My trick from 1968 still works flawlessly. By changing x=-t we have limes with t → + ∞ of (√(t^2-2t)) - t. After anti-rationalization we have limes t → + ∞ -2t/(√(t^2-2t) + t)=-1😎
I like your transformation in the 't' world a lot more. I must admit that I got confused when he took out the negative before he applied the limit value of 'x'. I felt that there might be a mistake without doing this. Big thanks for that comment!
What happened to x/2 You said something about if x gets large it will be zero but I still don't understand how you were be able to just ignore it like that
I don't think I agree. -2x/(x-Sqrt[x^2 + 2x]) = -2x/{x*(1 - Sqrt[1+ 2/x])} ~ as x=> -infinity -2/{1-(1+1/x)} ~2x => -infinity. Please let me know where I went wrong?
@@PrimeNewtonsAn alternative is to write everything down and save as pdf for which you provide a link. I’m a new subscriber and I really like your examples. Math window is also pretty good. The techer showed how to approximate a root to which I responded that I came up with a neat way at the tender age of 14. It bothered me tremendously that my cheap calculator could do something that I couldn’t understand. It works for any integer root and even decimal roots as long as the reciprocal of the decimal part is an integer. Funny thing is that I have never seen this method presented anywhere!
so that is the failed logic. just because the limit is approaching a negative term does not mean that we are to use the negative version with absolute value. You are introducing an extra negative term that is taken care of by using -infinity. if you do not like working with the negative infinity because out intuition is bad just do a change of variable with -x and the limit will go to positive infinity. I do not think you will get the answer you provided.
@@PrimeNewtons so you are introducing a negative when it does not belong. have you tested the answers with a graph or putting in test values. I think you might be surprised at the results as they will not match what you got.
@@PrimeNewtons I would also say that the limit is going towards something as the variable means nothing. We can just as easily replace x with any other variable using change of variable, so it the limit that is approaching something and not the variable.
Kenneth you arent making mathematical sense. You can only know what the limit is approaching after the calculations. He changed it to negative due to x approaching negative infinity
@@LiamFranklinFarang that is adding in an extra negative. The issue is that absolute value has a sharp turn in the graph, so it is not differentiable. We can only find anti-derivatives for things that are differentiable.
