Find the limit $ \lim_{x \to \infty} \frac{x^{3}}{x^{ln(x)}} $

Question

I want to find the limit of:
$ \lim_{x \to \infty} \frac{x^{3}}{x^{ln(x)}} $

I already know that it has to be $0$, since $x^{ln(x)}$ grows faster than $x^3$, but I don't know how to get there myself (the steps I have to make). Thank you for your help. $:)$

Answer 1

\begin{align*} \lim_{x \rightarrow \infty} \frac{x^3}{x^{\ln x}} &= \lim_{x \rightarrow \infty} \frac{x^{3 - \ln(x)}}{1} \\ &= \lim_{x \rightarrow \infty} x^{3 - \ln(x)} \\ &= \lim_{x \rightarrow \infty} \left( \mathrm{e}^{\ln x} \right) ^{3 - \ln(x)} \\ &= \lim_{x \rightarrow \infty} \mathrm{e}^{(\ln x)(3 - \ln x)} \text{.} \end{align*} Let $u = \ln x$, so we treat $u$ as an implicit function of $x$. Notice that $\lim_{x \rightarrow \infty} u = \infty$. Then \begin{align*} \lim_{x \rightarrow \infty} (\ln x)(3 - \ln x) &= \lim_{x \rightarrow \infty} u(3 - u) \\ &= \lim_{x \rightarrow \infty} (-u^2 + 3u) \\ &= -\infty \text{, so} \\ \lim_{x \rightarrow \infty} \mathrm{e}^{(\ln x)(3 - \ln x)} &= 0\text{.} \end{align*}

Answer 2

Just recall $\lim_{x\to\infty}\frac{x^{a(x)}}{x^{b(x)}}=0$ if $b$ is eventually larger than $a$.

Eventually means: Given some $x^*\in\mathbb{R}$, for all $x\geq x^*$ we have $b(x)\geq a(x).$

Answer 3

This is $$\lim_{x\to\infty} x^{3-\ln(x)}=\lim_{x\to\infty}\exp\left((3-\ln(x))\ln(x)\right)=...$$

Answer 4

Here is a possible method given that your are familiar with the Limit Chain Rule.

$\lim_{x \rightarrow \infty} \frac{x^{3}}{x^{\ln (x)}}$

1. Simply using exponent rules: $\lim_{x \rightarrow \infty} x^{3 - \ln(x)}$

2. Apply the Limit Chain Rule*: $\lim_{x \rightarrow \infty} (e^{(3 - \ln(x))(\ln(x))})$

3. Direct Substitution: $\lim_{x \rightarrow \infty} (3 - \ln(x))(\ln(x)) = - \infty$

*Recall the Limit Chain Rule says if $\lim_{u \rightarrow b} f(u) = L$ and $\lim_{x \rightarrow b} g(x) = b$, and $f(x)$ is continuous at $x = b$, then $\lim_{x \rightarrow a} f(g(x)) = L$ (Symbolab definition of Limit Chain Rule).

Answer 5

Set $x=e^y$, and consider $y \rightarrow \infty$.

$\dfrac{e^{3y}}{(e^y)^y}= \dfrac{e^{3y}}{e^{y^2}}=e^{3y-y^2}=$

$ e^{-(y-3/2)^2+9/4}= e^{9/4}e^{-z^2}$,

where $z:=y-3/2$.

Take the limit $z \rightarrow \infty$.