求解第24題

t (a), explain wh 2+x-6 x-2 lim (x + 3) imit, if it exists. 12. lim x² - 4x x-1 x²-3x - 4 14. lim 2x² + 3x + 1 2 x-1 X² 2x - 3 16. lim 18. lim h→0 20. lim X→-1 22. lim u-2 24. lim t-0 (2 + h) - 8 h 3x - 4 28. lim h→0 x² + 2x + 1 .4. - 1 26. lim x--4 (1 t 4u + 1 u - 2 3 t² + t x²+9-5 x + 4 1 (x + h)² h 1 X Illustrate tion of th 33. If 4x 34. If 2x - 35. Prove 36. Prove 37-42= explain 37. lim 39. 41. 43

請問(b)的答案為什麼是1呢(´・ω・`)

[3x - 1, 2x² + 1₂ / (a) lim f(x) 17. f(x): X x-0- f(x) = - (3x 0 1, x² + 1, x > 0 x ≤ 0 x > 0 x≤0 X (b) lim f(x) x→0+ -1 -0.1 -0.01 -0.001 0 ? X(c) lim f(x) x→0 0.001 0.01 0.1 12 1

第一行是題目 第二行是給的詳解 我從第一步就不懂 為什麼要假設成L?🤨 後面也看不懂QWQ

Question Limit. 4. Find the humber c such that lim 3x²+CX+c+3 X-7-2 X²+X-2 exists and evaluate this limit. lim A = Suppose X+2+x²+x-2² = L. We have that lim (3x+exto+ 3) = lim (2xxxx112 x (x²7x²)) X7-2 x=-2 X+X-2 3x²+x+c+3 = ( Lim Xxx-9²³) x (&im₂ (X+X-2)) = Lx0=0 X-7-2 => 3x(-2)² + cx (-2) + C+3=15-C=O=> C=15

來個高手幫忙解解~

試求下列積分式:「 x³ - 1 x² - 2x dx ; x³ - 1 3-2 x² - 2x · = ax+b+ + X d x-2

主要問題是“移項”，左邊1、2是已知，3是給的答案，右邊是我整理的部分開頭，但我整理不出答案3的樣式，想問的有1.詳細的公式移項 2.這是哪部分的課程，我得去讀一讀，謝謝

P₁ P₂ E₁ E₂ P₁ + P₂ =P 2 E, P P₁ = E₁ + E₂ 3 Ez P P₂ = E₁ + E₂ P₁ = ER - P₂ - EP 1 EB₁ + E₂P₁ = P

不知怎算？

3. fxe dx dy

請問第41題這種類型的題目要怎麼解🙏🙏

41-44 Use implicit differentiation to find az/ax and az/ay. .2 41. x² + 2y² + 3z² = 1 (42. x² - y² + z² – 2z = 4 (43) e² = xyz 44.) yz + x ln y = z²

請問第六和第十題如何計算？ 大學微積分 Integration by parts

S. cite the formula for integration by parts. plain how you would choose u and du when using the egration by parts formula. Illustrate your answer with ar Exercises es 1-28, find each indefinite integral. 32. x dx -X 35. F. et dx - tc th - x)² dx 36. F. tc 1) ex dx th - 2. { xe* dx 4. S 6xe* | * dx 6. [(e-* + x) dx 8. S (x – 3)e% de 10. S x(x + 4)-2 dx . 14. S x? In 2x dx 16. Vx In x dx + 1)-32 dx 37. F f f E-5 dx 12. 3x V2x + 3 dx 38. F f f 2x dx 39. V af x dx 1

#離散數學 #集合運算和集合論定律 請問(d)畫線地方是怎麼變的 用的什麼集合論定律？

U (0 DU00B) d) Ā U BU CAOBOC) Lal An (B-A)-An (BĀ) = BOIAA) - BOO = Ø (b) (AB) UCABA CODUCĀB) = (AOBU (Ā OB) Absorption Law = (AVĀ), B = UnB - B (C) (A-B) U CAOB)= (A BUCAOB) = AO (BUB) = AO U = A dĀ UBU CAOBOC)= (AOBULAB) 07] = [[ABU CAO B)]n[LADB) U C] = (AO BUT = ĀU BUC

求這兩題的解法還有完整過程

0 A. . In x dx 1 + x2 0 X B. Sexy dx ex – 1 -