Which is asymptotically larger $n \log(n)$ or $n$ analytically? If we take the ratio we get $\log(n)$ and if we set $\lim_{n\to\infty}\log(n)$ the value of the limit is $\infty$. Then how can we tell which is asymptotically larger?
$n \log(n)$ is intuitively asymptotically larger than $n$. But how can it be proved analytically?
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When you divide two equatipns you get $log(n)$ in numerator. Now for big values of n ofcourse $log(n)>1$ thus its proved. Also for small $n$ the equation $log(n)\approx n$ so $nlog(n)$ becomes $\approx n^2$ while it is easily observed that $nlog(n)>n$ for big values or you can also sketch their graphs place and you will get relation as $nlog(n)>=n$
If you dig into the definition of $\lim_{x\to\infty}f(x)=\infty$, you can see that it means for all $M$, there exists $x_M$ such that $f(x)\ge M$ for $x\ge x_M$.
In terms of comparing $n$ and $n\log n$, we can apply this reasoning to get some fairly explicit bounds. Let $M$ be some (large, if you'd like) positive number - then, whenever $n\ge e^M$, we have that $\log n \ge M$, and thus that $n\log n \ge Mn$.
So, for any $M>0$, $n\log n \ge Mn$ eventually. Hopefully this goes some way to justifying that $n\log n$ dominates $n$ by some margin.