Show that $\lim\limits_{x\rightarrow 0}f(x)=1$

Question

Suppose a function $f:(-a,a)-\{0\}\rightarrow(0,\infty)$ satisfies $\lim\limits_{x\rightarrow 0}\left(f(x)+\frac{1}{f(x)}\right)=2$. Show that $$\lim\limits_{x\rightarrow 0}f(x)=1$$

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Let $\epsilon>0$ , then there exists a $\delta>0$ such that $$\left(f(x)+\frac{1}{f(x)}\right)-2<\epsilon\;\;\;\text{and}\;\;\;|x|<\delta$$

Then

\begin{align} &\left(f(x)+\frac{1}{f(x)}\right)-2\\ =&\left(f(x)-1\right) +\left(\frac{1}{f(x)}-1\right)<\epsilon\tag{1} \end{align}

Squaring $(1)$ both sides gives $$\left(\left(f(x)-1\right) +\left(\frac{1}{f(x)}-1\right)\right)^2<\epsilon^2\tag{2}$$

Since $(f(x)-1)^2\leq\left(\left(f(x)-1\right) +\left(\frac{1}{f(x)}-1\right)\right)^2$, by using $(2)$, $(f(x)-1)^2<\epsilon^2\Rightarrow f(x)-1<\epsilon$; therefore, as $\epsilon$ is arbitrary, $\lim\limits_{x\rightarrow 0}f(x)=1$

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Can someone give me a hint to do this question without using epsilon-delta definition? Thanks

Answer 1

$$ a + \frac 1 a = \left( a - 2 + \frac 1 a \right) +2 = \left( \sqrt a - \frac 1 {\sqrt a} \right)^2 + 2 = \text{square} + 2. $$ So this is $\ge 2$ unless the square is $0$.

$a+ \dfrac 1 a$ cannot get close to $2$ unless $a$ gets close to $1$.

Even if you allow complex numbers (so that the inequality above doesn't hold), we have $$ a + \frac 1 a = 2 \Longrightarrow a^2 + 1 = 2a $$ and that is a quadratic equation whose only solution is $a=1$ (a double root).

Answer 2

Define $h(u) = \frac{u + \sqrt{u^2 - 4}}{2}$ (this is the inverse function of $x \mapsto x + \frac{1}{x}$). Note that $h(2) = 1$ and that $h$ is continuous at $u = 2$. Thus,

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

which implies that

$$\lim_{x \to 0} \left( f(x) - \frac{1}{f(x)} \right) = 0. $$

Combining both results together we have

$$ 2 = 2 + 0 = \lim_{x \to 0} \left( f(x) + \frac{1}{f(x)} \right) + \lim_{x \to 0} \left( f(x) - \frac{1}{f(x)} \right) = \lim_{x \to 0} 2f(x) $$

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

Answer 3

Since $f$ is positive and $f + (1/f)$ tends to $2$ as $x \to 0$, it follows that $f$ is bounded and away from $0$ as $x \to 0$ . Therefore it follows that $\sqrt{f(x)}$ is bounded and away from $0$ as $x \to 0$. Now we can see that $$f(x) + \frac{1}{f(x)} = \left(\sqrt{f(x)} - \frac{1}{\sqrt{f(x)}}\right)^{2} + 2$$ and therefore $$\frac{(f(x) - 1)^{2}}{f(x)} \to 0$$ so that $(f(x) - 1)^{2} \to 0$ and therefore $|f(x) - 1| \to 0$ and then $f(x) \to 1$.

Answer 4

Generalization: Suppose $f:(-a,a) \setminus \{0\} \to (0,\infty)$ and $g: (0,\infty)\to \mathbb R.$ Let $y_0> 0.$ Assume $g$ is strictly decreasing on $(0,y_0],$ and strictly increasing on $[y_0,\infty).$ If $\lim_{x\to 0} g(f(x)) = g(y_0),$ then $\lim_{x\to 0} f(x) = y_0.$

This applies to the problem at hand by letting $g(y) = y + 1/y$ with $y_0 = 1.$

Sketch of proof: Suppose $\limsup_{x\to 0} f(x) > y_0.$ Then there exists $z>y_0$ and a sequence $x_n \to 0$ such that $f(x_n)> z$ for all $n.$ This implies $g(f(x_n))> g(z)$ for all $n.$ Because $g(z) > g(y_0),$ this is a contradiction. Thus $\limsup_{x\to 0} f(x) \le y_0.$ Same idea if $\liminf_{x\to 0} f(x) < y_0.$ Thus

$$y_0 \le \liminf_{x\to 0} f(x)\le \limsup_{x\to 0} f(x)\le y_0,$$

giving the result.