Evalutating $\lim \limits_{ x \to \infty} (x+1)^k - (x)^k$ , given $0<k<1$ [duplicate]

I Tried evaluating as follows but got stuck with the uncertainty $( 0·\infty ) $

What I did was :

$\lim \limits_{ x \to \infty} (x+1)^k - (x)^k = \lim \limits_{ x \to \infty} (x)^k·(\frac{(x+1)^k}{(x)^k} - 1) = \lim \limits_{ x \to \infty} (x)^k·((\frac{x+1}{x})^k - 1) = $
$\lim \limits_{ x \to \infty} (x)^k·((1+\frac{1}{x})^k - 1) $

$\lim \limits_{ x \to \infty} (x)^k·(e^{ln(1+\frac{1}{x})^k} - 1) = (e^{k·ln(\lim \limits_{ x \to \infty}(1+\frac{1}{x})} - 1) ·\lim \limits_{ x \to \infty} (x)^k = (e^{k·ln(1)} - 1) ·\infty $ = $(e^{0} - 1) ·\infty = (1 - 1) ·\infty $ = \

⇒ $0 ·\infty $

( I Also Tried using the Squeeze theorem but all upper functions I tried went to $\infty$ )

➔ Using $k=\frac{1}{2}$ and then $\lim \limits_{ x \to \infty} \sqrt{(x+1)} - \sqrt{(x)}$ :

I found that $\lim \limits_{ x \to \infty} \frac{1}{\sqrt{(x+1)} + \sqrt{(x)}}$ = $\lim \limits_{ x \to \infty} \frac{1}{\infty} = 0$ ,

However , I'm struggling to do the same algorithm with $\lim \limits_{ x \to \infty} \sqrt[q]{(x+1)} - \sqrt[q]{(x)}$ : Assuming $k=\frac{1}{q} $ ; $q \in (\mathbb{R}^+/ 0)$

(meaning $\lim \limits_{ x \to \infty} (x+1)^k - (x)^k$ = $\lim \limits_{ x \to \infty} \sqrt[q]{(x+1)} - \sqrt[q]{(x)}$)

I would love to get some help on how to evalutate the limit !

You can just rewrite the expression and use L'Hôpital's rule:

$$ \lim_{x\to +\infty}\left((x+1)^k-x^k\right)=\lim_{x\to +\infty}\dfrac{(1+1/x)^k-1}{1/x^k}=\lim_{x\to +\infty}\dfrac{(-k/x^2)(1+1/x)^{k-1}}{-k x^{-k-1}}=\lim_{x\to +\infty}(x+1)^{k-1} = 0 $$

Use Taylor polynomial: as $h \to 0$, we have $(1+h)^k = 1+kh+o(h)$.
As $x\to\infty$, $$ (x+1)^k = x^k\left(1+\frac{1}{x}\right)^k = x^k\left(1+\frac{k}{x}+o(1/x)\right) = x^k+kx^{k-1}+o(x^{k-1}) \\ (x+1)^k - x^k = kx^{k-1}+o(x^{k-1}) $$ Now $k<1$ so $k-1 < 0$ and thus $\lim_{x\to\infty} x^{k-1} = 0$. Conclude $$ (x+1)^k - x^k \to 0\quad\text{ as }x\to\infty $$ or more pricisely $$ (x+1)^k - x^k \sim kx^{k-1}\quad\text{ as }x\to\infty $$ Example: $$ \sqrt{x+1}-\sqrt{x} \sim \frac{1}{2\sqrt{x}}\quad\text{ as }x\to\infty $$