I absolutely adore this man and his ability to teach. I LOVE solving problems like this because I know how to do them now, thanks to him. He makes math so comedic, natural, interesting, and engaging. Thanks so much!
That last 10 seconds of explaining when to apply the negative sign in a square root situation as the x approaches the negative infinity will be a huge help. Not that it's all that difficult to explain it to yourself over and over again, but in a testing environment, a rule that you can rely on is very comforting.
Thank you so much, you’re amazing. I was struggling for the last few hours wondering why my answers were wrong only because of the sign. My answers were right but the signs were off. Thank you for everything. I appreciate you and this video so much. I’m a high school student taking calculus. I just subscribed to you. Have a wonderful day. 😊
I'm always queasy working with square roots, but this teacher made me look at things differently - it makes so much sense now! Right after he solved that first equation I seriously wanted to jump up and start clapping.
but to remove all confusion, put t=minus x. so we hv lim t tending to +infinity of (root(9t^6+t))/-t^3+ 6 which ultimately= -1. in second case it is (root(9t^4+t))/ t^2+6 wbich ultimataly becomes +1.
Could you simply argue that, for the cubic, any negative value of x will always give a positive number under the square root in the numerator. By taking the sqrt you would take the principle root, thus making the numerator positive overall. The denominator is clearly negative (for sufficiently large x), so you know the limit will be a negative? By the same logic, the squared limit is positive over positive, therefore, the limit is also positive.
Hey, your videos are amazing. How do you think of making playlists like curriculum for those who just started to learn calculus. I can see that there is already a playlists for calculus, but it looks like all mixed up.
Embora o resultado estejja correto, a explicação esta muito confusa. O resultado pode ser obtido naturalmente se fizermos uma mudança de variável x = - u, assim o primeiro limite ficará lim[(9x^6-x)^(1/2)/(x^3 +6)] para x tendendo para menos infinito = lim[(9(-u)^6 -(-u))^(1/2)/((-u)^3 +6)]= lim[(9u^6+u)^(1/2)/(6-u^3)] = lim[(9u^6/u^6 +u/u^6)^(1/2)/(6/u^3 - 1)] = lim[(9+1/u^5)^(1/2)/(6/u^3 -1)] = (9 + 0)^(1/2)/(0-1) = 3/(-1) = -3, quando u tende para + infinito
@@InfoMedia1403 why would it be -1? Even if x is negative, it would be a negative number divided by another negative number, which will always become positive
I absolutely adore this man and his ability to teach. I LOVE solving problems like this because I know how to do them now, thanks to him. He makes math so comedic, natural, interesting, and engaging. Thanks so much!
omg, I would have never thought I would binge-watch the calculus video. I owe you some of my tuition
This is the funniest comment ever!
That last 10 seconds of explaining when to apply the negative sign in a square root situation as the x approaches the negative infinity will be a huge help. Not that it's all that difficult to explain it to yourself over and over again, but in a testing environment, a rule that you can rely on is very comforting.
First time visiting your channel, is like it was made just for me
Thank you so much, you’re amazing. I was struggling for the last few hours wondering why my answers were wrong only because of the sign. My answers were right but the signs were off. Thank you for everything. I appreciate you and this video so much. I’m a high school student taking calculus. I just subscribed to you. Have a wonderful day. 😊
I'm glad the video helped. I appreciate the feedback.
This explanation is far clearer than the old one. Thank you.
Glad it was helpful!
I wish you were my teacher in my school days. You're really a wonderful teacher. Well done!!
when you are good, you are good. no tutor on youtube is like you
I saw lots of videos in this case but you are the best thank you.
Thank you
I'm always queasy working with square roots, but this teacher made me look at things differently - it makes so much sense now! Right after he solved that first equation I seriously wanted to jump up and start clapping.
You've put a smile on my face. Thank you!
My pleasure!
long life to you my perfect teacher
This guy is very amazing one , thank you for your great job you have supported me in the world of mathematics 🙏🙏🙏🙏🙏🙏
Wow! I'm glad my videos helped.
This vedio is so beneficial to me🖤thanks for making such a outstanding vedio❤
Glad you liked it
Very good. Thanks Sir
I freaking love learning from Will Smith ;)
I love the thumbnail for this video 🤪😵💫🤓
Thanks. My daughter took the photo.
U r a good guy,
dude thanks you just make everything make sense
طريقة جيدة الشرح
Very clear👍
Glad to hear that
Thank you so much! Your work means the world to me!
The best❤
You're the best ❤ thanks
The left problem , second line ??? Factorize the x^3 as you explained at the beginning-> that why (-x)^3 -> it is clearer for students !
Nice video!
Helloooooo Ambika
Nice Job
You found me here!
but to remove all confusion, put t=minus x. so we hv lim t tending to +infinity of
(root(9t^6+t))/-t^3+ 6 which ultimately= -1.
in second case it is (root(9t^4+t))/ t^2+6 wbich ultimataly becomes +1.
you are great!
Amen!
Could you simply argue that, for the cubic, any negative value of x will always give a positive number under the square root in the numerator. By taking the sqrt you would take the principle root, thus making the numerator positive overall. The denominator is clearly negative (for sufficiently large x), so you know the limit will be a negative?
By the same logic, the squared limit is positive over positive, therefore, the limit is also positive.
Thank you professor
Loved this
Very helpful !
Hey, your videos are amazing. How do you think of making playlists like curriculum for those who just started to learn calculus. I can see that there is already a playlists for calculus, but it looks like all mixed up.
you're my herooooo❤❤❤
can you please make a video of the Reimann integers and the mean value theorem? thank you! Your videos are very helpful
Riemann Integral/Sum: Right/Left Endpoints, Upper Sum, Lower Sum and Midpoint Estimate
He has a video coverin Reimann integrals:
ua-cam.com/video/qdifmP6K_j4/v-deo.html
thank you!
Great!.. Thanks!
=> must be replaced by =
waw
Nice 👍
Damn!🔥
Embora o resultado estejja correto, a explicação esta muito confusa.
O resultado pode ser obtido naturalmente se fizermos uma mudança de variável
x = - u, assim o primeiro limite ficará lim[(9x^6-x)^(1/2)/(x^3 +6)] para x tendendo para menos infinito = lim[(9(-u)^6 -(-u))^(1/2)/((-u)^3 +6)]= lim[(9u^6+u)^(1/2)/(6-u^3)] = lim[(9u^6/u^6 +u/u^6)^(1/2)/(6/u^3 - 1)] = lim[(9+1/u^5)^(1/2)/(6/u^3 -1)] = (9 + 0)^(1/2)/(0-1) = 3/(-1) = -3, quando u tende para + infinito
Why doesn’t x^3 / x^3 at 6:42 equal -1?
I think I explained in the video. That is 1
@@PrimeNewtonsyou didn't explain it. And even if you do, it's not clear
@@InfoMedia1403 why would it be -1? Even if x is negative, it would be a negative number divided by another negative number, which will always become positive
No it is not true😮 at first question you add minas, but the second you do not add minas. although the both of question have even power.
Hope you understand that math is not about whether you agree or not. It's about whether you are correct. Never stop learning!
@@PrimeNewtons but still I do not know. thank you.🥺🥰
-> , trying to find that one. =>?. Found.
I was taught 9/9=1. Gotta re-learn.😂😢😢
Absolute value is fascinating.
This still confuses me.
Life saver 🛟
373 videos is alot to digest, could you be so kind to take only the essential theory videos of calculus and compile them in a single playlist?