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Tugas matematika kalkulus
- 1. 1 Group : 1 ( Page 1-8 Kalkulus ) Nama : 1. Azhari Rahman 2. Muhammad Pachroni Suryana 3. Yudiansyah Kelas : 1EA Latihan 1.1 Hitunglahhasil dari f(X), jika x menyatakan nilai a dan b. untuk c, buatlah sebuahobservasi dari hasil a dan b. 1. f(x)= 𝒙+𝟐 𝒙−𝟓 a. x= 3,001 Solusi : f(3,001)= 𝒙+𝟐 𝒙−𝟓 f(x) = 𝟑,𝟎𝟎𝟏+𝟐 𝟑,𝟎𝟎𝟏−𝟓 =− 𝟓,𝟎𝟎𝟏 𝟏,𝟗𝟗𝟗 = - 2,501 b. x= 2,99 Solusi : f(2,99)= 𝒙+𝟐 𝒙−𝟓 f(x)= 𝟐,𝟗𝟗+𝟐 𝟐,𝟗𝟗−𝟓 =− 𝟒,𝟗𝟗 𝟐,𝟎𝟏 = - 2,482 c. observasi ? Terlepas dari ituketikax mendekati hasil 3,ketika f(x) mendekati hasil dari -2.5 2. f(x)= 𝒙−𝟓 𝟒𝒙 a. x= 1,002 Solusi :
- 2. 2 f(1,002)= 𝒙−𝟓 𝟒𝒙 f(x)= 𝟏,𝟎𝟎𝟐−𝟓 𝟒(𝟏,𝟎𝟎𝟐) =− 𝟑,𝟗𝟗𝟖 𝟒,𝟎𝟎𝟖 = - 0,997 b. x= ,993 Solusi : f(,993)= 𝒙−𝟓 𝟒𝒙 f(x)= ,𝟗𝟗𝟑−𝟓 𝟒(,𝟗𝟗𝟑) =− 𝟒,𝟎𝟎𝟕 𝟑,𝟗𝟕𝟐 = - 1,008 c. observasi ? terlepas dari ituketikax mendekati hasil 1, ketikaf(x) mendekati hasil dari -1. 3. f(x)= 𝟑𝒙 𝒙 𝟐 a. x= ,001 Solusi : f(.001)= 𝟑𝒙 𝒙 𝟐 f(x)= 𝟑(,𝟎𝟎𝟏) ,𝟎𝟎𝟏 𝟐 = 𝟎,𝟎𝟎𝟎𝟎𝟎𝟑 ,𝟎𝟎𝟏 = 0,003 b. x= -,001 Solusi : f(-,001)= 𝟑𝒙 𝒙 𝟐 f(x)= 𝟑(−,𝟎𝟎𝟏) −,𝟎𝟎𝟏 𝟐 =− 𝟎,𝟎𝟎𝟎𝟎𝟎𝟑 ,𝟎𝟎𝟏 = - 0,003 c. observasi ? terlepas dari ituketikax mendekati hasil 0, berarti f(x) tidak mendekati hasil tetap
