Masih sering bingung dengan limit trigonometri lanjutan? Yuk, simak penjelasan lengkapnya lewat video yang ada di sini. Setelahnya, kamu juga bisa mengerjakan latihan soal yang telah disediakan untuk mengasah kemampuan belajarmu.
Di sini, kamu akan belajar tentang Limit Trigonometri Lanjutan melalui video yang dibawakan oleh Bapak Anton Wardaya. Kamu akan diajak untuk memahami materi hingga metode menyelesaikan soal.
Selain itu, kamu juga akan mendapatkan latihan soal interaktif dalam 3 tingkat kesulitan (mudah, sedang, sukar). Dengan begitu, kamu bisa langsung mempraktikkan materi yang telah dijelaskan.
Sekarang, kamu bisa mulai belajar dengan 3 video dan 3 set latihan soal yang ada di halaman ini. Apabila materi ini berguna, bagikan ke teman atau rekan kamu supaya mereka juga mendapatkan manfaatnya.
Kamu dapat download modul & contoh soal serta kumpulan latihan soal lengkap dalam bentuk pdf pada list dibawah ini:
Contoh Soal Limit Trigonometri Lanjutan (1)
Contoh Soal Limit Trigonometri Lanjutan (2)
Contoh Soal Limit Trigonometri Lanjutan (3)
Latihan Soal Limit Trigonometri Lanjutan (Mudah)
Ringkasan kuis
0 dari 5 pertanyaan telah diselesaikan
Pertanyaan:
- 1
- 2
- 3
- 4
- 5
Informasi
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
Anda harus masuk atau mendaftar untuk memulai kuis.
Anda harus menyelesaikan kuis dibawah ini, untuk memulai kuis ini:
Hasil
Hasil
0 dari 5 pertanyaan terjawab dengan benar
Waktu yang telah berlalu
Kategori
- Tidak Berkategori 0%
- 1
- 2
- 3
- 4
- 5
- Terjawab
- Tinjau
-
Pertanyaan ke 1 dari 5
1. Pertanyaan
$\underset{x\rightarrow0}{lim}\frac{tanx}{sec^{2}x-1}=…$
BetulIngat bentuk identitas Trigonometri $1+tan^{2}=sec^{2}x$
$\underset{x\rightarrow0}{lim}\frac{tanx}{sec^{2}x-1}$$=\underset{x\rightarrow0}{lim}\frac{tanx}{tanx}$$=1$
SalahIngat bentuk identitas Trigonometri $1+tan^{2}=sec^{2}x$
$\underset{x\rightarrow0}{lim}\frac{tanx}{sec^{2}x-1}$$=\underset{x\rightarrow0}{lim}\frac{tanx}{tanx}$$=1$
-
Pertanyaan ke 2 dari 5
2. Pertanyaan
$\underset{x\rightarrow1}{lim}\frac{sin\left(1-\frac{1}{x}\right)cos\left(1-\frac{1}{x}\right)}{x-1}=…$
Betul$\underset{x\rightarrow1}{lim}\frac{sin\left(1-\frac{1}{x}\right)cos\left(1-\frac{1}{x}\right)}{x-1}$$=\underset{x\rightarrow1}{lim}\frac{sin\left(1-\frac{1}{x}\right)cos\left(1-\frac{1}{x}\right)}{x\left(1-\frac{1}{x}\right)}$
