$
• $$\lim_{x\rightarrow0}f(x)=1$$
• 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 $x\to 0$. – DanielWainfleet Sep 1 '20 at 3:46