www.jagostat.com

www.jagostat.com

Website Belajar Matematika & Statistika

Website Belajar Matematika & Statistika

Bahas Soal Matematika   »   Limit dan Kekontinuan   ›  Contoh Soal Limit Trigonometri dengan Pemfaktoran
Joki Tugas Matematika Murah, Hanya Rp10-50 Ribu. Hub. WA: 0812-5632-4552

Contoh Soal Limit Trigonometri dengan Pemfaktoran

Supaya makin paham dengan pembahasan mengenai limit fungsi trigonometri, Anda perlu sering latihan mengerjakan soal. Dalam artikel ini kita akan membahas 20 contoh soal limit trigonometri yang penyelesaiannya perlu menggunakan pemfaktoran.

Sebelum kita masuk ke pembahasan dari masing-masing soal tersebut, ada baiknya Anda pahami beberapa teorema penting terkait limit trigonometri berikut ini. Kita akan sering menggunakan teorema ini untuk menyelesaikan limit fungsi trigonometri.

Teorema Limit Trigonometri
Contoh 1: SPMB 2005

Nilai \( \displaystyle \lim_{x \to 1} \ \frac{(x^2+x-2) \sin (x-1)}{x^2 - 2x + 1} = \cdots \)

  1. 4
  2. 3
  3. 0
  4. \( -\frac{1}{4} \)
  5. \( -\frac{1}{2} \)
Pembahasan »
\begin{aligned} \lim_{x \to 1} \ \frac{(x^2+x-2) \sin (x-1)}{x^2 - 2x + 1} &= \lim_{x \to 1} \ \frac{(x+2)(x-1) \sin(x-1)}{(x-1)(x-1)} \\[8pt] &= \lim_{x \to 1} \ \frac{(x+2)\sin(x-1)}{(x-1)} \\[8pt] &= \lim_{x \to 1} \ (x+2) \cdot \lim_{x \to 1} \ \frac{\sin(x-1)}{(x-1)} \\[8pt] &= (1+2) \cdot 1 \\[8pt] &= 3 \end{aligned}

Jawaban B.

Contoh 2: EBTANAS SMA IPA 1998

Nilai \( \displaystyle \lim_{x \to 5} \ \frac{(4x-10) \sin (x-5)}{x^2-25} = \cdots \)

  1. -3
  2. -1
  3. 1
  4. 2
  5. 4
Pembahasan »
\begin{aligned} \lim_{x \to 5} \ \frac{(4x-10) \sin (x-5)}{x^2-25} &= \lim_{x \to 5} \ \frac{(4x-10) \sin (x-5)}{(x-5)(x+5)} \\[8pt] &= \lim_{x \to 5} \ \frac{(4x-10)}{(x+5)} \cdot \lim_{x \to 5} \ \frac{\sin (x-5)}{(x-5)} \\[8pt] &= \frac{4(5)-10}{5+5} \cdot 1 = \frac{20-10}{10} = \frac{10}{10} \\[8pt] &= 1 \end{aligned}

Jawaban C.

Contoh 3: UM UGM 2004

Nilai \( \displaystyle \lim_{x \to 1} \ \frac{\tan (x-1) \sin(1-\sqrt{x})}{x^2-2x+1} = \cdots \)

  1. -1
  2. \( -\frac{1}{3} \)
  3. 0
  4. \( -\frac{1}{2} \)
  5. 1
Pembahasan »
\begin{aligned} \lim_{x \to 1} \ \frac{\tan (x-1) \sin(1-\sqrt{x})}{x^2-2x+1} &= \lim_{x \to 1} \ \frac{\tan (x-1) \sin(1-\sqrt{x})}{-(x-1)(1-x)} \\[8pt] &= \lim_{x \to 1} \ \frac{\tan (x-1) \sin(1-\sqrt{x})}{-(x-1)(1+\sqrt{x})(1-\sqrt{x})} \\[8pt] &= \lim_{x \to 1} \ \frac{1}{-(1+\sqrt{x})} \cdot \lim_{x \to 1} \ \frac{\tan(x-1)}{(x-1)} \cdot \lim_{x \to 1} \ \frac{\sin(1-\sqrt{x})}{(1-\sqrt{x})} \\[8pt] &= \frac{1}{-(1+\sqrt{1})} \cdot 1 \cdot 1 = -\frac{1}{2} \end{aligned}

Jawaban D.

