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I tried solving this question :$\lim \limits_{x \rightarrow \infty} \sqrt[]{x^2+2x+3}-\sqrt[]{x^2-x+5}.$ but got nowhere

how ever it made me wonder if $$\lim \limits_{x \rightarrow \infty} \sqrt[x]{x^x+2x^{x-1}+3}-\sqrt[x]{x^x-x^{x-1}+5}.$$ has limit, I tried some of the tricks from tjat question but not getting anywhere so far.

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jimjim
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  • Perhaps exploiting the binomial expansion theorem. – Simply Beautiful Art Oct 01 '16 at 12:41
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    Limited expansions work like a charl to solve this, as always... For every polynomial $P$, $$\sqrt[x]{x^x+P(x)}=x\sqrt[x]{1+x^{-x}P(x)}=x\exp(x^{-1}\log(1+x^{-x}P(x)))$$ and, for every polynomial $P$, $$x^{-x}P(x)=O(x^{-1})$$ hence $|\log(1+x^{-x}P(x))|=O(x^{-1})$ and $$\sqrt[x]{x^x+P(x)}=x(1+O(x^{-2}))=x+o(1)$$ Using this for $P(x)=2x+3$ and for $P(x)=-x+5$ shows that $$\lim \limits_{x \rightarrow \infty} \sqrt[x]{x^x+2x^{x-1}+3}-\sqrt[x]{x^x-x^{x-1}+5}=0$$ – Did Oct 01 '16 at 12:43
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    The same approach applies to the revised version of the question. Any trouble applying it? If so, just whistle... – Did Oct 01 '16 at 12:45
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    *Typo: charl > charm. – Did Oct 01 '16 at 12:48
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    Looks like the accepted answer here contains all the info you need: http://math.stackexchange.com/a/45985/5189 – Myself Oct 01 '16 at 13:01
  • @Myself : not suprised, was thinking Did's comment has more than all i can ask for. – jimjim Oct 01 '16 at 13:56
  • This might be one of those situations where proving more is actually easier: Show that for avery functions $f$ and $g$ such that $$\lim_{x\to\infty}\frac{f(x)}{x^x}=\lim_{x\to\infty}\frac{g(x)}{x^x}=0$$ one has $$\lim_{x\to\infty}\sqrt[x]{x^x+f(x)}-\sqrt[x]{x^x+g(x)}=0$$ – Did Oct 01 '16 at 14:01
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    And more generally... Show that for avery functions $f$ and $g$ such that there exists $\alpha>-1$ and $\beta>-1$ such that $$\lim_{x\to\infty}\frac{f(x)}{x^x}=\alpha\qquad\lim_{x\to\infty}\frac{g(x)}{x^x}=\beta$$ one has $$\lim_{x\to\infty}\sqrt[x]{x^x+f(x)}-\sqrt[x]{x^x+g(x)}=\log\frac{1+\alpha}{1+\beta}$$ – Did Oct 01 '16 at 14:04

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