I have the following given as an answer to a questions, but I don't understand how I get $1+ \log n$ in the second step. I think I am to use L'Hospital's, but deriving the logarithm is $1/(n\ln(b))$, so I don't know why I'm keeping $\log(n)$. I know it's probably something basic, but what am I missing about that step?
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You can also rewrite it as: $$\lim_{n \rightarrow \infty} \frac{n^{1,001}}{n \log{n}} = \lim_{n \rightarrow \infty} \frac{n^{0,001}}{\log{n}} = \lim_{n \rightarrow \infty} \frac{\frac{n^{-0,999}}{0,001}}{\frac{1}{n}} = \lim_{n \rightarrow \infty} \frac{n^{0,001}}{0,001} = \infty$$ which might be a little clearer.
The second step is the application of l'Hopital Rule and precisely the derivative of the denominator
$$(x\log x)'=x(\log x)'+(x)'\cdot\log x=x\frac1x+1\cdot \log x=1+\log x$$
For a direct approch without l'Hopital, note that for $x\to \infty$ $\forall a>0$
$$\frac{x^a}{\log x}\to +\infty$$
indeed set $x=e^y\to +\infty$ with $y\to +\infty$
$$\frac{x^a}{\log x}=\frac{e^{ay}}{y}\to+\infty$$