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Bahas Soal Matematika   »   Limit dan Kekontinuan   ›  Contoh Soal Limit Trigonometri Kelas 12
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Contoh Soal Limit Trigonometri Kelas 12

Contoh 1: SNMPTN 2010

Nilai \( \displaystyle \lim_{x \to 0} \ \frac{\sqrt{4x}}{\sqrt{\sin 2x}} = \cdots \)

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

Jawaban A.

Contoh 2: UN SMA IPA 2010

Nilai \( \displaystyle \lim_{x \to 0} \ \frac{\sin x + \sin 5x}{6x} = \cdots \)

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

Jawaban B.

Contoh 3: UM UGM 2006

Nilai \( \displaystyle \lim_{x \to 0} \ \left( \frac{1}{x} - \frac{1}{x \cos x} \right ) = \cdots \)

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

Jawaban C.

Contoh 4: UN SMA IPA 2004

Nilai \( \displaystyle \lim_{x \to -2} \ \frac{(x+6) \sin (x+2)}{x^2-3x-10} = \cdots \)

  1. -4/3
  2. -4/7
  3. -2/5
  4. 0
  5. 1
Pembahasan »
\begin{aligned} \lim_{x \to -2} \ \frac{(x+6) \sin (x+2)}{x^2-3x-10} &= \lim_{x \to -2} \ \frac{(x+6) \sin (x+2)}{(x-5)(x+2)} \\[8pt] &= \lim_{x \to -2} \ \frac{(x+6)}{(x-5)} \cdot \lim_{x \to -2} \ \frac{\sin (x+2)}{(x+2)} \\[8pt] &= \frac{-2+6}{-2-5} \cdot 1 \\[8pt] &= -\frac{4}{7} \end{aligned}

Jawaban B.

Contoh 5: SPMB 2005

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

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

Jawaban A.

Contoh 6: EBTANAS SMA IPA 1996

Nilai \( \displaystyle \lim_{x \to 0} \ \frac{\sin 4x + \sin 2x}{3x \cos x} = \cdots \)

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

Jawaban E.

Contoh 7: EBTANAS SMA IPA 2001

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

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

Jawaban D.

Contoh 8: UN SMA IPA 2007

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

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

Jawaban D.

Contoh 9: SPMB 2006

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

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

Jawaban A.

Contoh 10: UM UGM 2005

Nilai \( \displaystyle \lim_{x \to 0} \ \frac{x \tan 5x}{\cos 2x - \cos 7x} = \cdots \)

  1. 2/9
  2. 1/9
  3. 0
  4. -1/9
  5. -2/9
Pembahasan »
\begin{aligned} \lim_{x \to 0} \ \frac{x \tan 5x}{\cos 2x - \cos 7x} &= \lim_{x \to 0} \ \frac{x \tan 5x}{-2 \sin \left( \frac{2x+7x}{2} \right) \sin \left( \frac{2x-7x}{2} \right)} \\[8pt] &= \lim_{x \to 0} \ \frac{x \tan 5x}{-2 \sin \left( \frac{9}{2}x \right) \sin \left( -\frac{5}{2} x \right)} \\[8pt] &= \lim_{x \to 0} \ \frac{x \tan 5x}{-2 \sin \left( \frac{9}{2}x \right) \left( -\sin (\frac{5}{2} x) \right)} \\[8pt] &= \lim_{x \to 0} \ \frac{x \tan 5x}{2 \sin \left( \frac{9}{2}x \right) \left( \sin (\frac{5}{2} x) \right)} \\[8pt] &= \frac{1}{2} \cdot \lim_{x \to 0} \ \frac{x}{\sin \left( \frac{9}{2}x \right)} \cdot \lim_{x \to 0} \ \frac{\tan 5x}{\sin (\frac{5}{2} x)} \\[8pt] &= \frac{1}{2} \cdot \frac{1}{\frac{9}{2}} \cdot \frac{5}{\frac{5}{2}} = \frac{1}{2} \cdot \frac{2}{9} \cdot 2 \\[8pt] &= \frac{2}{9} \end{aligned}

Jawaban A.

Contoh 11: UM UGM 2013

Nilai \( \displaystyle \lim_{x \to 0} \ \frac{1-\cos^3 x}{x \tan x} = \cdots \)

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

Jawaban D.

Contoh 12: SPMB 2006

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

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

Jawaban E.

Contoh 13: UM STIS 2011

Jika \( \displaystyle \lim_{x \to 0} \ \frac{x^a \sin^4 x}{\sin^6 x} = 1\), maka nilai \(a\) adalah...

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

Agar nilai limit fungsi di atas adalah 1, maka nilai \(a\) harus 2.

Jawaban B.

Contoh 14: SNMPTN 2012

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

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

Jawaban E.

