Limit of sequence with floor function

Question

$$ \mbox{How do I compute}\quad \lim_{n \to \infty}% \left\lfloor\sqrt[4]{\vphantom{\Large A}n^{4} + 4n^{3}}-n\right\rfloor\ {\large ?} $$

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

Answer 1

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 $n$.

Answer 2

HINT:

Use $$a^4-b^4=(a-b)(a^3+a^2b+ab^2+b^3)$$

$$(n^4+4n^3)^{\frac14}-(n^4)^{\frac14}=\frac{n^4+4n^3-n^4}{(n^4+4n^3)^{\frac34}+(n^4+4n^3)^{\frac24}n+(n^4+4n^3)^{\frac14}n^{^2}+n^3}$$

Now divide the numerator & the denominator by $n^3$

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Alternatively, set $\frac1n=h$

$$\lim_{n\to\infty }[\sqrt[4]{n^{4}+4n^{3}}-n]=\lim_{h\to0^+}\frac{(1+4h)^{\frac14}-1}h$$

Now applying Taylor's expansion $$(1+4h)^{\frac14}=1+\frac h4+O(h^2)$$

Answer 3

If you just extract "$n$" from the radical, it becomes $n \left(1 + \frac{4}{n}\right)^{1/4}$. This is always greater than "$n$" but the term $1 + \frac{4}{n}$ is smaller and smaller and goes to zero when "$n$" increases more and more. So, the radical becomes closer and closer to "$n$" and so the limit is zero as shown in the previous answers.