4

fine limit :

$$ \lim_{x \to \infty } ( \sqrt[3]{4 x^{a} + x^{2} } - \sqrt[3]{ x^{a} + x^{2} } )^{x-[x]} $$

such that : $$ a \in (0,2)$$

and :

$[x]: \ \ $ floor function

My Try :

$$f(x):=( \sqrt[3]{4 x^{a} + x^{2} } - \sqrt[3]{ x^{a} + x^{2} } )^{x-[x]}$$

$$\ln f(x)=(x-[x])\ln(\sqrt[3]{4 x^{a} + x^{2} } - \sqrt[3]{ x^{a} + x^{2} } )$$

Now ?please help

Almot1960
  • 4,782
  • 16
  • 38
  • Your try is not sufficient. You should show what approach you are trying (something like a particular theorem, result you wish to use). Moreover the answer depends on value of $a$. In particular if $a=4/3$ then the answer is $1$ otherwise the limit does not exist. – Paramanand Singh May 21 '17 at 17:09
  • @ParamanandSingh. Please write complete. thank you . – Almot1960 May 21 '17 at 17:47

2 Answers2

1

The expression under limit operation say $F(x) $ is of the form $\{f(x) \} ^{g(x)} $ and clearly $f(x) >0$ for all $x>0$ and $0\leq g(x) <1$ for all $x>0$. It is now clear that $F(x) $ lies between $1$ and $f(x) $ and it takes the value $1$ when $x$ is a positive integer. Moreover if $x$ is near a positive integer and less than it then $F(x) $ is near $f(x) $. Hence it is clear that if $F(x) $ tends to a limit as $x\to\infty$ then it must be $1$ and moreover this will happen if and only if $f(x) \to 1$ as $x\to\infty$.

Now it is easy to analyze $f(x) $ which is of the form $p-q$. Multiplying it by $(p^{2}+pq+q^{2})$ and dividing it by the same quantity we see that $f(x) $ is expressed as a fraction with numerator $p^{3}-q^{3}=3x^{a}$. Further if divide the numerator and denominator of $f(x) $ by $x^{4/3}$ then the denominator tends to $3$ and the numerator is $3x^{a-4/3}$. It follows that $f(x) \to 1$ if and only if $a=4/3$. Hence the desired limit is equal to $1$ if $a=4/3$ and the limit does not exist if $a\neq 1$.

Paramanand Singh
  • 87,309
1

$\begin{array}\\ ( \sqrt[3]{4 x^{a} + x^{2} } - \sqrt[3]{ x^{a} + x^{2} } )^{x-[x]} &=(x^{2/3} \sqrt[3]{1+4 x^{a-2}} - x^{2/3}\sqrt[3]{1+ x^{a-2} } )^{x-[x]}\\ &=x^{2(x-[x])/3} (\sqrt[3]{1+4 x^{a-2}} - \sqrt[3]{1+ x^{a-2} } )^{x-[x]}\\ &=x^{2(x-[x])/3} (1+4 x^{a-2}/3+O(x^{a-3}) - (1+ x^{a-2}/3+O(x^{a-3})) )^{x-[x]}\\ &=x^{2(x-[x])/3} ( x^{a-2}+O(x^{a-3} ))^{x-[x]}\\ &=x^{(x-[x])(2/3+a-2)} ( 1+O(x^{-1} ))^{x-[x]}\\ &=x^{(x-[x])(a-4/3)} ( 1+O(x^{-1} ))^{x-[x]}\\ \end{array} $

As Paramanand Singh's answer states, if $a = 4/3$, this is $( 1+O(x^{-1} ))^{x-[x]} $ which goes to $1$.

If $a \ne 4/3$, this goes to $x^{(x-[x])(a-4/3)} $. This goes from $1$ (when $x = [x]$) to $x^{a-4/3}$ (when $x-[x] \approx 1$). Therefore, when $a \ne 4/3$, the limit does not exist.

marty cohen
  • 107,799
  • 1
    Thanks for adding those calculations. It's a tough job to type all that $\mathrm\LaTeX $ on smartphone so I avoided it in my answer. +1 – Paramanand Singh May 22 '17 at 05:54
  • Real keyboards and copy&paste rule! – marty cohen May 22 '17 at 06:00
  • Get a bluetooth keyboard. I like this one for Android: https://www.amazon.com/iClever-Portable-Tri-folding-Bluetooth-keyboard/dp/B01MTVC775/ref=sr_1_3?ie=UTF8&qid=1495432859&sr=8-3&keywords=iclever+keyboard – marty cohen May 22 '17 at 06:01
  • Will look at Bluetooth keyboards options available in India. The one you mention looks very stylish. – Paramanand Singh May 22 '17 at 06:35
  • That one is very good for Android phones and regular computers since it has a trackpad. If you want one just for iPhone, save money by getting one without the trackpad. – marty cohen May 22 '17 at 06:40