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.
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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!
Just screenshoted in case my brain works
I wish you were my teacher. Thank you so much for this clear explanation!
i was super confused when I got a question like this on my homework and kept getting it wrong, but you taught me how to do it! thank you so much ❤️
Glad I could help!
YOUR METHODS ARE REALLY HELPFUL. MY CALCULATIONS ARE GETTING BETTER 🎉
Great to hear!
From the moment I started watching these videos on TH-cam I started liking maths again, the problem is that i already failed in the national exam 🙂✨
There's always a future and a hope
@@PrimeNewtonsthanks 🙏♥️✨
This was so helpful, I appreciate it 🙏🙏🙏
Fantastic video. Never felt lost
the best explanation I have ever watched
Wow! Thank you
youre the best!!!!!!!!
I am really getting you clearly
"Those who stop learning have stopped living." Nice
You just gained another subscriber🔥🙌
Instead of deviding by x I personally prefere taking |x| outside the square root. I think it is simpler. Thank you for your interesting content.
Thank you so much for your explanation!! It is extremely helpful!!!
Very nice limit - thanks a lot👍
Thank u so much
Thx so much this really helped me, can you explain about limit as x approaches to negative zero too sir
I'll see what I can do
Thank you!!
i like the cut of this guys jib
Well explained 🙏
Good work
Thank you.
Nice explanation!
Very good. Thanks 👍
Thanks!
Thank you 😊
Welcome!
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
Why do we make the radical negative but not everything else divided by x?
It is called the Conjugate. You change the middle sign. If it was negative, you make it positive.
Hi, I just watched you vid and would like to get more explanations on why the denominator root get's a minus sign infront
tnx so much
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?
Does this also apply if the infinity is at positive?
Yes
But I would like your help in vectors as well
If you email me the problem I may be able to help. Primenewtons@gmail.com
Awesome!
Thx
Very nice
For the mistake of root x sq equals mod x we can substitute x equals minus x in the question
Good. But 2/x as x goes to negative infinity is undefined.
Again!!
Yes!
Iam 60 years ago. l follow you .are you africain or americain
Nigerian
Wow
I would have rewritten the limit like this
x(1+sqrt(1+2/x)) and I would have got -inf
Same result in wolfram alpha
The lettering are too tiny for people viewing it on phone. Zoom your write-ups
Thanks for the suggestion. I believe newer videos have improved zoom. Please check them out and let me know what you think.
@@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.
Not that the limit is approaching negative infinity, but that x is approaching negative infinity. And yes, that's the only reason.
@@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.