$=\underset{x\rightarrow1}{lim}\frac{cos\left(1-\frac{1}{x}\right)}{x}\cdot\frac{sin\left(1-\frac{1}{x}\right)}{\left(1-\frac{1}{x}\right)}$
$=\underset{x\rightarrow1}{lim}\frac{cos\left(1-\frac{1}{x}\right)}{x}\cdot\underset{x\rightarrow1}{lim}\frac{sin\left(1-\frac{1}{x}\right)}{\left(1-\frac{1}{x}\right)}$
$=\frac{cos0^{0}}{1}\cdot1=1$
Salah$\underset{x\rightarrow1}{lim}\frac{sin\left(1-\frac{1}{x}\right)cos\left(1-\frac{1}{x}\right)}{x-1}$$=\underset{x\rightarrow1}{lim}\frac{sin\left(1-\frac{1}{x}\right)cos\left(1-\frac{1}{x}\right)}{x\left(1-\frac{1}{x}\right)}$
$=\underset{x\rightarrow1}{lim}\frac{cos\left(1-\frac{1}{x}\right)}{x}\cdot\frac{sin\left(1-\frac{1}{x}\right)}{\left(1-\frac{1}{x}\right)}$
$=\underset{x\rightarrow1}{lim}\frac{cos\left(1-\frac{1}{x}\right)}{x}\cdot\underset{x\rightarrow1}{lim}\frac{sin\left(1-\frac{1}{x}\right)}{\left(1-\frac{1}{x}\right)}$
$=\frac{cos0^{0}}{1}\cdot1=1$
-
Pertanyaan ke 3 dari 5
3. Pertanyaan
$\underset{x\rightarrow0}{lim}\frac{sin3x-sin3x\, cos2x}{4x^{3}}=…$
Betul$\underset{x\rightarrow0}{lim}\frac{sin3x-sin3x\, cos2x}{4x^{3}}$$=\underset{x\rightarrow0}{lim}\frac{sin3x(1-cos2x)}{4x^{3}}$
$=\underset{x\rightarrow0}{lim}\frac{sin3x\left(2sin^{2}x\right)}{4x^{3}}$
$=\underset{x\rightarrow0}{lim}\frac{2}{4}\cdot\frac{sin3x}{x}\cdot\frac{sinx}{x}\cdot\frac{sinx}{x}$
$=\frac{2}{4}\cdot3\cdot1\cdot1=\frac{3}{2}$
Salah$\underset{x\rightarrow0}{lim}\frac{sin3x-sin3x\, cos2x}{4x^{3}}$$=\underset{x\rightarrow0}{lim}\frac{sin3x(1-cos2x)}{4x^{3}}$
$=\underset{x\rightarrow0}{lim}\frac{sin3x\left(2sin^{2}x\right)}{4x^{3}}$
$=\underset{x\rightarrow0}{lim}\frac{2}{4}\cdot\frac{sin3x}{x}\cdot\frac{sinx}{x}\cdot\frac{sinx}{x}$
$=\frac{2}{4}\cdot3\cdot1\cdot1=\frac{3}{2}$
-
Pertanyaan ke 4 dari 5
4. Pertanyaan
$\underset{x\rightarrow0}{lim}\frac{x(cos^{2}6x-1)}{sin3x\cdot tan^{2}2x}=…$
Betul$\underset{x\rightarrow0}{lim}\frac{x(cos^{2}6x-1)}{sin3x\cdot tan^{2}2x}=\underset{x\rightarrow0}{lim}\frac{x(-sin^{2}6x)}{sin3x\cdot tan^{2}2x}$
$=\underset{x\rightarrow0}{lim}\left(\frac{-x}{sin3x}\right).\left(\frac{sin6x}{tan2x}\right).\left(\frac{sin6x}{tan2x}\right)$
$=\underset{x\rightarrow0}{lim}\left(\frac{-x}{sin3x}\right).\underset{x\rightarrow0}{lim}\left(\frac{sin6x}{tan2x}\right).\underset{x\rightarrow0}{lim}\left(\frac{sin6x}{tan2x}\right)$
$=\frac{-1}{3}.\frac{6}{2}.\frac{6}{2}=-3$