Contoh 15: SBMPTN 2013

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

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

Jawaban D.

Contoh 16:

Nilai \( \displaystyle \lim_{x\to 0} \ \frac{\sin (4x^2)}{x^2+\tan^2 3x} = \cdots \)

  1. 1/5
  2. 2/5
  3. 3/5
  4. 4/5
  5. 1
Pembahasan »
\begin{aligned} \lim_{x\to 0} \ \frac{\sin (4x^2)}{x^2+\tan^2 3x} &= \lim_{x\to 0} \ \frac{\sin (4x^2)}{x^2+\tan^2 3x} \times \frac{\frac{1}{x^2}}{\frac{1}{x^2}} \\[8pt] &= \lim_{x\to 0} \ \frac{ \frac{\sin (4x^2)}{x^2}}{1+\frac{\tan^2 3x}{x^2}} \\[8pt] &= \frac{\displaystyle \lim_{x\to 0} \ \frac{\sin (4x^2)}{x^2}}{\displaystyle \lim_{x\to 0} \ 1 + \lim_{x\to 0} \ \frac{\tan^2 3x}{x^2}} \\[8pt] &= \frac{4}{1+3^3} = \frac{4}{10}= \frac{2}{5} \end{aligned}

Jawaban B.

Contoh 17:

Nilai \( \displaystyle \lim_{x\to 0} \ \frac{\tan(2x^3)}{x^3+\sin^3 4x} = \cdots \)

  1. 5/65
  2. 4/65
  3. 3/65
  4. 2/65
  5. 1/65
Pembahasan »
\begin{aligned} \lim_{x\to 0} \ \frac{\tan(2x^3)}{x^3+\sin^3 4x} &= \lim_{x\to 0} \ \frac{\tan(2x^3)}{x^3+\sin^3 4x} \times \frac{\frac{1}{x^3}}{\frac{1}{x^3}} \\[8pt] &= \lim_{x\to 0} \ \frac{ \frac{\tan(2x^3)}{x^3}}{1+\frac{\sin^3 4x}{x^3}} \\[8pt] &= \frac{\displaystyle \lim_{x\to 0} \ \frac{\tan(2x^3)}{x^3}}{\displaystyle \lim_{x\to 0} \ 1 + \lim_{x\to 0} \ \frac{\sin^3 4x}{x^3}} \\[8pt] &= \frac{2}{1+4^3} = \frac{2}{1+64}= \frac{2}{65} \end{aligned}

Jawaban D.

Contoh 18:

Nilai \( \displaystyle \lim_{x\to 45^0} \ \frac{\cos 2x}{1-\tan x} = \cdots \)

  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
Pembahasan »
\begin{aligned} \lim_{x\to 45^0} \ \frac{\cos 2x}{1-\tan x} &= \lim_{x\to 45^0} \ \frac{\cos^2 x - \sin^2 x}{\frac{\cos x - \sin x}{\cos x}} \\[8pt] &= \lim_{x\to 45^0} \ \frac{(\cos^2 x - \sin^2 x) \cos x}{\cos x - \sin x} \\[8pt] &= \lim_{x\to 45^0} \ \frac{(\cos x - \sin x) (\cos x + \sin x) \cos x}{\cos x - \sin x} \\[8pt] &= \lim_{x\to 45^0} \ (\cos x + \sin x) \cos x \\[8pt] &= (\cos 45^0 + \sin 45^0) \cos 45^0 \\[8pt] &= \left( \frac{1}{2} \sqrt{2} + \frac{1}{2} \sqrt{2} \right) \frac{1}{2} \sqrt{2} \\[8pt] &= 1 \end{aligned}

Jawaban A.

Contoh 19:

Nilai \( \displaystyle \lim_{x\to 0} \ \frac{x^2 \tan 2x}{x-x \cos 4x} = \cdots \)

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

Jawaban D.

Contoh 20:

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

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

Jawaban C.

Contoh 21:

Nilai \( \displaystyle \lim_{x\to 0} \ \frac{1-\cos 4x}{x \sin x} = \cdots \)

  1. 1
  2. 2
  3. 4
  4. 8
  5. 16
Pembahasan »
\begin{aligned} \lim_{x\to 0} \ \frac{1-\cos 4x}{x \sin x} &= \lim_{x\to 0} \ \frac{1-\cos (2 \cdot 2x)}{x \sin x} \\[8pt] &= \lim_{x\to 0} \ \frac{2 \sin^2 2x}{x \sin x} \\[8pt] &= 2 \cdot \lim_{x\to 0} \ \frac{\sin 2x}{x} \cdot \lim_{x\to 0} \ \frac{\sin 2x}{\sin x} \\[8pt] &= 2 \cdot 2 \cdot 2 \\[8pt] &= 8 \end{aligned}

Jawaban D.

