I'm only familiar with limits as they approach a number. Thanks
Q questions77 New member Joined Sep 13, 2020 Messages 2 Sep 13, 2020 #1 I'm only familiar with limits as they approach a number. Thanks
pka Elite Member Joined Jan 29, 2005 Messages 11,971 Sep 13, 2020 #2 questions77 said: View attachment 21596 Click to expand... Note that as \(x\to \infty\) then \(\dfrac{1}{x}\to 0\). \(x-\sqrt{x^2+7x}=\dfrac{-7x}{x+\sqrt{x^2+7x}}=\)\(\dfrac{-7}{{1}+\sqrt{1+\tfrac{7}{x}}}\)
questions77 said: View attachment 21596 Click to expand... Note that as \(x\to \infty\) then \(\dfrac{1}{x}\to 0\). \(x-\sqrt{x^2+7x}=\dfrac{-7x}{x+\sqrt{x^2+7x}}=\)\(\dfrac{-7}{{1}+\sqrt{1+\tfrac{7}{x}}}\)