Salah$\underset{x\rightarrow0}{lim}\frac{x(cos^{2}6x-1)}{sin3x\cdot tan^{2}2x}=\underset{x\rightarrow0}{lim}\frac{x(-sin^{2}6x)}{sin3x\cdot tan^{2}2x}$
$=\underset{x\rightarrow0}{lim}\left(\frac{-x}{sin3x}\right).\left(\frac{sin6x}{tan2x}\right).\left(\frac{sin6x}{tan2x}\right)$
$=\underset{x\rightarrow0}{lim}\left(\frac{-x}{sin3x}\right).\underset{x\rightarrow0}{lim}\left(\frac{sin6x}{tan2x}\right).\underset{x\rightarrow0}{lim}\left(\frac{sin6x}{tan2x}\right)$
$=\frac{-1}{3}.\frac{6}{2}.\frac{6}{2}=-3$
-
Pertanyaan ke 5 dari 5
5. Pertanyaan
$\underset{x\rightarrow1}{lim}\frac{1-cos^{2}(x-1)}{4(x^{2}-2x+1)}=…$
Betul$\underset{x\rightarrow1}{lim}\frac{1-cos^{2}(x-1)}{4(x^{2}-2x+1)}$$=\underset{x\rightarrow1}{lim}\frac{2sin^{2}(x-1)}{4(x-1)(x-1)}$
$=\underset{x\rightarrow1}{lim}\left(\frac{1}{2}\right).\left(\frac{sin(x-1)}{(x-1)}\right)\left(\frac{sin(x-1)}{(x-1)}\right)$
$=\frac{1}{2}\cdot1\cdot1=\frac{1}{2}$
Salah$\underset{x\rightarrow1}{lim}\frac{1-cos^{2}(x-1)}{4(x^{2}-2x+1)}$$=\underset{x\rightarrow1}{lim}\frac{2sin^{2}(x-1)}{4(x-1)(x-1)}$
$=\underset{x\rightarrow1}{lim}\left(\frac{1}{2}\right).\left(\frac{sin(x-1)}{(x-1)}\right)\left(\frac{sin(x-1)}{(x-1)}\right)$
$=\frac{1}{2}\cdot1\cdot1=\frac{1}{2}$
Latihan Soal Limit Trigonometri Lanjutan (Sedang)
Ringkasan kuis
0 dari 5 pertanyaan telah diselesaikan
Pertanyaan:
- 1
- 2
- 3
- 4
- 5
Informasi
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
Anda harus masuk atau mendaftar untuk memulai kuis.
Anda harus menyelesaikan kuis dibawah ini, untuk memulai kuis ini:
Hasil
Hasil
0 dari 5 pertanyaan terjawab dengan benar
Waktu yang telah berlalu
Kategori
- Tidak Berkategori 0%
- 1
- 2
- 3
- 4
- 5
- Terjawab
- Tinjau
-
Pertanyaan ke 1 dari 5
1. Pertanyaan
$\underset{x\rightarrow\frac{1}{8}\pi}{lim}\frac{sin^{2}2x-cos^{2}2x}{sin2x-cos2x}=…$
BetulIngat bentuk $a^{2}-b^{2}=(a-b)(a+b)$
$\underset{x\rightarrow\frac{1}{8}\pi}{lim}\frac{sin^{2}2x-cos^{2}2x}{sin2x-cos2x}$$=\underset{x\rightarrow\frac{1}{8}\pi}{lim}\frac{(sin2x-cos2x)(sin2x+cos2x)}{sin2x-cos2x}$$=\underset{x\rightarrow\frac{1}{8}\pi}{lim}sin2x+cos2x$
$=\left(sin2\cdot\frac{1}{8}\pi+cos2\cdot\frac{1}{8}\pi\right)$
$=sin\frac{1}{4}\pi+cos\frac{1}{4}\pi$
$=\frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{2}$
$=\sqrt{2}$
SalahIngat bentuk $a^{2}-b^{2}=(a-b)(a+b)$
$\underset{x\rightarrow\frac{1}{8}\pi}{lim}\frac{sin^{2}2x-cos^{2}2x}{sin2x-cos2x}$$=\underset{x\rightarrow\frac{1}{8}\pi}{lim}\frac{(sin2x-cos2x)(sin2x+cos2x)}{sin2x-cos2x}$$=\underset{x\rightarrow\frac{1}{8}\pi}{lim}sin2x+cos2x$
