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Given this question $$\lim_{x\to{\infty}} (x-\sqrt{x^2+x})$$

Find the limit

Work so far...:

$$\lim_{x\to{\infty}} \frac{x^2-x^2-x}{x+\sqrt{x^2+x}}$$ $$\lim_{x\to{\infty}} \frac{-x}{x+\sqrt{x^2+x}} $$

Would the easiest way to proceed to be dividing bottom and top by $x$?

If so then is this attempt correct?:

$$\lim_{x\to{\infty}} \frac{-1}{1+\sqrt{x+\frac1x}}$$ $$\lim_{x\to{\infty}}\frac{-1}{1+\sqrt\infty} $$

Something is wrong there..

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    $\begingroup$ One of the easiest ways. $\endgroup$ – Daniel Fischer Jul 7 '15 at 9:47
  • $\begingroup$ When you divide by $x$ the denominator, you must carry $1/x^2$ inside the square root, so the denominator is $1+\sqrt{1+\frac{1}{x}}$. $\endgroup$ – egreg Jul 7 '15 at 10:53
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Yes $$\lim_{x\to{\infty}} \frac{-x}{x+\sqrt{x^2+x}} =\lim_{x\to{\infty}} \frac{-1}{1+\sqrt{1+1/x}} =\cdots$$


Alternatively set $1/x=h$ to get

$$\lim_{x\to{\infty}} (x-\sqrt{x^2+x})=\lim_{h\to0^+}\dfrac{1-\sqrt{1+h}}h$$

$$=\lim_{h\to0^+}\dfrac{1-(1+h)}h\cdot\dfrac1{\lim_{h\to0^+}(1+\sqrt{1+h})}=\cdots$$

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  • $\begingroup$ how are you getting $\sqrt{1+\frac1x}$ on the denominator? $\endgroup$ – Panthy Jul 7 '15 at 9:51
  • $\begingroup$ @Panthy, $x^2+x=x^2\left(1+\dfrac1x\right)$ $\endgroup$ – lab bhattacharjee Jul 7 '15 at 9:52
  • $\begingroup$ Ah i see. Thank you for your help! $\endgroup$ – Panthy Jul 7 '15 at 9:53

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