$$
\mbox{How do I compute}\quad
\lim_{n \to \infty}%
\left\lfloor\sqrt[4]{\vphantom{\Large A}n^{4} + 4n^{3}}-n\right\rfloor\ {\large ?}
$$
I know that could use squeeze theorem, but for that I would need to find two sequences, which I don't know how to find. For floor function I know that I can somehow use $x – 1 < \left\lfloor x\right\rfloor \leq x$.
Thank you
Best Answer
Let $n$ be a positive integer. Clearly $$ n^4 < n^4 + 4 n^3 < n^4 + 4 n^3 + 6 n^2 + 4 n + 1 = (n+1)^4. $$ Therefore $$ n < \sqrt[4]{n^4 + 4 n^3} < n + 1. $$ Thus $$ 0 < \sqrt[4]{n^4 + 4 n^3} - n < 1. $$ Hence $$ \left\lfloor \sqrt[4]{n^4 + 4 n^3} - n \right\rfloor = 0 $$ for all positive integers $n$.