Resolving $\lim\limits_{x \to \infty} (x-{\sqrt x})$ I'm relearning limits and I'm stuck at an exercise. I have to resolve the following limit

$$\lim\limits_{x \to \infty} (x-{\sqrt x})$$

If I use $\infty$ instead of $x$, it'll be $\infty - \infty$ and it gets undefined. I tried rewriting the square root to $x^{1/2}$ but then it remains undefined.
How can I resolve this kind of problem?
 A: Note that $x-\sqrt x=\sqrt x\bigl(\sqrt x-1\bigr)$. So, $\lim_{x\to\infty}x-\sqrt x=\infty$.
A: Hint :
$$(x- \sqrt{x}) = \left(x - \sqrt{x^2\frac{1}{x}}\right) = \left( x - |x|\sqrt{\frac{1}{x}}\right)$$
But since $x \to \infty$, then $x>0$ and $|x| = x$. Thus :
$$(x-\sqrt{x}) = \dots = x \left( 1 - \sqrt{\frac{1}{x}}\right)$$
Can you conclude what happens when $x \to \infty$ now ?
A: Obviously, the second term is neglectable compared to the first. A way to enhance this is by pulling the first term as a factor.
$$\lim\limits_{x \to \infty} (x-{\sqrt x})=\lim\limits_{x \to \infty}x\lim\limits_{x \to \infty}\left(1-\dfrac1{\sqrt x}\right).$$

The factorization by @José is even more convincing.
A: Let $\sqrt x=\dfrac1h>0$
$$\lim_{h\to0^+}\dfrac{1-h}{h^2}=?$$
A: For $x>4$, $\sqrt x < \frac x 2$, and thus $-\sqrt x > -\frac x 2$. So $x-\sqrt x > x-\frac x 2=\frac x 2$. Since $x - \sqrt x $ is bounded below by an expression that goes to $\infty$, it goes to $\infty$ as well.
A: Here it is an alternative approach
\begin{align*}
x-\sqrt{x} = (x-\sqrt{x})\times\frac{x+\sqrt{x}}{x+\sqrt{x}} = \frac{x^{2}-x}{x+\sqrt{x}} = \frac{x-1}{1 + \displaystyle\frac{1}{\sqrt{x}}}
\end{align*}
Since
\begin{align*}
\lim_{x\rightarrow+\infty} \left(1 + \frac{1}{\sqrt{x}}\right) = 1\quad\text{and}\quad\lim_{x\rightarrow+\infty}(x-1) = +\infty
\end{align*}
We conclude the given limit diverges to $+\infty$.
