There was a "typo" in the last two lines on your chalkboard. You left out the "x^2" term. Nevertheless, the answer to the original limit is correct. As an aside, could you do one more squeeze theorem example? Thanks!
I am a 9th grader & really love math......... I was pretty much onto calculus, but only this was what was troubling me, even when I understood Jacobian........ Thank you so much sir for explaining this topic 😊
One small remark. When wrote the second inequality (with e) you said you did not have to change the inequality signs because e to x is positive. The real reason is not that but the fact that e to x is a growing function. Also e to negative x is always positive but it is a diminishing function, and in that case you would need to change the inequality signs.
Take the natural log of this expression: limit as x→0 of 2*ln(x)+sin(1/x) is -∞ since |sin(1/x)|≤1. Thus, e^(-∞)=0, so the limit of x^2*e^sin(1/x) t as x→0 is 0.
Sir I have a doubt that if we put a value near 0 like 0.000001 for x and see we find that the graph of x^2 approaches very fast towards 0 and the value of e^ sin (1/x) is always finite because we can use the thita braking method of 90 multiple and adding the rest part thus in this way we can find the value of it as 0 easily
@@PrimeNewtons yes actually I am in class 9 and I use my father's Google account so yes I think it was new to me so thankyou sir for letting me know about such a good theory😁
Solution: one function is bounded, the other approaches zero. Hence so does the product of the two functions. All these words and equations are unnecessary. Learning math is, in part, learning to be concise.
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There was a "typo" in the last two lines on your chalkboard. You left out the "x^2" term. Nevertheless, the answer to the original limit is correct. As an aside, could you do one more squeeze theorem example? Thanks!
Oh my! Yes, it was a major typo. I hope it doesn't spoil the purpose of the video.
@@PrimeNewtons It didn't spoil it for me. You just forgot to say "come on" and then add the x^2 you inadvertently omitted 😉
Dw king nothing as spoiled, love your content as always :D@@PrimeNewtons
Ive never seen someone discover peoples sandwich preferences while doing math. Love the explanation though!
I am a public investigator. I see things.
For the first time I understood the squeeze theorem. Thank you!
Thank you for your help, you are a great teacher.
Literally the best math TH-camr out there fr
Omg I finally understand how to construct the squeeze theorem inequality to solve the limit. Thanks a lot
Got hungry after this masterpiece
Amazing. As someone who has not taken calculus yet, this makes so much sense to me. Thank you from Canada.
You're very welcome!
The analogy is really crazy , great video. Very well explain thanks man
Indeed you are powerful, thank you very much you have made the understand easy
From Zambia
Thy voice is interesting, it is calm and enthusiastic simultaneously. Great video, fam!
You really make my day, now, I have understood squeeze theory
I am a 9th grader & really love math......... I was pretty much onto calculus, but only this was what was troubling me, even when I understood Jacobian........ Thank you so much sir for explaining this topic 😊
Thanks a bunch!! I almost completly forgot about it.
Thank you ❤
Wow I love math so much ❤❤❤
2:32 This is unintentionally really funny.
Oxford Mathematic will be call you one day for lectures. AL PAZA
One small remark. When wrote the second inequality (with e) you said you did not have to change the inequality signs because e to x is positive. The real reason is not that but the fact that e to x is a growing function. Also e to negative x is always positive but it is a diminishing function, and in that case you would need to change the inequality signs.
Yes. My problem is that I have many assumptions as I speak. Thank you 😊
Never stop learning ❤❤
You are a master who gives easy explanations ❤
What would change if it was x insted of x^2 ?
Hello, I am a student from Jordan. I liked your explanation, but I would like to ask if you can explain Advance Calculus??
ooh and the partial please answer me???
@@ckgcckc200 i do not think he does higher level university mathematics
In Tenet there was a temporal pincer movement, this is mathematical pincer movement
Take the natural log of this expression: limit as x→0 of 2*ln(x)+sin(1/x) is -∞ since |sin(1/x)|≤1.
Thus, e^(-∞)=0, so the limit of x^2*e^sin(1/x) t as x→0 is 0.
4:52 some people are vegetarians....
In the last steps x^2 was omitted from x^2 * e^sin (1/x). Was this by accident?
Yes. It was a major typo.
@@PrimeNewtons Nevertheless, fascinating video!
great calc 1 video as always! i don’t think i can ever eat your sandwiches though 😅
Sir I have a doubt that if we put a value near 0 like 0.000001 for x and see we find that the graph of x^2 approaches very fast towards 0 and the value of e^ sin (1/x) is always finite because we can use the thita braking method of 90 multiple and adding the rest part thus in this way we can find the value of it as 0 easily
That is not the purpose of the video.
@@PrimeNewtons oh sorry you were trying to make people understand about a new theory sorry
In India mainly we do maths like this which I did
@@shamicray Squeeze theorem is not new. Is it new to you? Some call it sandwich theorem.
@@PrimeNewtons yes actually I am in class 9 and I use my father's Google account so yes I think it was new to me so thankyou sir for letting me know about such a good theory😁
For exp(sin(1/x)) is bounded and x^2 converges to 0 as x -> 0. Therefore, the limit of the product is 0 as x -> 0.
For -1
Cool thrm
Or, the theorem of two polices! 😀😉
I like weep cream on my sandwich 😅
Solution: one function is bounded, the other approaches zero. Hence so does the product of the two functions.
All these words and equations are unnecessary. Learning math is, in part, learning to be concise.
You have forgot it x^2 to the e^[sin(1/x)]! 😀😉