Why does $\log^{100}(n)$ grow asymptotically slower than $n^{7/8}$ More formally:
How do I compute the following limit:
$$
\lim _{x \rightarrow \infty} \frac{\log ^{100}(x)}{x^{7 / 8}}
$$
I would start using l'Hôpital's rule and notice the pattern, is there some other procedure? (without knowing it)
 A: Squaring $x$ multiplies the numerator by $2^{100}$ and the denominator by $x^{7/8}$. Since $2^{100}$ is a constant but $x^{7/8}$ is unbounded, there exists an $x$ for which $x^{7/8}>2^{100}$, i.e. the denominator grows faster than the numerator. Thus the limit is $0$.
A: It is standard that, for any $\alpha,\beta>0$, $\;\ln^\alpha(x)=o\bigl(x^\beta\bigr)$ as $x\to\infty$.
This is very simple to prove: set $u=x^{\tfrac \beta\alpha}$. We have
$$\frac{\ln^\alpha(x)}{x^\beta}=\frac{\ln^\alpha\Bigl(u^{\tfrac\alpha\beta}\Bigr)}{u^\alpha}=\underset{\strut\text{constant}}{\biggl(\frac\alpha\beta\biggr)^{\mkern-5mu\alpha}} \biggl(\underbrace{\frac{\ln u}u}_{{}\to 0}\biggr)^{\mkern-5mu\alpha}.$$
A: Let set $f(x)=\dfrac{\ln(x)^{100}}{x^{\frac 78}}$ it is not to difficult to see that $f\searrow$ at infinity.
Indeed $f'(x)=-\dfrac{\overbrace{\ln(x)^{99}}^{>0}(\overbrace{7\ln(x)-800}^{>0})}{\underbrace{x^{\frac{15}8}}_{>0}}<0\quad$ for $x$ large enough.
In particular since $f$ is continuous and $f(x)>0$ for $x>1$ we can conclude that $f$ is bounded. $$\exists M>0\text{ : }|f(x)|<M\quad\forall x>1$$
Now we just use apply to $x=u^2$ and use the logarithm multiplicative to additive property to show the denominator is negligible.
$$|f(x)|=|f(u^2)|=\left|\dfrac{2^{100}\ln(u)^{100}}{u^{\frac 78}u^{\frac 78}}\right|=\left|\dfrac{2^{100}f(u)}{u^{\frac 78}}\right|\le2^{100}M\dfrac 1{|u|^{\frac 78}}\to 0$$
Naturally, you can easily transpose the proof for any suitable $\, \alpha,\beta\, $ and $\, \dfrac{\ln(x)^\alpha}{x^\beta}$
A: Taking the logarithm of the numerator and denominator gives
$$ \frac{\log \left( \,(\log x)^{100}\,\right)}{\log \left(x^{\frac{7}{8}}\right)} \;\;= \;\; \frac{100 \log \log x}{\frac{7}{8}\log x} \;\; = \;\; \frac{800}{7} \cdot \frac{\log \log x}{\log x} \;\; \rightarrow \;\; 0, $$
which suggests the limit will be $0.$ However, this process can change the value of a limit, as can be seen by considering $\lim\limits_{x \rightarrow \infty} \frac{2x + 42}{3x + 42}$ — the limit is $\frac{2}{3}$ before taking logarithms of the numerator and denominator, and the limit is $\frac{\log 2}{\log 3}$ after taking logarithms of the numerator and denominator. Although I believe this process will NOT change the value of the limit (for positive numerators and positive $\neq 1$ denominators) if the original limit is zero or infinite, we can circumvent this issue by rewriting in such a way that the original numerator and denominator are not changed, but still have present the simplifying aspect of taking a logarithm. This can be done by applying TWO successive operations, taking a logarithm followed by an exponentiation:
$$ \frac{(\log x)^{100}}{x^{\frac{7}{8}}} \;\;= \;\; \frac{\exp \left[ \log \left( \,(\log x)^{100}\,\right)\right]}{\exp \left[\log \left(x^{\frac{7}{8}}\right)\right]} \;\;= \;\; \frac{\exp \left[100 \log \log x \right]}{\exp \left[ \frac{7}{8}\log x \right]}$$
$$ = \;\; \exp \left[100\log \log x \; - \; \frac{7}{8}\log x  \right] $$
Since we clearly have $\;100\log \log x - \frac{7}{8}\log x \; \rightarrow \; -\infty,\;$ and $\exp(\rightarrow -\infty) \rightarrow 0,$ it follows that the original quotient approaches $0.$
A: Using
$$ \ln(1+u)<u, u>0$$
one has, for $x>1$,
$$ 0<\frac{\ln^{100}x}{x^{7/8}}=\bigg(\frac{\ln^{\frac12}x}{x^{175}}\bigg)^{200}=\bigg(\frac{\ln^{\frac12}[1+(x-1)]}{x^{175}}\bigg)^{200}<\bigg(\frac{(x-1)^{\frac12}}{x^{175}}\bigg)^{200}<\frac1{x^{174\cdot200}}$$
from which it is easy to obtain
$$ \lim_{x\to\infty}\frac{\ln^{100}x}{x^{7/8}}=0.$$
