# 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?

• √x(√x-1) >√x for x >4.Hence ? – Peter Szilas May 17 '19 at 18:52

Note that $$x-\sqrt x=\sqrt x\bigl(\sqrt x-1\bigr)$$. So, $$\lim_{x\to\infty}x-\sqrt x=\infty$$.

• How did you get to sqrt(x) * (sqrt(x)-1)? – Novac May 17 '19 at 18:56
• Factoring out $\sqrt{x}$. – KM101 May 17 '19 at 18:58
• Got it, thank you for your help! – Novac May 17 '19 at 19:02

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 ?

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.

Let $$\sqrt x=\dfrac1h>0$$

$$\lim_{h\to0^+}\dfrac{1-h}{h^2}=?$$

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.

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 $$+\infty$$.