$=\left(sin2\cdot\frac{1}{8}\pi+cos2\cdot\frac{1}{8}\pi\right)$
$=sin\frac{1}{4}\pi+cos\frac{1}{4}\pi$
$=\frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{2}$
$=\sqrt{2}$
-
Pertanyaan ke 2 dari 5
2. Pertanyaan
$\underset{h\rightarrow0}{lim}\frac{sin\left(\frac{1}{3}\pi+h\right)-sin\frac{1}{3}\pi}{h}=…$
Betul$\underset{h\rightarrow0}{lim}\frac{sin\left(\frac{1}{3}\pi+h\right)-sin\frac{1}{3}\pi}{h}$$=\underset{h\rightarrow0}{lim}\frac{2cos\frac{1}{2}\left(\frac{1}{3}\pi+h+\frac{1}{3}\pi\right)sin\frac{1}{2}\left(\frac{1}{3}\pi+h-\frac{1}{3}\pi\right)}{h}$
$=\underset{h\rightarrow0}{lim}\frac{2cos\left(\frac{1}{3}\pi+\frac{h}{2}\right)sin\frac{1}{2}h}{h}$
$=\underset{h\rightarrow0}{lim}\left[2cos\left(\frac{1}{3}\pi+\frac{h}{2}\right)\right]\cdot\frac{sin\frac{1}{2}h}{h}$
$=\underset{h\rightarrow0}{lim}\left[2cos\left(\frac{1}{3}\pi+\frac{h}{2}\right)\right]\cdot$$\underset{h\rightarrow0}{lim}\frac{sin\frac{1}{2}h}{h}$
$=2cos\frac{1}{3}\pi\cdot\frac{1}{2}$
$=2\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{2}$
Salah$\underset{h\rightarrow0}{lim}\frac{sin\left(\frac{1}{3}\pi+h\right)-sin\frac{1}{3}\pi}{h}$$=\underset{h\rightarrow0}{lim}\frac{2cos\frac{1}{2}\left(\frac{1}{3}\pi+h+\frac{1}{3}\pi\right)sin\frac{1}{2}\left(\frac{1}{3}\pi+h-\frac{1}{3}\pi\right)}{h}$
$=\underset{h\rightarrow0}{lim}\frac{2cos\left(\frac{1}{3}\pi+\frac{h}{2}\right)sin\frac{1}{2}h}{h}$
$=\underset{h\rightarrow0}{lim}\left[2cos\left(\frac{1}{3}\pi+\frac{h}{2}\right)\right]\cdot\frac{sin\frac{1}{2}h}{h}$
$=\underset{h\rightarrow0}{lim}\left[2cos\left(\frac{1}{3}\pi+\frac{h}{2}\right)\right]\cdot$$\underset{h\rightarrow0}{lim}\frac{sin\frac{1}{2}h}{h}$
$=2cos\frac{1}{3}\pi\cdot\frac{1}{2}$
$=2\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{2}$
-
Pertanyaan ke 3 dari 5
3. Pertanyaan
Jika $\underset{x\rightarrow0}{lim}\frac{tanx}{x}=1$, maka $\underset{x\rightarrow a}{lim}\frac{x-a}{3x-3a+tan(x-a)}=…$
Betul$\underset{x\rightarrow a}{lim}\frac{x-a}{3x-3a+tan(x-a)}$$=\underset{x\rightarrow a}{lim}\frac{x-a}{3(x-a)+tan(x-a)}$
$=\underset{x\rightarrow a}{lim}\left(\frac{1}{\frac{3(x-a)+tan(x-a)}{x-a}}\right)$
$=\underset{x\rightarrow a}{\lim}\left(\frac{1}{3+\frac{\tan(x-a)}{x-a}}\right)$
$=\frac{1}{3+\underset{x\rightarrow a}{\lim}\frac{\tan(x-a)}{x-a}}$