Contoh 4: SPMB 2005

Nilai \( \displaystyle \lim_{x \to 2} \ \frac{\tan (2-\sqrt{2x})}{x^2 - 2x} = \cdots \)

  1. \( \frac{1}{4} \)
  2. \( \frac{1}{8} \)
  3. \( 0 \)
  4. \( -\frac{1}{6} \)
  5. \( -\frac{1}{4} \)
Pembahasan »
\begin{aligned} \lim_{x \to 2} \ \frac{\tan (2-\sqrt{2x})}{x^2 - 2x} &= \lim_{x \to 2} \ \frac{\tan (-\sqrt{2} \ (\sqrt{x}-\sqrt{2}))}{ x \ (\sqrt{x}+\sqrt{2})(\sqrt{x}-\sqrt{2}) } \\[8pt] &= \lim_{x \to 2} \ \frac{\tan (-\sqrt{2} \ (\sqrt{x}-\sqrt{2}))}{(\sqrt{x}-\sqrt{2}) } \cdot \lim_{x \to 2} \ \frac{1}{ x \ (\sqrt{x}+\sqrt{2})} \\[8pt] &= -\sqrt{2} \cdot \frac{1}{2(\sqrt{2} + \sqrt{2})} = -\sqrt{2} \cdot \frac{1}{4\sqrt{2}} \\[8pt] &= - \frac{1}{4} \end{aligned}

Jawaban E.

Contoh 5:

Nilai \( \displaystyle \lim_{x \to 0} \ \frac{\sin (x-1)}{x^2+x-2} = \cdots \)

Pembahasan »
\begin{aligned} \lim_{x \to 0} \ \frac{\sin (x-1)}{x^2+x-2} &= \lim_{x \to 0} \ \frac{\sin (x-1)}{(x+2)(x-1)} \\[8pt] &= \lim_{x \to 0} \ \frac{1}{(x+2)} \cdot \lim_{x \to 0} \ \frac{\sin (x-1)}{(x-1)} \\[8pt] &= \frac{1}{0+2} \cdot 1 \\[8pt] &= \frac{1}{2} \end{aligned}
Contoh 6:

Nilai \( \displaystyle \lim_{x \to a} \ \frac{\sin (x-a)}{x^2-a^2} = \cdots \)

Pembahasan »
\begin{aligned} \lim_{x \to a} \ \frac{\sin (x-a)}{x^2-a^2} &= \lim_{x \to a} \ \frac{\sin (x-a)}{(x-a)(x+a)} \\[8pt] &= \lim_{x \to a} \ \frac{\sin (x-a)}{(x-a)} \cdot \lim_{x \to a} \ \frac{1}{(x+a)} \\[8pt] &= 1 \cdot \frac{1}{a+a} \\[8pt] &= \frac{1}{2a} \end{aligned}
Contoh 7:

Nilai \( \displaystyle \lim_{x\to 1} \ \frac{\tan(1-x)}{x^3-1} = \cdots \)

  1. \( \frac{1}{3} \)
  2. \( -\frac{1}{3} \)
  3. \( 1 \)
  4. \( -1 \)
  5. \( \frac{1}{2} \)
Pembahasan »
\begin{aligned} \lim_{x\to 1} \ \frac{\tan(1-x)}{x^3-1} &= \lim_{x\to 1} \ \frac{\tan(1-x)}{(x-1)(x^2+x+1)} \\[8pt] &= \lim_{x\to 1} \ \frac{\tan(1-x)}{-(1-x)(x^2+x+1)} \\[8pt] &= \lim_{x\to 1} \ \frac{\tan(1-x)}{(1-x)} \cdot \lim_{x\to 1} \ \frac{1}{-(x^2+x+1)} \\[8pt] &= 1 \cdot \frac{1}{-(1^2 + 1 + 1)} = - \frac{1}{3} \end{aligned}

Jawaban B.

Contoh 8: UM UNDIP 2010

Nilai \( \displaystyle \lim_{x \to -1} \frac{\sin (1-x^2) \cos (1-x^2)}{x^2-1} = \cdots \)

  1. 1
  2. -1
  3. 2
  4. -2
  5. 0
Pembahasan »
\begin{aligned} \lim_{x \to -1} \frac{\sin (1-x^2) \cos (1-x^2)}{x^2-1} &= \lim_{x \to -1} \frac{\sin (1-x^2) \cos (1-x^2)}{-(1-x^2)} \\[8pt] &= \lim_{x \to -1} \frac{\sin (1-x^2)}{-(1-x^2)} \cdot \lim_{x \to -1} \cos (1-x^2) \\[8pt] &= \frac{1}{-1} \cdot \cos 0 = -1 \end{aligned}

Jawaban B.

