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Let $f:(a-\epsilon,a+\epsilon)\to(0,\infty).$

If $$\lim \limits_{x \to a}\left(f(x)+\frac{1}{f(x)}\right)=2,$$ prove that $$\lim \limits_{x \to a}f(x)=1.$$

I'm trying to get something using $\epsilon-\delta$ definition but I'm stuck . How to use $(a-\epsilon,a+\epsilon)$ information I'm wondering .

$$~~$$ An answer

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Milan
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  • @Fimpellizieri no – Milan Jan 03 '20 at 21:42
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    It makes sense to speak of the limit at $a$ only if $f$ is defined in a neighborhood of $a$, so the information $(a-\epsilon,a+\epsilon)$ is not particularly useful besides making sure that trying to compute limits makes sense in the first place. – Fimpellizzeri Jan 03 '20 at 21:49
  • Then i am out of ideas. Maybe you are right, it should have been said assuming the limit exists. – Milan Jan 03 '20 at 21:51

2 Answers2

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The assumption implies that $$\lim_{x\to a} \left(\sqrt{f(x)}-\frac{1}{\sqrt{ f(x)}}\right)^2=0,$$ hence $$\lim_{x\to a} \sqrt{f(x)}-\frac{1}{\sqrt{ f(x)}}=0.$$ Similarly, $$\lim_{x\to a} \left(\sqrt{f(x)}+\frac{1}{\sqrt{ f(x)}}\right)^2=4,$$ hence $$\lim_{x\to a} \sqrt{f(x)}+\frac{1}{\sqrt{f(x)}}=2.$$ Summing up and dividing by two we obtain $$\lim_{x\to a} \sqrt{f(x)}=1,$$ hence $$\lim_{x\to a} f(x)=1.$$

timon92
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You want to find a local inverse around 1 to $g(x)=x+\frac{1}{x}$ by solving quadratic formula. Then use the continuity of said inverse to find the epsilon and delta you need.

Call this inverse $h$. Then, $\lim_{x\to a}f(x)+\frac{1}{f(x)}=\lim_{x\to a}g(f(x))$

So, by continuity of $h$, $$1=h(2)=h(\lim_{x\to a}g(f(x)))=\lim_{x\to a}f(x)$$

Ciarán Ó Raghaillaigh
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