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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

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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. \( -\frac{1}{2} \)
  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
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
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.

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Penulis: Tju Ji Long · Statistisi

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