In this video I explained how to use the negative sign when x approaches negative infinity. i got a copyright claim on the old music I used. I am working on cleaning it up. Watch this newer version • Limit at infinity (of ...
The first video I find that addresses this type of limit and it is absolutely amazing, never seen someone explain math so well and with such positive energy
Thank you for doing this, you don't know how much it means to people that are trying to their best to study and have a full time job, We really appreciate it.
This is || absolutely the best math explanation video I have ever watched, I don't understand why it isn't always taught in common language like this, you have earned my subscription.
Math major student from Philippines here, thank you so much for your thorough discussion, it is way easier to comprehend. I love to learn more about your teaching esp. in calculus and I am hopeful that you will help me. Ty
Thank you, I use videos to help me understand my math homework outside of class, I watched multiple videos before this that wasn't helping me understand how to do the problems. Glad I found yours!
u are so amazing honestly its ur first video I watched and I learned so much like the absolute value I used to square -2 and I forget the negative sign
The last part you said, actually hit me. Thank you! You don't know how much it means to me to hear that, maybe someday, I can finally get that -3! T^T.
Thank you for uploading this video. It helped me better understand limits at negative infinity of rational functions involving radicals. Sometimes I just hate negatives in solving lol.
Thanks again for a good job and a nice problem. I still wonder about he minus (-) sign thought. it is x that will go to negative infinity, but why would that necessarily mean that the limit of the whole function should be negative too. Can't you have a function where the variable goes in one direction while the whole function goes into another? thanks again and keep up the great work.
If the graph for that limit approaches -3, what if x=-(6)⅓? Wouldn't it be undefined? The line for the graph (forgot what its called) pauses at before -(6)⅓ and continues after -(6)⅓, so the line can't be continous like what you showed. Good explanation tho.
The final result is ok , but there is one logic error about x^3 = squart (x^6) when x < 0 . Because the left is negative value but the right is positive value.
This is second video. Formalism is good. But you must write exactly what you present verbally. Also you will see that there is a difference between variables and examples |2| is only 2 but you say that there are two option +2 and -2. Also in other presentation samething is not suncronized beteen write and presentation that is a description not transposed in write. However the central element is verry important and the fact you underline the esențial of using |x| is important. This will help the students. tthe
I dont know if you still read comments, but: at 7:30 you put a - in front of the sqareroot. But why? Since you work with x^6, does this nor count as a positive number? Also you cancel x^6/x^6, so you have "+1" anyways?
so x^6/x^6 while x aproaches -infinity, basically means positive number/ positive number is a negative number? im super confused here how positive number divided on positive number gives you a negative number
@@PrimeNewtons the definition of sq. root is √(𝑥²)=|𝑥|, so if we take the sq. root of (-𝑥)² i.e. √{(-𝑥)²} it will be |-𝑥|=𝑥, there's no connection whether 𝑥→±∞,it will always be positive
You logic for sqrt(9) is not correct Sqrt(9) always equal 3 not -3 We get the nagetive sign not from the square root rather than from x^3 Because the limit is inf/-inf If we replace the x^3 by x^4 and x^6 with x^8 in the question, we will get 3 not -3 Becase the limit will be inf/inf
Be careful! This explanation is relevant to Radical expressions containing exponents that are odd multiples of 2 (for example x^6, x^10, x^14 , etc.). If the highest exponent of the argument of the radical was an even multiple of 2 (for example x^4, x^8, etc.) , the negative will be irrelevant and the limit will be positive. I will provide another video to explain this.
@@odgarig8601 Yes, 1 is odd, so 2*1 is odd multiple of 2 (x^2), 2*2 is even multiple of 2 (x^4), and 2*3 is again odd (x^6) etc. But this is the squared version already. The original was x^3, and here the exponent needs to be odd for the video to be correct and the negative sign to be added. It only works for odd multiples of 2 because when you turn the x^3 into a square root of a square, you actually need absolute value signs around x^3 to do that. And since because x^3 is negative when x approaches negative infinity the |x^3| is -x^3 , and therefore x^3 = -|x^3|, which then you can square and square root, having x^3 = -|x^3| = -sqrt((x^3)^2) = -sqrt(x^6). But if you have even exponent (which when squared will be even multiple of 2), like x^4, it doesn't introduce the negative sign, because x^4 is still positive when x approaches negative infinity. Then you can just take the absolute value, and square and square root: x^4 = |x^4| = sqrt((x^4)^2) = sqrt(x^8).
I have been taking online calculus for the past month and have understood absolutely nothing, but stumbling across this video was the best thing that's ever happened to me this whole semester! Doing a similar problem along with your video for once made me feel like I could solve a problem without resorting to an online calculator. You've got a natural teaching talent, keep on inspiring people!
Student from Canada here. Thank you so much for this. You're a lifesaver. You put it in terms I could understand and now it all makes sense to me. Thank you so much.