$=\frac{1}{3+1}$
$=\frac{1}{4}$
Salah$\underset{x\rightarrow a}{lim}\frac{x-a}{3x-3a+tan(x-a)}$$=\underset{x\rightarrow a}{lim}\frac{x-a}{3(x-a)+tan(x-a)}$
$=\underset{x\rightarrow a}{lim}\left(\frac{1}{\frac{3(x-a)+tan(x-a)}{x-a}}\right)$
$=\underset{x\rightarrow a}{\lim}\left(\frac{1}{3+\frac{\tan(x-a)}{x-a}}\right)$
$=\frac{1}{3+\underset{x\rightarrow a}{\lim}\frac{\tan(x-a)}{x-a}}$
$=\frac{1}{3+1}$
$=\frac{1}{4}$
-
Pertanyaan ke 4 dari 5
4. Pertanyaan
$\underset{x\rightarrow0}{lim}\frac{cosx-cos2x}{x^{2}}=…$
Betul$\underset{x\rightarrow0}{lim}\frac{cosx-cos2x}{x^{2}}$$=\underset{x\rightarrow0}{lim}\frac{-2sin\frac{1}{2}(x+2x)\cdot sin\frac{1}{2}(x-2x)}{x^{2}}$
$=\underset{x\rightarrow0}{lim}\frac{-2sin\frac{3}{2}x\cdot sin\left(-\frac{1}{2}x\right)}{x^{2}}$
$=\underset{x\rightarrow0}{lim}\frac{2sin\frac{3}{2}x\cdot sin\left(\frac{1}{2}x\right)}{x^{2}}$
$=\underset{x\rightarrow0}{lim}2\cdot\frac{sin\frac{3}{2}x}{x}\cdot\frac{sin\frac{1}{2}x}{x}$
$=2\cdot\frac{3}{2}\cdot\frac{1}{2}=\frac{3}{2}$
Salah$\underset{x\rightarrow0}{lim}\frac{cosx-cos2x}{x^{2}}$$=\underset{x\rightarrow0}{lim}\frac{-2sin\frac{1}{2}(x+2x)\cdot sin\frac{1}{2}(x-2x)}{x^{2}}$
$=\underset{x\rightarrow0}{lim}\frac{-2sin\frac{3}{2}x\cdot sin\left(-\frac{1}{2}x\right)}{x^{2}}$
$=\underset{x\rightarrow0}{lim}\frac{2sin\frac{3}{2}x\cdot sin\left(\frac{1}{2}x\right)}{x^{2}}$
$=\underset{x\rightarrow0}{lim}2\cdot\frac{sin\frac{3}{2}x}{x}\cdot\frac{sin\frac{1}{2}x}{x}$
$=2\cdot\frac{3}{2}\cdot\frac{1}{2}=\frac{3}{2}$
-
Pertanyaan ke 5 dari 5
5. Pertanyaan
Diketahui $a=\underset{x\rightarrow0}{lim}\frac{\sqrt{1+x+x^{2}}-1}{x}$, maka nilai dari $\underset{x\rightarrow\left(a-\frac{1}{2}\right)}{lim}$$\frac{sin2x+cosx.sin2x}{tanx+\frac{3.sinx}{cos^{2}x}}=…$
Betul$\begin{aligned}a & =\underset{x\rightarrow0}{lim}\frac{\sqrt{1+x+x^{2}}-1}{x}\cdot\frac{\sqrt{1+x+x^{2}}+1}{\sqrt{1+x+x^{2}}+1}\\
& =\underset{x\rightarrow0}{lim}\frac{1+x+x^{2}-1}{x\left(\sqrt{1+x+x^{2}}+1\right)}\\
& =\underset{x\rightarrow0}{lim}\frac{x\left(x+1\right)}{x\left(\sqrt{1+x+x^{2}}+1\right)}\\
& =\underset{x\rightarrow0}{lim}\frac{x+1}{\left(\sqrt{1+x+x^{2}}+1\right)}\\
& =\frac{0+1}{\sqrt{1+0+0^{2}}+1}\\
& =\frac{1}{2}
\end{aligned}
$$\underset{x\rightarrow\left(a-\frac{1}{2}\right)}{lim}\frac{sin2x+cosx\cdot sin2x}{tanx+\frac{3\cdot sinx}{cos^{2}x}}$$=\underset{x\rightarrow0}{lim}\frac{sin2x(1+cosx)}{tanx(1+\frac{3}{cosx})}$
