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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?

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  • $\begingroup$ √x(√x-1) >√x for x >4.Hence ? $\endgroup$ – Peter Szilas May 17 '19 at 18:52
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Note that $x-\sqrt x=\sqrt x\bigl(\sqrt x-1\bigr)$. So, $\lim_{x\to\infty}x-\sqrt x=\infty$.

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

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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.

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Let $\sqrt x=\dfrac1h>0$

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

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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.

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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 $+\infty$.

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