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LIMIT FUNGSI TAN + RUMUS + TRIK CEPAT DAN PEMBAHASAN
- 1. LIMIT By SMAK SANTO PAULUS JEMBER STUDENTS
- 2. SMAK SANTO PAULUS JEMBER Guru yang aktif dan kreatif membuat artikel serta memberikan motivasi kepada para murid Website SMAK SANTO PAULUS JEMBER saintpauljember. sch.id
- 3. SMAK SANTO PAULUS JEMBER Keakraban murid dengan guru Pembelajaran jarak jauh(PJJ) SMAK SANTO PAULUS JEMBER
- 4. Bonitasya Enwi Jorah XIIA1/05 Bradley Xavier XIIA1/06 Cabrina Irmadela XIIA1/07 Celine Angelina XIIA1/08 KELOMPOK 2 SMAK SANTO PAULUS JEMBER
- 5. Limit fungsi lim 𝑥→0 tan 𝑥 𝑥 SMAK SANTO PAULUS JEMBER
- 6. 02 03 Bukti secara intuitif Melalui intuisi, berdasar bisikan (gerak) hati Soal dan jawaban Terkait teorema saat x→0 pada limit fungsi trigonometri Bukti secara teoritis Melalui teori-teori yang dijadikan landasan berpikir 01 SMAK SANTO PAULUS JEMBER
- 7. SMAK SANTO PAULUS JEMBER
- 8. 𝑡𝑎𝑛 0 = 0 x → 0 tan x → x lim 𝑥→0 𝑡𝑎𝑛 𝑥 𝑥 lim 𝑥→0 𝑥 𝑥 = 1 = SMAK SANTO PAULUS JEMBER
- 9. Bukti Secara Teoritis 02 SMAK SANTO PAULUS JEMBER
- 10. Luas OAC ≤ luas juring OAC ≤ luas OAB 𝟏 𝟐 . OA . CD ≤ 𝒙 𝟐𝝅 . 𝝅𝑹𝟐 ≤ 𝟏 𝟐 . 𝑶𝑨 . 𝑨𝑩 𝟏 𝟐 . 𝑹 . 𝑹 sin 𝒙 ≤ 𝟏 𝟐 𝒙𝑹𝟐 ≤ 𝟏 𝟐 . 𝑹 . 𝑹 tan 𝒙 sin 𝒙 ≤ 𝒙 ≤ tan 𝒙 01 02 03 04 SMAK SANTO PAULUS JEMBER
- 11. 1≤ 𝒙 sin 𝒙 ≤ 𝟏 cos 𝒙 dan cos 𝒙 ≤ sin 𝒙 𝒙 ≤ 𝟏 lim 𝒙→𝟎 𝟏 ≤ lim 𝒙→𝟎 𝒙 sin 𝒙 ≤ lim 𝒙→𝟎 𝟏 cos 𝒙 dan lim 𝒙→𝟎 cos 𝒙 ≤ lim 𝒙→𝟎 sin 𝒙 𝒙 ≤ lim 𝒙→𝟎 𝟏 1≤ lim 𝒙→𝟎 𝒙 sin 𝒙 ≤ 𝟏 dan 1≤ lim 𝒙→𝟎 sin 𝒙 𝒙 ≤ 𝟏 05 06 07 SMAK SANTO PAULUS JEMBER
- 12. lim 𝒙→𝟎 tan 𝒙 𝒙 = lim 𝒙→𝟎 sin 𝒙 𝒙 . 𝟏 cos 𝒙 = lim 𝒙→𝟎 sin 𝒙 𝒙 . lim 𝒙→𝟎 𝟏 cos 𝒙 = 1.1= 1 08 09 10 11 Jadi terbukti bahwa lim 𝒙→𝟎 sin 𝒙 𝒙 = lim 𝒙→𝟎 𝒙 sin 𝒙 = 𝟏 Type equation here. Jadi terbukti bahwa lim 𝒙→𝟎 tan 𝒙 𝒙 = 𝟏 Dengan cara yang sama didapat lim 𝒙→𝟎 𝒙 tan 𝒙 = 𝟏 SMAK SANTO PAULUS JEMBER
- 13. QUIZ TIME Are you ready?