$=\underset{x\rightarrow0}{lim}\frac{sin2x}{tanx}\cdot\underset{x\rightarrow0}{lim}\frac{1+cosx}{(1+\frac{3}{cosx})}$
$=2\cdot\frac{1+1}{1+\frac{3}{1}}=2\cdot\frac{2}{4}=1$
Salah$\begin{aligned}a & =\underset{x\rightarrow0}{lim}\frac{\sqrt{1+x+x^{2}}-1}{x}\cdot\frac{\sqrt{1+x+x^{2}}+1}{\sqrt{1+x+x^{2}}+1}\\
& =\underset{x\rightarrow0}{lim}\frac{1+x+x^{2}-1}{x\left(\sqrt{1+x+x^{2}}+1\right)}\\
& =\underset{x\rightarrow0}{lim}\frac{x\left(x+1\right)}{x\left(\sqrt{1+x+x^{2}}+1\right)}\\
& =\underset{x\rightarrow0}{lim}\frac{x+1}{\left(\sqrt{1+x+x^{2}}+1\right)}\\
& =\frac{0+1}{\sqrt{1+0+0^{2}}+1}\\
& =\frac{1}{2}
\end{aligned}
$$\underset{x\rightarrow\left(a-\frac{1}{2}\right)}{lim}\frac{sin2x+cosx\cdot sin2x}{tanx+\frac{3\cdot sinx}{cos^{2}x}}$$=\underset{x\rightarrow0}{lim}\frac{sin2x(1+cosx)}{tanx(1+\frac{3}{cosx})}$
$=\underset{x\rightarrow0}{lim}\frac{sin2x}{tanx}\cdot\underset{x\rightarrow0}{lim}\frac{1+cosx}{(1+\frac{3}{cosx})}$
$=2\cdot\frac{1+1}{1+\frac{3}{1}}=2\cdot\frac{2}{4}=1$
Latihan Soal Limit Trigonometri Lanjutan (Sukar)
Ringkasan kuis
0 dari 5 pertanyaan telah diselesaikan
Pertanyaan:
- 1
- 2
- 3
- 4
- 5
Informasi
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
Anda harus masuk atau mendaftar untuk memulai kuis.
Anda harus menyelesaikan kuis dibawah ini, untuk memulai kuis ini:
Hasil
Hasil
0 dari 5 pertanyaan terjawab dengan benar
Waktu yang telah berlalu
Kategori
- Tidak Berkategori 0%
- 1
- 2
- 3
- 4
- 5
- Terjawab
- Tinjau
-
Pertanyaan ke 1 dari 5
1. Pertanyaan
$\underset{x\rightarrow\frac{\pi}{4}}{lim}\frac{cos2x}{x-\frac{\pi}{4}}=…$
Betul$\underset{x\rightarrow\frac{\pi}{4}}{lim}\frac{cos2x}{x-\frac{\pi}{4}}$
Misalkan $y=x-\frac{\pi}{4}$, maka $x=y+\frac{\pi}{4}$, sehingga :
$\begin{aligned}\underset{y\rightarrow0}{lim}\frac{cos2\left(y+\frac{\pi}{4}\right)}{y} & =\underset{y\rightarrow0}{lim}\frac{cos\left(2y+\frac{\pi}{2}\right)}{y}\\
& =\underset{y\rightarrow0}{lim}\frac{-sin2y}{y}\\
& =-2.
\end{aligned}
$Salah$\underset{x\rightarrow\frac{\pi}{4}}{lim}\frac{cos2x}{x-\frac{\pi}{4}}$
Misalkan $y=x-\frac{\pi}{4}$, maka $x=y+\frac{\pi}{4}$, sehingga :
$\begin{aligned}\underset{y\rightarrow0}{lim}\frac{cos2\left(y+\frac{\pi}{4}\right)}{y} & =\underset{y\rightarrow0}{lim}\frac{cos\left(2y+\frac{\pi}{2}\right)}{y}\\
& =\underset{y\rightarrow0}{lim}\frac{-sin2y}{y}\\
& =-2.
