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Clearly, $\infty -\infty \not =0$

I have a feeling the squeeze theorem can be applied here, but I am not sure how to write the required terms separately. Can I get a hint?

Aditya
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    See https://math.stackexchange.com/q/256806/42969 or https://math.stackexchange.com/q/2038809/42969 for a similar problem. Those solutions can be applied here as well. – Martin R Dec 01 '20 at 14:22
  • Have you thought of writing it as $(a-b)(a+b)=(a^2-b^2)$? I suggest trying this things might cancel – Jose M Serra Dec 01 '20 at 14:22
  • If you follow the question by @MartinR, and trying that Difference of Squares it might be worth while. – Jose M Serra Dec 01 '20 at 14:24
  • @EnlightenedFunky That wouldn't be a good idea, as it's not correct. And the correct $(a-b)(a+b)=a^2-b^2$ doesn't help, much. –  Dec 01 '20 at 14:25
  • @ProfessorVector You get the expression down to $((n+1)^{\frac{1}{3}}-(n-1)^\frac{1}{3})((n+1)^{\frac{1}{3}}+(n-1)^\frac{1}{3})$ – Jose M Serra Dec 01 '20 at 14:29
  • How about difference of cubes? – Empy2 Dec 01 '20 at 14:32

6 Answers6

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Let $f:[n-1,n+1]\to\mathbb{R}$ and $f(x)=x^{\frac{2}{3}}$. Then from Mean value theorem (Lagrange’a) exist $\xi\in (n-1,n+1)$ such that:

$$\frac{f(n+1)-f(n-1)}{n+1-(n-1)}=f'(\xi)$$

but (check it out) $$\frac{2}{3 \sqrt[3]{n+1} }\le f'(\xi)\le \frac{2}{3 \sqrt[3]{n-1} } $$

so

$$\frac{4}{3 \sqrt[3]{n+1} }\le f(n+1)-f(n-1 )\le \frac{4}{3 \sqrt[3]{n-1} } $$

$$\frac{4}{3 \sqrt[3]{n+1} }\le (n+1)^{\frac{2}{3}}-(n-1 )^{\frac{2}{3}}\le \frac{4}{3 \sqrt[3]{n-1} } $$

now You can aplay Squeeze theorem.

Marek Kryspin
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$n=\frac{1}{t}$ $$\lim_{t\to 0}\frac{{(1+t)}^{2/3}-{(1-t)}^{2/3}}{t^{2/3}}$$ Now apply L-Hospitals rule.. to get $$\lim_{t\to 0}\left(\frac{t}{\sqrt[3]{1+t}}+\frac{t}{\sqrt[3]{1-t}}\right)=?$$

Albus Dumbledore
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$$(n+1)^{2/3} -(n-1)^{2/3}=\frac{(n+1)^2-(n-1)^2}{(n+1)^{4/3}+(n+1)^{2/3}(n-1)^{2/3}+(n-1)^{4/3}} \\=\frac{n}{n^{4/3}}\frac4{(1+n^{-1})^{4/3}+(1-n^{-2})^{2/3}+(1-n^{-1})^{4/3}}\to 0$$

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Your expression almost looks like the symetric derivative of $f(x)=x^{2/3} $, so you have :

$$(n+1)^{2/3}-(n-1)^{2/3}\sim_{\infty}2\times\frac{2}{3}n^{-1/3} $$

Therefore, $\lim_{n\to\infty}(n+1)^{2/3}-(n-1)^{2/3}=0 $

edit : $$(n+1)^{2/3}-(n-1)^{2/3}= n^{2/3}\left(\left(1+\frac{1}{n}\right)^{2/3} -\left(1-\frac{1}{n}\right)^{2/3}\right)\\ =n^{2/3}\left(1+\frac{2}{3n}+o\left(\frac{1}{n}\right) -1+\frac{2}{3n}+o\left(\frac{1}{n}\right)\right)\\ =\frac{4}{3}n^{-1/3}+o\left(n^{-1/3}\right) $$

NHL
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    "Almost looks like" is very far from the idea of a mathematical proof. –  Dec 01 '20 at 14:29
  • "almost looks like" as intuition, I'll edit it using taylor expansion, it works – NHL Dec 01 '20 at 14:29
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Set $\dfrac1n=y\implies y\to0^+$ to find

$$F=\lim_{y\to0^+}\dfrac{(1+y)^{2/3}-(1-y)^{2/3}}y$$

Now rationalize the numerator using $a^3-b^3=(a-b)\cdot (?)$ to find

$$F=\lim_{y\to0^+}\dfrac{(1+y)^2-(1-y)^2}y\cdot\lim_{y\to0^+}\dfrac1{(1+y)^{4/3}+(1+y)^{2/3}(1-y)^{2/3}+(1+y)^{4/3}}=?$$

Albus Dumbledore
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lab bhattacharjee
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Using Taylor expansion

$$(1+\epsilon)^{2\over 3}= 1+{2\over 3}\epsilon +o(\epsilon)$$

We can write with $\epsilon ={1\over n}$

$$(n+1)^{2\over 3}-(n-1)^{2\over 3}=n^{2\over 3}\left(1+{2\over 3}\epsilon-1+{2\over 3}\epsilon+o(\epsilon)\right)={4\over 3}n^{-{1\over 3}}+o\left(n^{-{1\over 3}}\right)$$

And this quantity at $+\infty$ has limit $0$

marwalix
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