compute the limit lim x→∞ (x^2+x)^0.5 − (x^2 − x)^0.5

[hndalama]

lim x→∞ (x²+x)^0.5 − (x² − x)^0.5

My attempt is to try using L'Hopital's rule by converting it to a fraction by multiplying it by [(x²+x)^0.5 + (x² − x)^0.5] / [(x²+x)^0.5 + (x² − x)^0.5]
when i do this I get 2x / [(x²+x)^0.5 + (x² − x)^0.5] but using L'Hopital's rule seems to make the limit more complicated instead of less.

 

[skaa]

\(\displaystyle \lim_{x\to \infty}\frac{2x}{\sqrt{x^2+x}+\sqrt{x^2+x}}=\lim_{x\to \infty}\frac{2x}{\sqrt{x}(\sqrt{x+1}+\sqrt{x+1})}=\lim_{x\to \infty}\frac{2x}{\sqrt{x}(\sqrt{x}+\sqrt{x})}=\lim_{x\to \infty}\frac{2x}{\sqrt{x}\cdot 2\sqrt{x}}=1\)

 

[hndalama]

\(\displaystyle \lim_{x\to \infty}\frac{2x}{\sqrt{x^2+x}+\sqrt{x^2+x}}=\lim_{x\to \infty}\frac{2x}{\sqrt{x}(\sqrt{x+1}+\sqrt{x+1})}=\lim_{x\to \infty}\frac{2x}{\sqrt{x}(\sqrt{x}+\sqrt{x})}=\lim_{x\to \infty}\frac{2x}{\sqrt{x}\cdot 2\sqrt{x}}=1\)

Thanks for your help

 

[Deleted member 4993]

The denominator is not (x²+x)^{0.5 +} (x² + x)^0.5 It is (x²+x)^{0.5 +} (x² - x)^0.5

It is the same process. You can go ahead and get the correct result and sahre your answer with us.