- 3. 3 Latihan 1.2 Carilahpersamaan dari limit berikut ataumenunjukkan keberadaan bebas. 1. 𝐥𝐢𝐦 𝒙→𝟑 𝒙 𝟐−𝟒 𝒙+𝟏 Solusi : 𝐥𝐢𝐦 𝒙→𝟑 𝒙 𝟐−𝟒 𝒙+𝟏 = 𝐥𝐢𝐦 𝒙→𝟑 𝒙 𝟐−𝟒 𝐥𝐢𝐦 𝒙→𝟑 𝒙+𝟏 = 𝟓 𝟒 2. 𝐥𝐢𝐦 𝒙→𝟐 𝒙 𝟐−𝟗 𝒙−𝟐 Solusi : 𝐥𝐢𝐦 𝒙→𝟐 𝒙 𝟐−𝟗 𝒙−𝟐 = −𝟓 𝟎 = ~ 3. 𝐥𝐢𝐦 𝒙→𝟏 √𝒙 𝟑 + 𝟕 Solusi : 𝐥𝐢𝐦 𝒙→𝟏 √𝒙 𝟑 + 𝟕 = √ 𝟖 = 2√ 𝟐 4. 𝐥𝐢𝐦 𝒙→𝝅 ( 𝟓𝒙 𝟐 + 𝟗) Solusi : 𝐥𝐢𝐦 𝒙→𝝅 ( 𝟓𝒙 𝟐 + 𝟗) = 𝟓𝝅 𝟐 + 𝟗 5. 𝐥𝐢𝐦 𝒙→𝟎 𝟓−𝟑𝒙 𝒙+𝟏𝟏 Solusi : 𝐥𝐢𝐦 𝒙→𝟎 𝟓−𝟑𝒙 𝒙+𝟏𝟏 = 𝐥𝐢𝐦 𝒙→𝟎 𝟓−𝟑𝒙 𝐥𝐢𝐦 𝒙→𝟎 𝒙+𝟏𝟏 = 𝟓 𝟏𝟏
- 4. 4 6. 𝐥𝐢𝐦 𝒙→𝟎 𝟗+𝟑𝒙 𝟐 𝒙 𝟑+𝟏𝟏 Solusi : 𝐥𝐢𝐦 𝒙→𝟎 𝟗+𝟑𝒙 𝟐 𝒙 𝟑+𝟏𝟏 = 𝐥𝐢𝐦 𝒙→𝟎 𝟗+𝟑𝒙 𝟐 𝐥𝐢𝐦 𝒙→𝟎 𝒙 𝟑+𝟏𝟏 = 𝟗 𝟏𝟏 7. 𝐥𝐢𝐦 𝒙→𝟏 𝒙 𝟐−𝟐𝒙+𝟏 𝒙 𝟐−𝟏 Solusi: 𝐥𝐢𝐦 𝒙→𝟏 𝒙 𝟐−𝟐𝒙+𝟏 𝒙 𝟐−𝟏 = 𝐥𝐢𝐦 𝒙→𝟏 (𝒙−𝟏)(𝒙−𝟏) ( 𝒙−𝟏)(𝒙+𝟏) = 𝐥𝐢𝐦 𝒙→𝟏 𝒙−𝟏 𝒙+𝟏 = 𝟎 𝟐 = 0 8. 𝐥𝐢𝐦 𝒙→𝟒 𝟔−𝟑𝒙 𝒙 𝟐−𝟏𝟔 Solusi : 𝐥𝐢𝐦 𝒙→𝟒 𝟔−𝟑𝒙 𝒙 𝟐−𝟏𝟔 = 𝐥𝐢𝐦 𝒙→𝟒 𝟔−𝟑𝒙 (𝒙−𝟒)(𝒙+𝟒) = − 𝟔 𝟎 = ~ 9. 𝐥𝐢𝐦 𝒙→−𝟐 √𝟒𝒙 𝟑 + 𝟏𝟏 Solusi : 𝐥𝐢𝐦 𝒙→−𝟐 √𝟒𝒙 𝟑 + 𝟏𝟏= √−𝟑𝟐 + 𝟏𝟏 = √ 𝟐𝟏 10. 𝐥𝐢𝐦 𝒙→−𝟔 𝟖−𝟑𝒙 𝒙−𝟔 Solusi : 𝐥𝐢𝐦 𝒙→−𝟔 𝟖−𝟑𝒙 𝒙−𝟔 = 𝐥𝐢𝐦 𝒙→−𝟔 𝟖−𝟑𝒙 𝐥𝐢𝐦 𝒙→−𝟔 𝒙−𝟔 = − 𝟐𝟔 𝟏𝟐
- 5. 5 = − 𝟏𝟑 𝟔 Latihan 2.1 Tentukanlahlimit berikut. 1. 𝐥𝐢𝐦 𝒙→𝟑 𝒙−𝟑 𝒙 𝟐+𝒙−𝟏𝟐 Solusi : 𝐥𝐢𝐦 𝒙→𝟑 𝒙−𝟑 𝒙 𝟐+𝒙−𝟏𝟐 = 𝐥𝐢𝐦 𝒙→𝟑 𝒙−𝟑 ( 𝒙−𝟑)(𝒙+𝟒) = 𝐥𝐢𝐦 𝒙→𝟑 𝟏 (𝒙+𝟒) = 𝟏 𝟕 2. 