\end{aligned}
$ -
Pertanyaan ke 2 dari 5
2. Pertanyaan
$\underset{x\rightarrow\pi}{lim}\frac{1+cosx}{\left(x-\pi\right)^{2}}=…$
Betul$\underset{x\rightarrow\pi}{lim}\frac{1+cosx}{\left(x-\pi\right)^{2}}$
misalkan $y=x-\pi$, maka $x=y+\pi$
substitusikan $x=y+\frac{\pi}{4}$ ke pers limit diatas, sehingga diperoleh :
$\underset{y\rightarrow0}{lim}\frac{1+cos\left(y+\pi\right)}{\left(y+\pi-\pi\right)^{2}}$$=\underset{y\rightarrow0}{lim}\frac{1-cosy}{y^{2}}$
$=\underset{y\rightarrow0}{lim}\frac{2\cdot sin^{2}\frac{1}{2}y}{y^{2}}$
$=\underset{y\rightarrow0}{lim}\frac{1}{2}\cdot\frac{sin^{2}\frac{1}{2}y}{\frac{1}{4}y^{2}}$
$=\frac{1}{2}\cdot\underset{y\rightarrow0}{lim}\left(\frac{sin\frac{1}{2}y}{\frac{1}{2}y}\right)^{2}$
$=\frac{1}{2}\cdot1=\frac{1}{2}$
Salah$\underset{x\rightarrow\pi}{lim}\frac{1+cosx}{\left(x-\pi\right)^{2}}$
misalkan $y=x-\pi$, maka $x=y+\pi$
substitusikan $x=y+\frac{\pi}{4}$ ke pers limit diatas, sehingga diperoleh :
$\underset{y\rightarrow0}{lim}\frac{1+cos\left(y+\pi\right)}{\left(y+\pi-\pi\right)^{2}}$$=\underset{y\rightarrow0}{lim}\frac{1-cosy}{y^{2}}$
$=\underset{y\rightarrow0}{lim}\frac{2\cdot sin^{2}\frac{1}{2}y}{y^{2}}$
$=\underset{y\rightarrow0}{lim}\frac{1}{2}\cdot\frac{sin^{2}\frac{1}{2}y}{\frac{1}{4}y^{2}}$
$=\frac{1}{2}\cdot\underset{y\rightarrow0}{lim}\left(\frac{sin\frac{1}{2}y}{\frac{1}{2}y}\right)^{2}$
$=\frac{1}{2}\cdot1=\frac{1}{2}$
-
Pertanyaan ke 3 dari 5
3. Pertanyaan
$\underset{x\rightarrow\frac{\pi}{2}}{lim}\left(\frac{\pi}{2}-x\right)tanx=…$
Betul$\underset{x\rightarrow\frac{\pi}{2}}{lim}\left(\frac{\pi}{2}-x\right)tanx$
Misal $u=\frac{\pi}{2}-x$$\rightarrow x=u-\frac{\pi}{2}$
$x\rightarrow\frac{\pi}{2}$, maka $u\rightarrow0$
persamaan limit diatas menjadi :$\begin{aligned}\end{aligned}
$$\underset{u\rightarrow0}{lim}u.tan(u-\frac{\pi}{2})$$=\underset{u\rightarrow0}{lim}\frac{u}{cot(u-\frac{\pi}{2})}$
$=\underset{u\rightarrow0}{lim}\frac{u}{tan(u)}=1$
(ingat bentuk $\underset{x\rightarrow0}{lim}\frac{ax}{tanbx}=\frac{1}{b}$ )
Salah$\underset{x\rightarrow\frac{\pi}{2}}{lim}\left(\frac{\pi}{2}-x\right)tanx$
Misal $u=\frac{\pi}{2}-x$$\rightarrow x=u-\frac{\pi}{2}$
$x\rightarrow\frac{\pi}{2}$, maka $u\rightarrow0$
persamaan limit diatas menjadi :$\begin{aligned}\end{aligned}
$$\underset{u\rightarrow0}{lim}u.tan(u-\frac{\pi}{2})$$=\underset{u\rightarrow0}{lim}\frac{u}{cot(u-\frac{\pi}{2})}$
