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

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..

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

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
$$\lim_{x\to{\infty}} (x-\sqrt{x^2+x})=\lim_{h\to0^+}\dfrac{1-\sqrt{1+h}}h$$
$$=\lim_{h\to0^+}\dfrac{1-(1+h)}h\cdot\dfrac1{\lim_{h\to0^+}(1+\sqrt{1+h})}=\cdots$$
• how are you getting $\sqrt{1+\frac1x}$ on the denominator? – Panthy Jul 7 '15 at 9:51
• @Panthy, $x^2+x=x^2\left(1+\dfrac1x\right)$ – lab bhattacharjee Jul 7 '15 at 9:52