For this, I know you can use calculus techniques and set $\lim _{x\to \infty }\left(\frac{4n}{n^2}\right)$ and get 0. Though, I'm wanting to know how you can get a specific constant and k values.
Little O Computation: $4n\:\in \:o\left(n^2\right)$
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asymptotics
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2What is the definition of little o you use? – 2017-01-31
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0And what is $k$? – 2017-01-31
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0That limit is only 0 if n = 0. For any n > 0, it is 4/n > 0. – 2017-01-31
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0In other words, there is a typo - $x$ should be $n$ in the limit. – 2017-01-31
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Since you don't want to use the limit, we can in this case use the definition (see here for details): $$ f(n)=o(g(n)) \stackrel{\text{def}}\equiv \forall k>0\,\exists N\ge 0\, \forall n\ge N\,[\:f(n)\le k\cdot g(n)\:] $$ In other words, the definition is exactly the same as for big-O, except that the inequality eventually holds for all positive constant multiples $k$.
With this in mind, choose any positive $k$. Then to have $$ 4n\le k\:n^2 $$ we need the equivalent $$ \frac{4}{k}\le n $$ so for any positive $k$ we'll have $4n\le n^2$ for all $n\ge N = 4/k$.