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.
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.
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. 😊
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.
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.
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.
@@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
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
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
This is the quality calculus videos I have long been looking for all that years. You are a real asset, no doubt!
This explanation is far clearer than the old one. Thank you.
Glad it was helpful!
when you are good, you are good. no tutor on youtube is like you
You've put a smile on my face. Thank you!
My pleasure!
I saw lots of videos in this case but you are the best thank you.
Thank you
I wish you were my teacher in my school days. You're really a wonderful teacher. Well done!!
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.
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.
hi , its the 1st video that I saw in this channel, everything was clear , thank you for this big work!
long life to you my perfect teacher
This vedio is so beneficial to me🖤thanks for making such a outstanding vedio❤
Glad you liked it
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.
Thank you so much! Your work means the world to me!
dude thanks you just make everything make sense
U r a good guy,
Very clear👍
Glad to hear that
love from Bangladesh 🖤
I love the thumbnail for this video 🤪😵💫🤓
Thanks. My daughter took the photo.
I freaking love learning from Will Smith ;)
The best❤
U a like a prince of math
you're my herooooo❤❤❤
Nice video!
Helloooooo Ambika
The value under the radical will always be positive. It is the denominator the determines the sign of the limit.
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 !
Very good. Thanks Sir
Sir you are great ❤
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:
ruclips.net/video/qdifmP6K_j4/видео.html
thank you!
Nice Job
You found me here!
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.
Thank you professor
Very helpful !
you are great!
Amen!
طريقة جيدة الشرح
i like your vedios last ten seconds explained well before 10 sec i'm confused
Nice 👍
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.
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.
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
=> must be replaced by =
Loved this
thank you bro.
Great!.. Thanks!
Damn!🔥
waw
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
-> , trying to find that one. =>?. Found.
This still confuses me.
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.🥺🥰
I was taught 9/9=1. Gotta re-learn.😂😢😢
Absolute value is fascinating.
Life saver 🛟