How to find the limit of [sqrt (x^2+x) - 2 sqrt (x^2+2x) + x ]as x approaches infinity
T TSYUIA New member Joined Apr 4, 2020 Messages 1 Apr 4, 2020 #1 How to find the limit of [sqrt (x^2+x) - 2 sqrt (x^2+2x) + x ]as x approaches infinity
pka Elite Member Joined Jan 29, 2005 Messages 11,990 Apr 4, 2020 #2 TSYUIA said: How to find the limit of [sqrt (x^2+x) - 2 sqrt (x^2+2x) + x ]as x approaches infinity Click to expand... Note that as \(x\to\infty,\:\;x>0.\) Now \(\sqrt{x^2+x}-\sqrt{x^2+2x}=\dfrac{-x}{\sqrt{x^2+x}+\sqrt{x^2+2x}}\)
TSYUIA said: How to find the limit of [sqrt (x^2+x) - 2 sqrt (x^2+2x) + x ]as x approaches infinity Click to expand... Note that as \(x\to\infty,\:\;x>0.\) Now \(\sqrt{x^2+x}-\sqrt{x^2+2x}=\dfrac{-x}{\sqrt{x^2+x}+\sqrt{x^2+2x}}\)
Steven G Elite Member Joined Dec 30, 2014 Messages 14,596 Apr 4, 2020 #3 The function that the OP is taking the limit of is [math]\sqrt{x^2+x} - 2\sqrt{x^2+2x} + x[/math]