Contoh 9: SIMAK UI 2012

Nilai \( \displaystyle \lim_{x \to 1} \frac{\sin 2(x-1)}{(x^2-2x+1) \cot \frac{1}{2}(x-1)} = \cdots \)

  1. 1/4
  2. 1/2
  3. 1
  4. 2
  5. 4
Pembahasan »
\begin{aligned} \lim_{x \to 1} \ \frac{\sin 2(x-1)}{(x^2-2x+1) \cot \frac{1}{2}(x-1)} &= \lim_{x \to 1} \ \frac{\sin 2(x-1)}{(x-1)(x-1) \ \frac{\cos \frac{1}{2}(x-1) }{\sin \frac{1}{2}(x-1)} } \\[8pt] &= \lim_{x \to 1} \ \frac{\sin 2(x-1) \sin \frac{1}{2}(x-1)}{(x-1)(x-1) \ \cos \frac{1}{2}(x-1) } \\[8pt] &= \lim_{x \to 1} \ \frac{\sin 2(x-1)}{(x-1)} \cdot \lim_{x \to 1} \ \frac{\sin \frac{1}{2}(x-1)}{(x-1)} \cdot \lim_{x \to 1} \ \frac{1}{\cos \frac{1}{2}(x-1) } \\[8pt] &= 2 \cdot \frac{1}{2} \cdot \frac{1}{\cos 0} = 2 \cdot \frac{1}{2} \cdot \frac{1}{1} = 1 \end{aligned}

Jawaban C.

Contoh 10: SPMB 2006

Nilai \( \displaystyle \lim_{x \to 5} \ \frac{2x^3 - 20x^2 + 50x}{\sin^2 (x-5) \cos (2x-10)} = \cdots \)

  1. 0
  2. 1
  3. 5
  4. 10
  5. ∞
Pembahasan »
\begin{aligned} \lim_{x \to 5} \ \frac{2x^3 - 20x^2 + 50x}{\sin^2 (x-5) \cos (2x-10)} &= \lim_{x \to 5} \ \frac{2x \ (x^2 - 10x + 25)}{\sin^2 (x-5) \cos (2x-10)} \\[8pt] &= \lim_{x \to 5} \ \frac{2x \ (x-5)(x-5)}{\sin^2 (x-5) \cos (2x-10)} \\[8pt] &= \lim_{x \to 5} \ \frac{(x-5)(x-5)}{\sin^2 (x-5) } \cdot \lim_{x \to 5} \ \frac{2x}{\cos (2x-10)} \\[8pt] &= (1)^2 \cdot \frac{2(5)}{\cos 0} = 1 \cdot \frac{10}{1} = 10 \end{aligned}

Jawaban D.

Contoh 11: UM UGM 2016

Nilai \( \displaystyle \lim_{x \to 3} \ \frac{(x+6) \tan(2x-6)}{(x^2-x-6)} = \cdots \)

  1. \( -\frac{18}{5} \)
  2. \( -\frac{9}{5} \)
  3. \( \frac{9}{5} \)
  4. \( \frac{18}{5} \)
  5. \( \frac{27}{5} \)
Pembahasan »
\begin{aligned} \lim_{x \to 3} \ \frac{(x+6) \tan(2x-6)}{(x^2-x-6)} &= \lim_{x \to 3} \ \frac{(x+6) \tan 2(x-3)}{(x+2)(x-3)} \\[8pt] &= \lim_{x \to 3} \ \frac{(x+6)}{(x+2)} \cdot \lim_{x \to 3} \ \frac{\tan 2(x-3)}{(x-3)} \\[8pt] &= \frac{(3+6)}{(3+2)} \cdot 2 = \frac{9}{5} \cdot 2 = \frac{18}{5} \end{aligned}

Jawaban D.

Contoh 12: UM STIS 2017

Nilai \( \displaystyle \lim_{x \to 2} \frac{(x^2-5x-6) \sin 2(x-2)}{(x^2-x-2)} = \cdots \)

  1. -8
  2. -5
  3. -2
  4. 3/4
  5. 5
Pembahasan »
\begin{aligned} \lim_{x \to 2} \frac{(x^2-5x-6) \sin 2(x-2)}{(x^2-x-2)} &= \lim_{x \to 2} \frac{(x+1)(x-6) \sin 2(x-2)}{(x+1)(x-2)} \\[8pt] &= \lim_{x \to 2} \frac{(x-6) \sin 2(x-2)}{(x-2)} \\[8pt] &= \lim_{x \to 2} (x-6) \cdot \lim_{x \to 2} \frac{\sin 2(x-2)}{(x-2)} \\[8pt] &= (2-6) \cdot 2 = -8 \end{aligned}

Jawaban A.

