At the of 75 in a retired life I am enjoying your brilliant way of teaching mathematics; best wishes for your great work (from a senior citizen of India)
My brother you have explained the concept of squeeze theorem so simple thank so much pliz continue to make more videos you are a great tutor I have ever watched.
Apart from the clear way of your explanations you have a soothing voice which is just perfect. I am very glad I have found your videos. I taught math for many years.
It doesn't matter because if negative x^5 will be negative and so will change the inequality signs on both sides but that would then retain the original inequality just written from right to left and you can rewrite it. So, the short answer is that nothing will happen.
If it were not for the mandatory use of squeeze theorem, could you have said that sin(5/x) would go to an undetermined value between -1 and 1, which, when multiplied by x⁶ which would go to 0, would also become 0
I dont understand only one part though: How did you multiply the entire inequality with x^6 because thats basically 0? Or its not? Maybe its allowed because its very close to 0 from the right (positive) side? Please clarify on this.
The first example is quite common. Multiply x by 8 then multiply sin(8x) by 8. So the limit is 8. For the second, do the same think using sqrt t instead of 8. You should get 0 answer.
That first one requires a whole ‘nother one or two videos to explain. That second may require some dlc we learned in calc two…or you would at least have to come back to it after you learn derivatives and interpret it as one.
Ok for that second one you can do it with raw limits my argument is you can say sin x0+ sin x/sqrt(x) 0 x/sqrt x because your dividing more cookies by the same number of people Now we need a lower bound argument for the squeeze theorem. The limit is positive because positive/positive is positive for my lower bound I can just use the opposite lim x->0+ -x/sqrt(x) < L < lim x->0 x/sqrt(x). Since x/sqrt(x)=sqrt(x) -sqrt(0)0- sin(x)/sqrt(x) lim u->0+ sin(-u)/sqrt(-u) lim u->0+ -sin(u)/(i*sqrt(x)) Lim u->0+ (-1/i)*[sin(u)/sqrt(u)] Lim u-> 0+ (-1/i)*[0]= 0
![Limit of absolute value functions [ lim |x^3 - x|/(x^3 - |x|) as x goes to 0 ]](http://i.ytimg.com/vi/nJ9lEolfVpU/mqdefault.jpg)
If more professorstaught like you, the world would be a better place...love from south africa
Thank you
At the of 75 in a retired life I am enjoying your brilliant way of teaching mathematics; best wishes for your great work (from a senior citizen of India)
Many thanks!
The way you offer your classes is just fun, simple and fast.
"Those who stopped learning, stopped living."💯❤That's powerful sir. You are doing incredible work.
My brother you have explained the concept of squeeze theorem so simple thank so much pliz continue to make more videos you are a great tutor I have ever watched.
This solution was so understandable and clear. Thanks 🎉
Thank you so much for making this easier to solve and understandable
Apart from the clear way of your explanations you have a soothing voice which is just perfect. I am very glad I have found your videos. I taught math for many years.
Wow! Just exciting.
Thank you sir.
You are a great help, thank you
like.. thank u sm and i would be lost without u!!
big love from Iraq
Nice explanation. Easy to follow.
Great explanation.
You're just awesome my brother
THANK YOU FOR YOUR SUPPORT
Salute bro💯💯 i gained more in your tutorial
Great my brilliant sir.
We could use the sinu=u formula as you said to get 0
But I really like this method I learned it today from you !
Keep up the great work man ❤️🔥
Tomorrow will be my exam and i thaught squeeze theorem is easy enough but when i try to do questions i stuck up and then this video 😊 .
Thanks a lot
you have a beautiful hand writing
thank u very much 👍🙏
please do more!
Good one💥
Hey brother i have one question, what would happen if i replaced x^6 with an odd power? Thx for all your videos they really help me out a lot 🙏🙏
It doesn't matter because if negative x^5 will be negative and so will change the inequality signs on both sides but that would then retain the original inequality just written from right to left and you can rewrite it. So, the short answer is that nothing will happen.
that`s great🎉🎉🎉
you are brilliant 10q
Since you made a few videos using the squeeze theorem, can you please make up a video that proofs this theorem? That would be cool.
lovely lol i wish i was your student
Thanks for your lessons, why didn't you divide all sides by × from 5/× at the middle?
If it were not for the mandatory use of squeeze theorem, could you have said that sin(5/x) would go to an undetermined value between -1 and 1, which, when multiplied by x⁶ which would go to 0, would also become 0
Guy is damn smart.
Wow you are smart
❤❤
Thank you so helpful if it was x inplace of x^6 and the second question if x approaches 0 from the right or from the left could you please reply
All 3 answers are 0
@@PrimeNewtons so if the exponent of the x is odd it don't matter as all the inequalities will stay the same all all of them will flip right?
😊
What happen if x has an odd exponent ? do we assume negative or ? depend on what value x is approaching
I dont understand only one part though: How did you multiply the entire inequality with x^6 because thats basically 0? Or its not? Maybe its allowed because its very close to 0 from the right (positive) side? Please clarify on this.
GOOD SIR WHERE CAN WE FIND YOU MAYBE TO ADRESS SOME KIND OF QUESTIONS.
Message me on Twitter or email primenewtons @gmail.com
kindly do triagle of inequality
Hello Sir, I wanna if ever what if the power was odd , were we gonna change the sign since we didn't change because it was to the power of 6
It will adjust in the end even if it's odd. Watch the other squeeze Theorem video. I explained it there.
Sir can you pleasee make video on proving squeeze theorem and please make a video on Weisteress Thereom with proofs
What if we have a fraction for example the limit of sin(8x)/x as x approaches 0 or the limit of sin(t)/square root of t as t approaches 0
The first example is quite common. Multiply x by 8 then multiply sin(8x) by 8. So the limit is 8. For the second, do the same think using sqrt t instead of 8. You should get 0 answer.
That first one requires a whole ‘nother one or two videos to explain. That second may require some dlc we learned in calc two…or you would at least have to come back to it after you learn derivatives and interpret it as one.
Ok for that second one you can do it with raw limits my argument is you can say sin x0+ sin x/sqrt(x) 0 x/sqrt x because your dividing more cookies by the same number of people
Now we need a lower bound argument for the squeeze theorem. The limit is positive because positive/positive is positive for my lower bound I can just use the opposite
lim x->0+ -x/sqrt(x) < L < lim x->0 x/sqrt(x).
Since x/sqrt(x)=sqrt(x)
-sqrt(0)0- sin(x)/sqrt(x)
lim u->0+ sin(-u)/sqrt(-u)
lim u->0+ -sin(u)/(i*sqrt(x))
Lim u->0+ (-1/i)*[sin(u)/sqrt(u)]
Lim u-> 0+ (-1/i)*[0]= 0
2024 GANG