Contoh 22:

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

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

Jawaban D.

Contoh 23:

Nilai \( \displaystyle \lim_{x\to 0} \ \frac{\tan 2x \cdot \tan 3x}{3x^2} = \cdots \)

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

Jawaban D.

Contoh 24:

Nilai \( \displaystyle \lim_{x\to 0} \ \frac{\sin 7x + \tan 3x - \sin 5x}{\tan 9x - \tan 3x - \sin x} = \cdots \)

  1. 9
  2. 7
  3. 5
  4. 3
  5. 1
Pembahasan »
\begin{aligned} \lim_{x\to 0} \ \frac{\sin 7x + \tan 3x - \sin 5x}{\tan 9x - \tan 3x - \sin x} &= \lim_{x\to 0} \ \frac{\sin 7x + \tan 3x - \sin 5x}{\tan 9x - \tan 3x - \sin x} \times \frac{\frac{1}{x}}{\frac{1}{x}} \\[8pt] &= \lim_{x\to 0} \ \frac{ \frac{\sin 7x}{x} + \frac{\tan 3x}{x} - \frac{\sin 5x}{x} }{\frac{\tan 9x}{x} - \frac{\tan 3x}{x} - \frac{\sin x}{x}} \\[8pt] &= \frac{ \displaystyle \lim_{x\to 0} \ \frac{\sin 7x}{x} + \lim_{x\to 0} \ \frac{\tan 3x}{x} - \lim_{x\to 0} \ \frac{\sin 5x}{x} }{ \displaystyle \lim_{x\to 0} \ \frac{\tan 9x}{x} - \lim_{x\to 0} \ \frac{\tan 3x}{x} - \lim_{x\to 0} \ \frac{\sin x}{x} } \\[8pt] &= \frac{7 + 3 - 5}{9 - 3 - 1} = \frac{5}{5} = 1 \end{aligned}

Jawaban E.

Contoh 25:

Nilai \( \displaystyle \lim_{x\to 0} \ \frac{x \cos 2x}{\tan x - \sin 2x} = \cdots \)

  1. -1
  2. -1/2
  3. 0
  4. 1/2
  5. 1
Pembahasan »
\begin{aligned} \lim_{x\to 0} \ \frac{x \cos 2x}{\tan x - \sin 2x} &= \lim_{x\to 0} \ \frac{x \cos 2x}{\tan x - \sin 2x} \times \frac{\frac{1}{x}}{\frac{1}{x}} \\[8pt] &= \lim_{x\to 0} \ \frac{\cos 2x}{\frac{\tan x}{x} - \frac{\sin 2x}{x}} \\[8pt] &= \frac{ \displaystyle \lim_{x\to 0} \ \cos 2x }{ \displaystyle \lim_{x\to 0} \ \frac{\tan x}{x} - \lim_{x\to 0} \ \frac{\sin 2x}{x} } \\[8pt] &= \frac{\cos 0}{1 - 2} = \frac{1}{-1} = -1 \end{aligned}

Jawaban A.

Contoh 26: SIMAK UI 2009

Nilai \( \displaystyle \lim_{x \to 0} \ \left(\csc x - \frac{1}{x} \right) = \cdots \)

  1. ∞
  2. 2
  3. 0
  4. -2
  5. -∞
Pembahasan »
\begin{aligned} \lim_{x \to 0} \ \left(\csc x - \frac{1}{x} \right) &= \lim_{x \to 0} \ \left(\frac{1}{\sin x} - \frac{1}{x} \right) \\[8pt] &= \lim_{x \to 0} \ \frac{x-\sin x}{x \sin x} \quad (\text{gunakan aturan L'Hospital}) \\[8pt] &= \lim_{x \to 0} \ \frac{1-\cos x}{1 \cdot \sin x + x \cos x} \\[8pt] &= \lim_{x \to 0} \ \frac{\sin x}{2\cos x - x\sin x} \\[8pt] &= \frac{\sin 0}{2 \cos 0 - 0 \cdot \sin 0} \\[8pt] &= \frac{0}{2\cdot 1 - 0} = 0 \end{aligned}

Jawaban C.

Contoh 27: SBMPTN 2018

Nilai \( \displaystyle \lim_{x \to 3} \ \frac{\sin (2x-6)}{\sqrt{4-x}-1} = \cdots \)

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

Jawaban E.

Contoh 28: SBMPTN 2018

Nilai \( \displaystyle \lim_{x \to 2} \frac{\sin (2x-4)}{2-\sqrt{6-x}} = \cdots \)

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

Jawaban E.

Contoh 29: UM UNDIP 2010

Nilai \( \displaystyle \lim_{x \to y} \ \frac{\sin x - \sin y}{x-y} = \cdots \)

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

Jawaban E.

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Penulis: Tju Ji Long · Statistisi

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