Evaluate $\displaystyle\lim_{x\rightarrow +\infty} \left(\sqrt[n]{n + x} - \sqrt[n]{x}\right)$

Question

Is there a way to evaluate the following limit without resorting to power series expansion?

$$\displaystyle\lim_{x\rightarrow +\infty} \left(\sqrt[n]{n + x} - \sqrt[n]{x}\right), \quad\text{ $n$ is a positive integer greater than 1.}$$

I've attempted approaching the problem using Squeeze Theorem like this one but to no avail.

EDIT: Preferably, the solution should only involve inequalities with appealing to asymptotics or approximation.

Answer 1

By Bernoulli's inequality

$$\sqrt[n]{n + x} =\sqrt[n]{ x}\left(1+\frac n x\right)^{\frac 1n} \le \sqrt[n]{ x}\left(1+\frac 1n\frac n x\right)=\sqrt[n]{ x}\left(1+\frac 1 x\right)$$

then

$$0\le \sqrt[n]{n + x} - \sqrt[n]{x}\le \frac 1 {x^{1-\frac1n}} $$

Answer 2

As $x\to+\infty$,$$\left(\sqrt[n]{n + x} - \sqrt[n]{x}\right)=x^{1/n}\left(\left(1+\frac nx\right)^{1/n}-1\right)\sim x^{\frac1n-1}\to0.$$

Answer 3

Using $$ a^n-b^n=(a-b)\sum_{k=1}^{n-1}a^{n-k}b^k, \sqrt[n]{n+x}\ge\sqrt[n]{x} $$ one has $$0<\sqrt[n]{n + x} - \sqrt[n]{x}=\frac{\left(\sqrt[n]{n + x} - \sqrt[n]{x}\right)\sum_{k=1}^{n-1}\sqrt[n]{n+x}^{n-k}\sqrt[n]{x}^k}{\sum_{k=1}^{n-1}\sqrt[n]{n+x}^{n-k}\sqrt[n]{x}^k}\le\frac{n}{nx}=\frac1x $$ which implies the limit.