10 Challenging Limit Problems! Can you survive?😱(part1)🔎Riemann Sum, Squeeze Theorem, Floor Function
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- Опубликовано: 16 сен 2023
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影片做的活潑有創意,謝謝老師的分享,希望大家記得按讚多多支持喔!
感謝支持❤️
影片做得很棒👍
感謝支持❤️
Great video!
Thank you! ❤️
第三題過程這樣寫的話,閱卷老師可以直接給零分。floor function 不是連續函數,極限怎麼可以直接搬進去?
第一題直接開羅,第二題用夾,第三題先配方,第四題原本想說直接開羅,原來要先變成e,第五題用黎曼
第二題為什麼 {(1+3/x)^(x/3)}^(3(x-1)/x) 可以把base和exponent 分開take limit?
第三題直接把limit pass進去floor function 有點misleading,9減去那個limit應該還是9,所以找你的寫法那個floor function的結果(如果pass limit進去)應該還是9。更加準確的說法應該要考慮x在一個small enough neighborhood of 3, 然後bound 9-(x-3)^2 above strictly by 9然後 再lower bound by 8,這樣才能conclude floor function 是8吧。
感謝提醒🙏
計算題這樣寫確實不大嚴謹,我要表達的是可以初步用判斷得到答案是 8。
For the 5th question, I think it is possible to just integrate the top and bottom, as the error becomes infinitely small as n approaches infinity, so you will obtain (n^(3)/3)/((2n^(3/2)/3))^2 = 3/4.
The reason why the error will be very small is that when you draw the function x^2 and x^1/2, they will both go to infinity, and the error of the calculation is just the difference of area between the function and the function where x is an integer, it is hard to explain with words, but you will understand if you draw the graph yourself
For example, 1+2+3+...n=(n+1)n/2, and integral x dx = x²/2, well if you limit n to infinity it becomes n²/2 just like the integral
But I would admit your method is the bigger brain one
This idea is good. More precisely, simply divide both the numerator and denominator by n^3, and you'll obtain the Riemann sum in both the numerator and denominator.
( ∫ x² dx )/( ∫ x½ dx )² from 0 to 1
= (1/3)/(2/3)² = (1/3)/(4/9) = 3/4.
我以為第四題又要拋棄羅畢達了xd 難得看到羅畢達 第五題感覺是在考對summation 的基本概念夠不夠強跟熟練
羅畢達還是會給鏡頭的 XD
用遊戲王的音樂來解題 必須給讚
決鬥⚔️
窩似被遊戲王吸引來的😂
決鬥😆
你第一題懂夾擠定理,但後面卻說趨近極限忽略不計,這很奇怪,好像懂數學又不懂