$=\underset{u\rightarrow0}{lim}\frac{u}{tan(u)}=1$
(ingat bentuk $\underset{x\rightarrow0}{lim}\frac{ax}{tanbx}=\frac{1}{b}$ )
-
Pertanyaan ke 4 dari 5
4. Pertanyaan
$\underset{x\rightarrow0}{lim}\left(\frac{2}{x^{2}}-\frac{sin2x}{x^{2}\cdot tanx}\right)=…$
Betul$\underset{x\rightarrow0}{lim}\left(\frac{2}{x^{2}}-\frac{sin2x}{x^{2}\cdot tanx}\right)$$=\underset{x\rightarrow0}{lim}\left(\frac{2tanx-sin2x}{x^{2}\cdot tanx}\right)$
$=\underset{x\rightarrow0}{lim}\left(\frac{2\left\{ \frac{sinx}{cosx}\right\} -2\cdot sinx\cdot cosx}{x^{2}\cdot tanx}\right)$
$=\underset{x\rightarrow0}{lim}\left(\frac{2sinx\left\{ \frac{1}{cosx}-cosx\right\} }{x^{2}.tanx}\right)$
$=\underset{x\rightarrow0}{lim}\left(\frac{2sinx\left\{ \frac{1-cos^{2}x}{cosx}\right\} }{x^{2}.tanx}\right)$
$=\underset{x\rightarrow0}{lim}\left(\frac{2sinx.sin^{2}x}{x^{2}.tanx.cosx}\right)$
$=\underset{x\rightarrow0}{lim}\left(\frac{2}{cosx}\right)\cdot\left(\frac{sinx}{x}\right)\cdot\left(\frac{sinx}{x}\right)\left(\frac{sinx}{tanx}\right)$
$=\frac{2}{1}\cdot1\cdot1\cdot1=2.$
Salah$\underset{x\rightarrow0}{lim}\left(\frac{2}{x^{2}}-\frac{sin2x}{x^{2}\cdot tanx}\right)$$=\underset{x\rightarrow0}{lim}\left(\frac{2tanx-sin2x}{x^{2}\cdot tanx}\right)$
$=\underset{x\rightarrow0}{lim}\left(\frac{2\left\{ \frac{sinx}{cosx}\right\} -2\cdot sinx\cdot cosx}{x^{2}\cdot tanx}\right)$
$=\underset{x\rightarrow0}{lim}\left(\frac{2sinx\left\{ \frac{1}{cosx}-cosx\right\} }{x^{2}.tanx}\right)$
$=\underset{x\rightarrow0}{lim}\left(\frac{2sinx\left\{ \frac{1-cos^{2}x}{cosx}\right\} }{x^{2}.tanx}\right)$
$=\underset{x\rightarrow0}{lim}\left(\frac{2sinx.sin^{2}x}{x^{2}.tanx.cosx}\right)$
$=\underset{x\rightarrow0}{lim}\left(\frac{2}{cosx}\right)\cdot\left(\frac{sinx}{x}\right)\cdot\left(\frac{sinx}{x}\right)\left(\frac{sinx}{tanx}\right)$
$=\frac{2}{1}\cdot1\cdot1\cdot1=2.$
-
Pertanyaan ke 5 dari 5
5. Pertanyaan
$\underset{x\rightarrow\frac{\pi}{2}}{lim}\left[(1+cotx)^{tanx}\right]^{3}=…$
Betul$\underset{x\rightarrow\frac{\pi}{2}}{lim}\left[(1+cotx)^{tanx}\right]^{3}$$=\underset{x\rightarrow\frac{\pi}{2}}{lim}\left[\left(1+\frac{1}{tanx}\right)^{tanx}\right]^{3}$$=e^{3}$
Salah$\underset{x\rightarrow\frac{\pi}{2}}{lim}\left[(1+cotx)^{tanx}\right]^{3}$$=\underset{x\rightarrow\frac{\pi}{2}}{lim}\left[\left(1+\frac{1}{tanx}\right)^{tanx}\right]^{3}$$=e^{3}$