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$\begingroup$

find limit :

$$\lim_{ x \to -\infty }\frac{\sqrt[n]{x+1}+\sqrt[n]{x+2}-2\sqrt[n]{x-3}}{\sqrt[n]{x-1}+\sqrt[n]{x-2}-2\sqrt[n]{x+3}}=\;\; ? \;\quad \text {given }\,n \in \mathbb{N}, n>2 ,\text{odd}$$

I tried :

$$\lim_{ x \to -\infty }\frac{\sqrt[n]{x+1}+\sqrt[n]{x+2}-2\sqrt[n]{x-3}}{\sqrt[n]{x-1}+\sqrt[n]{x-2}-2\sqrt[n]{x+3}}=\frac{x+x-2x }{x+x-2x}=\frac{0}{0}$$

  • 3
    I don't understand your first equality on the second line. My suggestion to you is to try $n=3,5$ first. – 2017-01-14
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    for the nth root we have $$x\geq 3$$ how we can have $$x\to -\infty$$? – 2017-01-14
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    @Dr.SonnhardGraubner $ n \in odd$ – 2017-01-14
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    Set $x=-1/h$ and use Taylor's expansion – 2017-01-14

2 Answers 2

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If we apply binomial expansion, we get

$$L=\lim_{x\to\pm\infty}\frac{\frac9{nx^{1-1/n}}+\mathcal O\left(\frac1{x^{2-1/n}}\right)}{-\frac9{nx^{1-1/n}}+\mathcal O\left(\frac1{x^{2-1/n}}\right)}\to-1$$

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    so : $\quad\lim_{ x \to -\infty }\frac{\sqrt[n]{x+1}+\sqrt[n]{x+2}-2\sqrt[n]{x-3}}{\sqrt[n]{x-1}+\sqrt[n]{x-2}-2\sqrt[n]{x+3}}=-1 ?$ – 2017-01-14
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    @Almot1960 I would believe so. – 2017-01-14
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Substitute $x=-1/t$, so you get $$ \lim_{t\to0^+} \frac{\sqrt[n]{t-1}+\sqrt[n]{2t-1}-2\sqrt[n]{-3t-1}} {\sqrt[n]{-t-1}+\sqrt[n]{-2t-1}-2\sqrt[n]{3t-1}} = \lim_{t\to0^+} \frac{\sqrt[n]{1-t}+\sqrt[n]{1-2t}-2\sqrt[n]{1+3t}} {\sqrt[n]{1+t}+\sqrt[n]{1+2t}-2\sqrt[n]{1-3t}} $$ The Taylor expansion of the numerator is $$ 1-\frac{1}{n}t+1-\frac{2}{n}t-2-\frac{6}{n}t+o(t)=-\frac{9}{n}t+o(t) $$ The Taylor expansion of the denominator is…