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I would like to find or solve the limit of:

$$\lim_{n \to \infty} \frac{80^{(n+1)/4}}{37^{(2n+3)/4}}$$

My idea was somehow non-intuitive:

$$\lim_{n \to \infty} \frac{80^{(n+1)/4}}{37^{(2n+3)/4}} \leq \lim_{n \to \infty} \frac{3^{4(n+1)/4}} {6^{2(2n+3)/4}}= \lim_{n \to \infty}(\frac{3^{4n+4}}{6^{4n+6}})^{1/4} = \lim_{n \to \infty}(\frac{1^{4n+4}}{3^{4n+6}})^{1/4}=0$$

But it seems wrong somehow...

Mamba
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  • just to confirm the fractions are exponents, correct? – Chan Hunt Apr 06 '15 at 18:12
  • Why do you alternate in and out of MathJax the way you do? You have $\displaystyle\lim_{n\to\infty}$ in one set of MathJax tags, then a fraction following it in another, then the "equals" sign after that in another, etc. That is not proper use of MathJax. It can result in lack of proper spacing or alignment. ${}\qquad{}$ – Michael Hardy Apr 06 '15 at 18:41
  • The line that says $\displaystyle\lim_{n \to \infty} \frac{80^{(n+1)/4}}{37^{(2n+3)/4}} \leq \frac{3^{4(n+1)/4}} {6^{6(2n+3)/4}}= \frac{3^{4n+4}}{6^{12n+18}} = \frac{1^{4n+4}}{3^{12n+18}}=0$ might make sense if it said $\displaystyle\lim_{n \to \infty} \frac{80^{(n+1)/4}}{37^{(2n+3)/4}} \leq \lim_{n \to \infty}\frac{3^{4(n+1)/4}} {6^{6(2n+3)/4}}= \lim_{n \to \infty}\frac{3^{4n+4}}{6^{12n+18}} = \lim_{n \to \infty}\frac{1^{4n+4}}{3^{12n+18}}=0$. ${}\qquad{}$ – Michael Hardy Apr 06 '15 at 18:43
  • Yes of course I'm sorry.... – Mamba Apr 06 '15 at 19:18

2 Answers2

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You're on the right track but I have issues with some details:

$$\lim_{n \to \infty} \frac{80^{(n+1)/4}}{37^{(2n+3)/4}} \leq \lim_{n \to \infty}\frac{3^{4(n+1)/4}} {6^{6(2n+3)/4}} $$

The part above would be correct if you had $6^2$ instead of $6^6$. The point is $6^2<37$, so $\dfrac 1{37}<\dfrac 1{6^2}$.

$$\lim_{n \to \infty}\frac{3^{4(n+1)/4}} {6^{2(2n+3)/4}}= \lim_{n \to \infty}\frac{3^{(4n+4)/4}}{6^{(4n+6)/4}} = \lim_{n\to\infty} \frac{3^{n+1}}{6^{n+(3/2)}} \le \lim_{n \to \infty}\frac{3^{n+1}}{6^{n+1}}$$

Notice above I have "$\le$" rather than "$=$" for a reason.

$$ =\lim_{n\to\infty} \left( \frac 1 2 \right)^{n+1} = 0. $$

Michael Hardy
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HINT. $$\frac{80^{\frac{n+1}{4}}}{37^{\frac{2n+3}{4}}}=\frac{\sqrt[4]{80}}{\sqrt[4]{37^3}}\frac{(\sqrt[4]{80})^{n}}{(\sqrt[4]{37^2})^{n}}=C \left(\frac{(\sqrt[4]{80})}{(\sqrt[4]{37^2})}\right)^4= C q^n$$ Therefore we calculate the limit: $$\lim C q^n$$ with $<q<1$. Result: $0$.

Mark
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