- 14. 𝐥𝐢𝐦 𝒙→𝟎 𝒙𝟐 𝐭𝐚𝐧 𝟐𝒙 𝒙 − 𝒙 𝐜𝐨𝐬 𝟒𝒙 SMAK SANTO PAULUS JEMBER
- 15. 𝒍𝒊𝒎 𝒙→𝟎 𝒙𝟐 𝒕𝒂𝒏 𝟐𝒙 𝒙 − 𝒙 𝒄𝒐𝒔 𝟒𝒙 = 𝐥𝐢𝐦 𝒙→𝟎 𝒙 . 𝒙 𝐭𝐚𝐧 𝟐𝒙 𝒙 . (𝟏 − 𝐜𝐨𝐬 𝟒𝒙) = 𝐥𝐢𝐦 𝒙→𝟎 𝒙 𝐭𝐚𝐧 𝟐𝒙 𝟏 −(𝟏 −𝟐 𝐬𝐢𝐧𝟐 𝟐𝒙) = 𝐥𝐢𝐦 𝒙→𝟎 𝒙 𝐭𝐚𝐧 𝟐𝒙 𝟐 𝐬𝐢𝐧𝟐 𝟐𝒙 = 𝐥𝐢𝐦 𝒙→𝟎 ( 𝟏 𝟐 . 𝒙 𝐬𝐢𝐧 𝟐𝒙 . 𝐭𝐚𝐧 𝟐𝒙 𝐬𝐢𝐧 𝟐𝒙 ) = 𝟏 𝟐 . 𝟏 𝟐 . 𝟏 = 𝟏 𝟒 𝐜𝐨𝐬 𝒂𝒙 = 𝟏 − 𝟐 𝐬𝐢𝐧𝟐 𝒂 𝟐 𝒙 𝐥𝐢𝐦 𝒙→𝟎 𝒂𝒙 𝐬𝐢𝐧 𝒃𝒙 = 𝐥𝐢𝐦 𝒙→𝟎 𝒕𝒂𝒏 𝒂𝒙 𝒔𝒊𝒏 𝒃𝒙 = 𝒂 𝒃 SMAK SANTO PAULUS JEMBER
- 16. 𝐥𝐢𝐦 𝒙→𝟎 𝐭𝐚𝐧 𝟐𝒙𝟑 𝒙𝟑 + 𝐬𝐢𝐧𝟑 𝟒𝒙 SMAK SANTO PAULUS JEMBER
- 17. 𝐥𝐢𝐦 𝒙→𝟎 𝐭𝐚𝐧 𝟐𝒙𝟑 𝒙𝟑 + 𝐬𝐢𝐧𝟑 𝟒𝒙 = 𝒍𝒊𝒎 𝒙→𝟎 𝐭𝐚𝐧 𝟐𝒙𝟑 𝒙𝟑 𝒙𝟑 𝒙𝟑 + 𝐬𝐢𝐧𝟑(𝟒𝒙) 𝒙𝟑 = 𝒍𝒊𝒎 𝒙→𝟎 𝐭𝐚𝐧 𝟐𝒙𝟑 𝒙𝟑 𝒙𝟑 𝒙𝟑+ 𝐬𝐢𝐧(𝟒𝒙) 𝒙 × 𝐬𝐢𝐧(𝟒𝒙) 𝒙 × 𝐬𝐢𝐧(𝟒𝒙) 𝒙 = 𝟐 𝟏+ 𝟒 ×𝟒 × 𝟒 = 𝟐 𝟔𝟓 𝐥𝐢𝐦 𝒙→𝟎 tan 𝑎𝑥3 𝒃𝒙𝟑 = 𝒂 𝒃 𝐥𝐢𝐦 𝒙→𝟎 sin 𝒂𝒙 𝒃𝒙 = 𝒂 𝒃 SMAK SANTO PAULUS JEMBER
- 18. 𝐥𝐢𝐦 𝒙→𝟎 𝒙 𝐭𝐚𝐧 𝟑𝒙 𝟏 − 𝒄𝒐𝒔𝟐𝟐𝒙 SMAK SANTO PAULUS JEMBER
- 19. SMAK SANTO PAULUS JEMBER
- 20. 𝐥𝐢𝐦 𝒙→𝟎 𝟏 − 𝒄𝒐𝒔𝟐 𝒙 𝒙𝟐 𝐭𝐚𝐧(𝒙 + 𝝅 𝟑 ) SMAK SANTO PAULUS JEMBER
- 21. SMAK SANTO PAULUS JEMBER
- 22. 𝐥𝐢𝐦 𝒙→ 𝝅 𝟐 (𝝅 − 𝟐) 𝐭𝐚𝐧 𝒙 Kategori sukar ! SOAL UM UGM 2018 SMAK SANTO PAULUS JEMBER
- 23. SMAK SANTO PAULUS JEMBER
- 24. Resources Buku cetak Matematika untuk siswa SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam https://mathcyber1997.com/soal-dan-pembahasan-super- lengkap-limit-fungsi-trigonometri/ https://www.defantri.com/2018/09/matematika-dasar-limit- fungsi-trigonometri.html?m=1 SMAK SANTO PAULUS JEMBER
- 25. CREDITS: This presentation template was created by Slidesgo, including icons by Flaticon, and infographics & images by Freepik Do you have any questions? Thanks Instagram: @jrhbn_ @cabrinastwnn @bradleyxavierr @celine.angelinaaa @santopaulusjember @osis_saintpaul SMAK SANTO PAULUS JEMBER