I figured out that $\lim_{n \to \infty}\frac{(3n^2−2n+1)\sqrt{5n−2}}{(\sqrt{n}−1)(1−n)(3n+2)} $ is $-\sqrt{5}$ but I don't know how to prove it.
Show that $\lim_{n \to \infty}\frac{(3n^2−2n+1)\sqrt{5n−2}}{(\sqrt{n}−1)(1−n)(3n+2)} = -\sqrt{5}$
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limits
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0Hint: Multiply the numerator and denominator by $\frac{1}{n^2\sqrt{n}}.$ Also, the answer should be $-\sqrt{5}$. – 2017-01-03
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0@LeonSot That's right. I will give it a try. Thanks! – 2017-01-03
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1If you don't know how to prove it, you haven't figured out yet. – 2017-01-03
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0A quick version: once verified that leading term have the same order, just look at the leading term coefficients. – 2017-01-03
2 Answers
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Multiplying numerator and denominator with $\frac{1}{n^{5/2}}$, we get
$$\lim_{n\to\infty}\frac{(3-\frac{2}{n}+\frac{1}{n^2})(\sqrt{5-\frac{2}{n}})}{(1-\frac{1}{n^{1/2}})(\frac{1}{n}-1)(3+\frac{2}{n})}=\frac{3\times \sqrt 5}{-1 \times 3}=-\sqrt5$$
Since $\lim_{n\to\infty}\frac{1}{n}=0.$
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The dominant terms in the numerator and denominator are both in $n^{5/2}$, hence the limit is finite. The corresponding coefficients are $3\sqrt5$ and $-3$. The ratio gives the answer.