Exploring the Limit of $\displaystyle \frac{\infty}{\infty}$

[karush]

$\displaystyle
L_b=\lim_{x \to \infty}
\left\{\frac{n^2}{2^n}\right\} \implies\frac{\infty}{\infty} \\
\text{take natural log of both sides} \\
\ln\left(L_b{}\right)=\lim_{x \to \infty}
\left\{\frac{2\ln\left({n}\right)}{n\ln\left({2}\right)}\right\} \\
\text{not sure?? } $

 

[alyafey22]

You need to prove that $2^n > n^2$ for large $n$.

 

[Greg]

Why take logs? Why not simply apply L'Hopital's rule to the original limit twice?

 

[karush]

Why take logs? Why not simply apply L'Hopital's rule to the original limit twice?

$\text{thusly..}$
$$\displaystyle
L_b=\lim_{x \to \infty}
\left\{\frac{n^2}{2^n}\right\} $$

$$\displaystyle
L'_b=\lim_{x \to \infty}
\left\{\frac{2n}{2^{n}\ln\left({2}\right)}\right\} $$

$$\displaystyle
L''_b=\lim_{x \to \infty}
\left\{\frac{2}{2^n\ln\left({2}\right)^2}\right\} $$
$x \to \infty$
$$L_b=0$$

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You need to prove that $2^n > n^2$ for large $n$.

Prove?

 

[alyafey22]

You can use the mathematical induction.

 

[karush]

whatever that is?

 

[HallsofIvy]

$\displaystyle
L_b=\lim_{x \to \infty}
\left\{\frac{n^2}{2^n}\right\} \implies\frac{\infty}{\infty} \\
\text{take natural log of both sides} \\
\ln\left(L_b{}\right)=\lim_{x \to \infty}
\left\{\frac{2\ln\left({n}\right)}{n\ln\left({2}\right)}\right\} \\
\text{not sure?? } $

The logarithm of $$\frac{n^2}{2^n}$$ is not $$\frac{2ln(n)}{nln(2)}$$. It is $$2ln(n)- n ln(2)$$.