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We're asked to find the limit (as $n$ goes to infinity) of the following sequence: $$a_n = \frac {\sqrt[3n]{4}\ -6 \ \sqrt[3n]{2} \ + \ 9}{\sqrt[2n]{9} \ - \ 4 \ \sqrt[2n]{3}+ 4} $$.

I thought that since the limit of the numerator exists, and since the limit of the denominator also exists and is non-zero, then the limit of the fraction should be $\frac{9}{4}$.

But apperently, the limit of this sequence is $4$ as $n \rightarrow +\infty$.

I don't understand why my approach is incorrect, nor why the limit of the sequence is $4$.

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3 Answers 3

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$$\lim_{n\to\infty}a_n=\lim_{n\to\infty}\left(\dfrac{2^{1/3n}-3}{3^{1/2n}-2}\right)^2=\left(\dfrac{2^0-3}{3^0-2}\right)^2=?$$

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Note that $$a^{0}=1$$ $$\quad{\lim_ {n\to \infty} \frac {\sqrt[3n]{4}\ -6 \ \sqrt[3n]{2} \ + \ 9}{\sqrt[2n]{9} \ - \ 4 \ \sqrt[2n]{3}+ 4}\\= \lim_ {n\to \infty} \frac {{4}^{\frac{1}{3n}}\ -6 \times{2}^{\frac{1}{3n}} \ + \ 9}{{9}^{\frac{1}{2n}} \ - \ 4 \times{3}^{\frac{1}{2n}}+ 4}\\= \frac {{4}^{0}\ -6 \times{2}^{0} \ + \ 9}{{9}^{0} \ - \ 4 \times{3}^{0}+ 4}=\\\frac{1-6+9}{1-4+4}}$$

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It's just $$\frac{1-6+9}{1-4+4}=4$$ because for example $4^{\frac{1}{3n}}\rightarrow4^0=1$.

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  • $\begingroup$ Oh my god, I'm so stupid, thank you $\endgroup$
    – Skyris
    Commented Jan 12, 2018 at 14:18
  • $\begingroup$ You are welcome! $\endgroup$ Commented Jan 12, 2018 at 14:19

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