Contoh 13: UM UGM 2017

Nilai \( \displaystyle \lim_{x \to -4} \ \frac{1-\cos(x+4)}{x^2+8x+16} = \cdots \)

  1. \( -2 \)
  2. \( -\frac{1}{2} \)
  3. \( \frac{1}{3} \)
  4. \( \frac{1}{2} \)
  5. \( 2 \)
Pembahasan »
\begin{aligned} \lim_{x \to -4} \ \frac{1-\cos(x+4)}{x^2+8x+16} &= \lim_{x \to -4} \ \frac{2\sin^2 \frac{1}{2}(x+4)}{(x+4)(x+4)} \\[8pt] &= 2 \cdot \lim_{x \to -4} \ \frac{\sin \frac{1}{2}(x+4)}{(x+4)} \cdot \lim_{x \to -4} \ \frac{\sin \frac{1}{2}(x+4)}{(x+4)} \\[8pt] &= 2 \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2} \end{aligned}

Jawaban D.

Contoh 14: UM UNDIP 2011

Nilai \( \displaystyle \lim_{x \to 1} \ \frac{(x^2+x-2) \sin(x^2-1)}{x^2-2x+1} = \cdots \)

  1. \( -4 \)
  2. \( -\frac{3}{4} \)
  3. \( \frac{1}{2} \)
  4. 3
  5. 6
Pembahasan »
\begin{aligned} \lim_{x \to 1} \ \frac{(x^2+x-2) \sin(x^2-1)}{x^2-2x+1} &= \lim_{x \to 1} \ \frac{(x+2)(x-1) \sin(x+1)(x-1)}{(x-1)(x-1)} \\[8pt] &= \lim_{x \to 1} \ \frac{(x+2)\sin(x+1)(x-1)}{(x-1)} \\[8pt] &= \lim_{x \to 1} \ (x+2) \cdot \lim_{x \to 1} \ \frac{\sin((x+1)(x-1))}{(x-1)} \\[8pt] &= \lim_{x \to 1} (x+2) \cdot (x+1) \\[8pt] &= (1+2)(1+1) = 3 \cdot 2 = 6 \end{aligned}

Jawaban E.

Contoh 15: UMPTN 1995

Nilai \( \displaystyle \lim_{x \to 0} \ \frac{(x^2-1) \sin 6x}{x^3+3x^2+2x} = \cdots \)

  1. -3
  2. -2
  3. 2
  4. 3
  5. 5
Pembahasan »
\begin{aligned} \lim_{x \to 0} \ \frac{(x^2-1) \sin 6x}{x^3+3x^2+2x} &= \lim_{x \to 0} \ \frac{(x+1)(x-1) \sin 6x}{x(x+1)(x+2)} \\[8pt] &= \lim_{x \to 0} \ \frac{(x-1) \sin 6x}{x(x+2)} \\[8pt] &= \lim_{x \to 0} \ \frac{(x-1)}{(x+2)} \cdot \lim_{x \to 0} \ \frac{\sin 6x}{x} \\[8pt] &= \frac{(0-1)}{(0+2)} \cdot 6 = \frac{-1}{2} \cdot 6 \\[8pt] &= -3 \end{aligned}

Jawaban A.

Contoh 16: UMPTN 1995

Nilai \( \displaystyle \lim_{x \to -2} \ \frac{1-\cos(x+2)}{x^2+4x+4} = \cdots \)

  1. 0
  2. 1/4
  3. 1/2
  4. 2
  5. 4
Pembahasan »
\begin{aligned} \lim_{x \to -2} \ \frac{1-\cos(x+2)}{x^2+4x+4} &= \lim_{x \to -2} \ \frac{2\sin^2 \frac{1}{2}(x+2)}{(x+2)^2} \\[8pt] &= 2 \cdot \left(\frac{1}{2} \right)^2 = 2 \cdot \frac{1}{4} \\[8pt] &= \frac{1}{2} \end{aligned}

Jawaban C.

