Squeeze/Sandwich Theorem

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In this video , I explained how to use the sandwich theorem. The key strategy is to find a part of the function that is bounded.

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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!

BartBuzz

For the first time I understood the squeeze theorem. Thank you!

Orillians

Ive never seen someone discover peoples sandwich preferences while doing math. Love the explanation though!

JourneyThroughMath

Thank you for your help, you are a great teacher.

AdityaDonkada

Literally the best math YouTuber out there fr

figgles

Omg I finally understand how to construct the squeeze theorem inequality to solve the limit. Thanks a lot

wilsonwu

Amazing. As someone who has not taken calculus yet, this makes so much sense to me. Thank you from Canada.

Zachary_Roemmich

The analogy is really crazy, great video. Very well explain thanks man

reallegendcode

You really make my day, now, I have understood squeeze theory

ChipiliroChinkudzu

Indeed you are powerful, thank you very much you have made the understand easy

WisdomMudenda-we

Thy voice is interesting, it is calm and enthusiastic simultaneously. Great video, fam!

PaulRodrichSancti

I am a 9th grader & really love I was pretty much onto calculus, but only this was what was troubling me, even when I understood Thank you so much sir for explaining this topic 😊

Subham-Kun

Thanks a bunch!! I almost completly forgot about it.

malikahashami

Thanks for the vid sir but plss i want to ask whether you can make a video of using squeeze theorem on normal limits of like algebraic or rational functions, i would really appreciate plss🙏 🙏

josephgyasiborr

2:32 This is unintentionally really funny.

idkgoodname

Hello, I am a student from Jordan. I liked your explanation, but I would like to ask if you can explain Advance Calculus??

tuqa

You are a master who gives easy explanations ❤

abdelbakib

Oxford Mathematic will be call you one day for lectures. AL PAZA

GPSPYHGPSPYH-dsgu

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.

harrikarri

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.

elmer

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