𝐥𝐢𝐦 𝒉→𝟎 (𝒙+𝒉) 𝟐−𝒙 𝟐 𝒉 Solusi: 𝐥𝐢𝐦 𝒉→𝟎 (𝒙+𝒉) 𝟐−𝒙 𝟐 𝒉 = 𝐥𝐢𝐦 𝒉→𝟎 𝒙 𝟐+𝟐𝒉𝒙+𝒉 𝟐−𝒙 𝟐 𝒉 = 𝐥𝐢𝐦 𝒉→𝟎 𝟐𝒉𝒙+𝒉 𝟐 𝒉 = 𝐥𝐢𝐦 𝒉→𝟎 𝒉(𝟐𝒙+𝒉) 𝒉 = 𝐥𝐢𝐦 𝒉→𝟎 𝟐𝒙 + 𝒉 = 2x 3. 𝐥𝐢𝐦 𝒙→𝟒 𝒙 𝟑−𝟔𝟒 𝒙 𝟐−𝟏𝟔 Solusi: 𝐥𝐢𝐦 𝒙→𝟒 𝒙 𝟑−𝟔𝟒 𝒙 𝟐−𝟏𝟔 = 𝐥𝐢𝐦 𝒙→𝟒 𝒙 𝟑−𝟔𝟒 𝒙 𝟐−𝟏𝟔 = 𝐥𝐢𝐦 𝒙→𝟒 (𝒙−𝟒)(𝒙 𝟐+𝟒𝒙+𝟏𝟔) ( 𝒙−𝟒)(𝒙+𝟒) = 𝐥𝐢𝐦 𝒙→𝟒 𝒙 𝟐+𝟒𝒙+𝟏𝟔 𝒙+𝟒 = 𝟒𝟖 𝟖 = 6
- 6. 6 4. Jika f(x) = 5x+8, tentukan 𝐥𝐢𝐦 𝒉→𝟎 𝒇( 𝒙+𝒉)−𝒇(𝒙) 𝒉 Solusi: 𝐥𝐢𝐦 𝒉→𝟎 𝒇( 𝒙+𝒉)−𝒇(𝒙) 𝒉 = 𝐥𝐢𝐦 𝒉→𝟎 (𝟓( 𝒙+𝒉)+𝟖)−(𝟓𝐱+𝟖) 𝒉 = 𝐥𝐢𝐦 𝒉→𝟎 (𝟓𝒙+𝟓𝒉+𝟖)−(𝟓𝐱+𝟖) 𝒉 = 𝐥𝐢𝐦 𝒉→𝟎 𝟓𝒉 𝒉 = ~ 5. 𝐥𝐢𝐦 𝒙→−𝟑 𝟓𝒙+𝟕 𝒙 𝟐−𝟑 Solusi : 𝐥𝐢𝐦 𝒙→−𝟑 𝟓𝒙+𝟕 𝒙 𝟐−𝟑 = 𝐥𝐢𝐦 𝒙→𝟑 𝟓𝒙+𝟕 𝐥𝐢𝐦 𝒙→𝟑 𝒙 𝟐−𝟑 = − 𝟖 𝟔 = - 𝟒 𝟑 6. 𝐥𝐢𝐦 𝒙→𝟐𝟓 √𝒙−𝟓 𝒙−𝟐𝟓 Solusi: 𝐥𝐢𝐦 𝒙→𝟐𝟓 √𝒙−𝟓 𝒙−𝟐𝟓 = 𝐥𝐢𝐦 𝒙→𝟐𝟓 (√𝒙−𝟓) (𝒙−𝟐𝟓) (√𝒙+𝟓) (√𝒙+𝟓) = 𝐥𝐢𝐦 𝒙→𝟐𝟓 𝒙−𝟐𝟓 𝒙√𝒙+ 𝟓𝒙−𝟐𝟓√𝒙−𝟏𝟐𝟓 = 𝟎 𝟎 = ~ 7. Jika g(x) =𝒙 𝟐 , tentukan 𝐥𝐢𝐦 𝒙→𝟐 𝒈( 𝒙)−𝒈(𝟐) 𝒙−𝟐 Solusi: 𝐥𝐢𝐦 𝒙→𝟐 𝒈( 𝒙)−𝒈(𝟐) 𝒙−𝟐 = 𝐥𝐢𝐦 𝒙→𝟐 𝒙 𝟐−𝟒 𝒙−𝟐 = 𝐥𝐢𝐦 𝒙→𝟐 (𝒙+𝟐)(𝒙−𝟐) 𝒙−𝟐 = 𝐥𝐢𝐦 𝒙→𝟐 𝒙 + 𝟐
- 7. 7 = 4 8. 𝐥𝐢𝐦 𝒙→𝟎 𝟐𝒙 𝟐− 𝟒𝒙 𝒙 Solusi: 𝐥𝐢𝐦 𝒙→𝟎 𝟐𝒙 𝟐− 𝟒𝒙 𝒙 = 𝐥𝐢𝐦 𝒙→𝟎 (𝟐𝒙− 𝟒)𝒙 𝒙 = 𝐥𝐢𝐦 𝒙→𝟎 ( 𝟐𝒙 − 𝟒) = -4 9. 𝐥𝐢𝐦 𝒓→𝟎 √𝒙+𝒓−√𝒙 𝒓 Solusi: 𝐥𝐢𝐦 𝒓→𝟎 √𝒙+𝒓−√𝒙 𝒓 = 𝐥𝐢𝐦 𝒓→𝟎 (√𝒙+𝒓−√ 𝒙 ) 𝒓 (√𝒙+𝒓+√ 𝒙) √𝒙+𝒓+√𝒙 = 𝐥𝐢𝐦 𝒓→𝟎 𝒙+𝒓−𝒙 𝒓(√𝒙+𝒓+√𝒙) = 𝟎 𝟎 = ~ 10.𝐥𝐢𝐦 𝒙→𝟒 𝒙 𝟑+𝟔 𝒙−𝟒 Solusi: 𝐥𝐢𝐦 𝒙→𝟒 𝒙 𝟑+𝟔 𝒙−𝟒 = 𝐥𝐢𝐦 𝒙→𝟒 𝒙 𝟑+𝟔 𝒙−𝟒 𝒙+𝟒 𝒙+𝟒 = 𝒙 𝟒+𝟒𝒙 𝟑+𝟔𝒙+𝟐𝟒 (𝒙−𝟒)(𝒙+𝟒) = 𝟕𝟎 𝟎 = ~ ==== Selesai ====