Contoh 17:

Nilai \( \displaystyle \lim_{x\to -2} \ \frac{(x^2-4) \tan(x+2)}{\sin^2 (x+2)} = \cdots \)

  1. -4
  2. -3
  3. 0
  4. 4
  5. ∞
Pembahasan »
\begin{aligned} \lim_{x\to -2} \ \frac{(x^2-4) \tan(x+2)}{\sin^2 (x+2)} &= \lim_{x\to -2} \ \frac{(x-2)(x+2) \tan(x+2)}{\sin^2 (x+2)} \\[8pt] &= \lim_{x\to -2} \ (x-2) \cdot \lim_{x\to -2} \ \frac{(x+2)}{\sin (x+2)} \cdot \lim_{x\to -2} \ \frac{\tan(x+2)}{\sin(x+2)} \\[8pt] &= (-2-2) \cdot 1 \cdot 1 \\[8pt] &= -4 \end{aligned}

Jawaban A.

Contoh 18:

Nilai \( \displaystyle \lim_{x \to 1} \ \frac{(x^2-1)\tan (2x-2)}{\sin^2 (x-1)} = \cdots \)

  1. 1
  2. 2
  3. 4
  4. 6
  5. 8
Pembahasan »
\begin{aligned} \lim_{x \to 1} \ \frac{(x^2-1)\tan (2x-2)}{\sin^2 (x-1)} &= \lim_{x \to 1} \ \frac{(x-1)(x+1) \tan 2(x-1)}{\sin^2 (x-1)} \\[8pt] &= \lim_{x \to 1} \ (x+1) \cdot \lim_{x \to 1} \ \frac{(x-1)}{\sin (x-1)} \cdot \lim_{x \to 1} \ \frac{\tan 2(x-1)}{\sin (x-1)} \\[8pt] &= (1+1) \cdot 1 \cdot 2 \\[8pt] &= 4 \end{aligned}

Jawaban C.

Contoh 19:

Nilai \( \displaystyle \lim_{x \to \frac{\pi}{8}} \ \frac{\sin^2 2x - \cos^2 2x}{\sin 2x - \cos 2x} = \cdots \)

  1. \( -\sqrt{2} \)
  2. \( -\frac{1}{2} \sqrt{2} \)
  3. \( 0 \)
  4. \( \frac{1}{2} \sqrt{2} \)
  5. \( \sqrt{2} \)
Pembahasan »
\begin{aligned} \lim_{x \to \frac{\pi}{8}} \ \frac{\sin^2 2x - \cos^2 2x}{\sin 2x - \cos 2x} &= \lim_{x \to \frac{\pi}{8}} \ \frac{(\sin 2x - \cos 2x)(\sin 2x + \cos 2x)}{\sin 2x - \cos 2x} \\[8pt] &= \lim_{x \to \frac{\pi}{8}} \ (\sin 2x + \cos 2x) \\[8pt] &= \sin \left(2 \cdot \frac{\pi}{8}\right) + \cos \left(2 \cdot \frac{\pi}{8}\right) \\[8pt] &= \sin \frac{\pi}{4} + \cos \frac{\pi}{4} \\[8pt] &= \frac{1}{2} \sqrt{2} + \frac{1}{2} \sqrt{2} = \sqrt{2} \end{aligned}

Jawaban E.

Contoh 20:

Nilai \( \displaystyle \lim_{x\to \frac{\pi}{2}} \ \frac{1-\sin^2 x}{(\sin^2 \frac{1}{2} x - \cos^2 \frac{1}{2} x)} = \cdots \)

  1. -2
  2. -1
  3. 0
  4. 1
  5. 2
Pembahasan »
\begin{aligned} \lim_{x\to \frac{\pi}{2}} \ \frac{1-\sin^2 x}{(\sin \frac{1}{2} x - \cos \frac{1}{2} x)^2} &= \lim_{x\to \frac{\pi}{2}} \ \frac{(1-\sin x)(1+\sin x)}{\sin^2 \frac{1}{2} x + \cos^2 \frac{1}{2} x - 2 \sin \frac{1}{2} x \cos \frac{1}{2} x} \\[8pt] &= \lim_{x\to \frac{\pi}{2}} \ \frac{(1-\sin x)(1+\sin x)}{1 - \sin x} \\[8pt] &= \lim_{x\to \frac{\pi}{2}} \ (1+\sin x) \\[8pt] &= 1 + \sin \frac{\pi}{2} \\[8pt] &= 1 + 1 = 2 \end{aligned}

Jawaban E.

Cukup sekian untuk artikel ini. Jika Anda merasa artikel ini bermanfaat, bantu klik tombol suka di bawah ini dan jika ada yang kurang jelas dari artikel ini silahkan tanyakan di kolom komentar. Terima kasih.

Penulis: Tju Ji Long · Statistisi

Artikel Terkait
img

Limit Matematika: Definisi, Metode, dan Contoh Soal