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I'm trying to evaluate this limit:

$$\lim_{x\to +\infty}[(x^2 +1)^{1/2} -2(x^3 + x)^{1/3} +(x^4 + x^2)^{1/4}]$$

Here's what I've tried:

$$ = \lim_{x\to +\infty} x\left((1+\frac{1}{x^2})^{1/2} - 2(1 + \frac{1}{x^2})^{1/3} +(1+\frac{1}{x^2})^{1/4} )\right) = +\infty \cdot 0$$

Which is undefined.

Faker123
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  • Have you made any headway? Have you seen that you can factor an $x$ out of every term? –  Oct 30 '16 at 16:36
  • yeah i tried it and it still +infinity x 0 which is undifined – Faker123 Oct 30 '16 at 16:37
  • OK, edit into your question that and anything else you tried (series expansions, attempts at finding bounds, etc) or else this question is likely to be put on hold as missing context. –  Oct 30 '16 at 16:38
  • FYI, I don't mind fixing your formatting, but if you're going to be asking more questions here, it might be beneficial for you to learn MathJax so that you can format your questions correctly yourself. –  Oct 30 '16 at 16:47
  • just put it as $(P(x)/Q(x))^n$ done. – Anonymous Oct 30 '16 at 16:51

3 Answers3

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The limit $$ \lim_{x\to \infty} (x^2+1)^{1/2}-2(x^3+x)^{1/3}+(x^4+x^2)^{1/4} $$ can be rewritten as $$ \lim_{x\to \infty} \frac{\sqrt{1+x^{-2}}-2\sqrt[3]{1+x^{-2}}+\sqrt[4]{1+x^{-2}}}{x^{-1}}. $$ Consider that, if $\alpha >0$, then as $x\to 0$ we have $$ (1+x)^\alpha=1+\alpha x + O(x^2). $$ It follows that $$ \lim_{x\to \infty} \frac{\left(1+\frac{1}{2}x^{-2}\right)-2\left(1+\frac{1}{3}x^{-2}\right)+\left(1+\frac{1}{4}x^{-2}\right)+O(x^{-4})}{x^{-1}} $$ is equal to $$ \lim_{x\to \infty} \frac{\frac{1}{12}x^{-2}+O(x^{-4})}{x^{-1}}=0. $$

Paolo Leonetti
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Consider $$A=(1+\frac{1}{x^2})^{1/2} - 2(1 + \frac{1}{x^2})^{1/3} +(1+\frac{1}{x^2})^{1/4} $$ and define $y=\frac 1{x^2}$ which makes $$A=(1+y)^{1/2} - 2(1 + y)^{1/3} +(1+y)^{1/4} $$ Now, use the generalized binomial theorem $$(1+y)^n=1+n y+\frac{n(n-1)}{2} y^2+\cdots$$ Apply to each piece; if no mistake, this would give $$A=\frac{y}{12}+\frac{y^2}{288}+\cdots=\frac{1}{12x^2}+\frac{1}{288x^4}+\cdots$$ making $$A x=\frac{1}{12x}+\frac{1}{288x^3}+\cdots$$

Edit

Just for the fun, use $x=10$; use your pocket calculator to get $A\approx 0.000833674$ while, just limited to the first term, the approximation gives $A\approx \frac 1{120}\approx 0.00833333$. Using the two terms $A\approx\frac{2401}{2880000}\approx 0.000833681$.

Claude Leibovici
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We have so much room here we can throw caution to the wind. All we need is: If $0< p \le 1$ and $h>0,$ then

$$1\le (1+h)^p \le 1+h.$$

So your second displayed expression is at least $x(1-2(1+1/x^2) + 1) = -2/x,$ and is no more than $x(1 + 1/x^2 -2 + 1 + 1/x^2) = 2/x.$ By the squeeze theorem the desired limit is $0.$

zhw.
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