prove:$n^{100}k^{n}=o((k+\varepsilon)^{n})$

Question

Let a constant $k>0$ and $\varepsilon > 0$

prove: $n^{100}k^{n}=o((k+\varepsilon)^{n})$

I understand why $(k+\varepsilon)^{n}$ "goes faster" to infinity than $n^{100}k^{n}$ but i cant understand how to prove it formally.

I tried to prove: $\lim{n\to\infty} \frac{n^{100}k^{n}}{(k+\varepsilon)^{n}} = 0$ but I didn't get anywhere. how can i prove it?

Answer 1

You need to prove that $$\lim_{n\to \infty} {n^{100}\over \left(1+{\varepsilon\over k}\right)^n}=0$$The task is very easily done by $100$ times applying the L^Hoptial's rule by taking $n$ continuously tending to $\infty$ since $1+{\varepsilon\over k}>1$.

Answer 2

The following is true for any $a>0$ and $b>1$: $$\lim_{n\to\infty}\frac{n^{a}}{b^n}=0$$ Then $$\frac{n^{100}k^n}{(k+\varepsilon)^n}=\frac{n^{100}}{\left(1+\frac{\varepsilon}{k}\right)^n}\to 0\;\text{as}\;n\to\infty $$ for any $\varepsilon>0$ using the fact above. See this question for various proofs of this fact. Note that this is a standard theorem.