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Website Belajar Matematika & Statistika
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Bahas Soal Matematika » Limit dan Kekontinuan › Contoh Soal Limit Trigonometri Bentuk Akar
Joki Tugas Matematika Murah, Hanya Rp10-50 Ribu. Hub. WA: 0812-5632-4552
Contoh Soal Limit Trigonometri Bentuk Akar
Contoh 1: SNMPTN 2010
Nilai \( \displaystyle \lim_{x \to 0} \ \frac{\sqrt{4x}}{\sqrt{\sin 2x}} = \cdots \)
- \( \sqrt{2} \)
- 1
- 1/2
- 1/4
- 0
\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:
Nilai \( \displaystyle \lim_{x\to 0} \ \frac{\sqrt{1-\cos x}}{x} = \cdots \)
- \( -\sqrt{2} \)
- \( -\frac{1}{2} \sqrt{2} \)
- 0
- \( \frac{1}{2}\sqrt{2} \)
- \( \sqrt{2} \)
\begin{aligned} \lim_{x\to 0} \ \frac{\sqrt{1-\cos x}}{x} &= \lim_{x\to 0} \ \frac{\sqrt{2\sin^2 \frac{1}{2}x}}{x} = \lim_{x\to 0} \ \sqrt{ \frac{2\sin^2 \frac{1}{2}x}{x^2} } \\[8pt] &= \sqrt{2} \ \lim_{x\to 0} \ \sqrt{ \frac{\sin^2 \frac{1}{2}x}{x^2} } \\[8pt] &= \sqrt{2} \cdot \sqrt{ \lim_{x\to 0} \ \frac{\sin^2 \frac{1}{2}x}{x^2} } \\[8pt] &= \sqrt{2} \cdot \sqrt{ \left(\frac{1}{2}\right)^2 } = \sqrt{2} \cdot \frac{1}{2} \\[8pt] &= \frac{1}{2}\sqrt{2} \end{aligned}
Jawaban D.
Contoh 3:
Nilai \( \displaystyle \lim_{x \to 0} \ \frac{1-\sqrt{\cos x}}{x^2} = \cdots \)
Pembahasan »\begin{aligned} \lim_{x \to 0} \ \frac{1-\sqrt{\cos x}}{x^2} &= \lim_{x \to 0} \ \frac{1-\sqrt{\cos x}}{x^2} \cdot \frac{1+\sqrt{\cos x}}{1+\sqrt{\cos x}} \\[8pt] &= \lim_{x \to 0} \ \frac{1 - \cos x}{x^2 (1+\sqrt{\cos x})} = \lim_{x \to 0} \ \frac{2 \sin^2 \frac{1}{2}x}{x^2 (1+\sqrt{\cos x})} \\[8pt] &= \lim_{x \to 0} \ \left( \frac{2}{1+\sqrt{\cos x}} \cdot \frac{\sin \frac{1}{2}x}{x} \cdot \frac{\sin \frac{1}{2}x}{x} \right) \\[8pt] &= \frac{2}{1+\sqrt{\cos 0}} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{2}{1+\sqrt{1}} \cdot \frac{1}{4} \\[8pt] &= \frac{1}{4} \end{aligned}
Contoh 4:
Nilai \( \displaystyle \lim_{x\to \pi} \ \frac{\sqrt{5+\cos x}-2}{(\pi-x)^2} = \cdots \)
- 1/10
- 1/8
- 1/3
- 1/2
- 1
\begin{aligned} \lim_{x\to \pi} \ \frac{\sqrt{5+\cos x}-2}{(\pi-x)^2} &= \lim_{x\to \pi} \ \frac{\sqrt{5+\cos x}-2}{(\pi-x)^2} \times \frac{\sqrt{5+\cos x}+2}{\sqrt{5+\cos x}+2} \\[8pt] &= \lim_{x\to \pi} \ \frac{(5+\cos x)-4}{(\pi-x)^2 (\sqrt{5+\cos x}+2)} \\[8pt] &= \lim_{x\to \pi} \ \frac{1 + \cos x}{(\pi-x)^2 (\sqrt{5+\cos x}+2)} \times \frac{1 - \cos x}{1 - \cos x} \\[8pt] &= \lim_{x\to \pi} \ \frac{1 - \cos^2 x}{(\pi-x)^2 (\sqrt{5+\cos x}+2)(1 - \cos x)} \\[8pt] &= \lim_{x\to \pi} \ \frac{\sin^2 x}{(\pi-x)^2 (\sqrt{5+\cos x}+2)(1 - \cos x)} \\[8pt] &= \lim_{x\to \pi} \ \frac{\sin^2 (\pi-x)}{(\pi-x)^2 (\sqrt{5+\cos x}+2)(1 - \cos x)} \\[8pt] &= \lim_{x\to \pi} \ \frac{\sin^2 (\pi-x)}{(\pi-x)^2} \cdot \lim_{x\to \pi} \ \frac{1}{(\sqrt{5+\cos x}+2)(1 - \cos x)} \\[8pt] &= 1 \cdot \frac{1}{(\sqrt{5+\cos \pi}+2)(1 - \cos \pi)} = \frac{1}{(\sqrt{5+(-1)}+2)(1 - (-1))} \\[8pt] &= \frac{1}{(\sqrt{4}+2)(2)} = \frac{1}{8} \end{aligned}
Jawaban B.
Contoh 5: SBMPTN 2013
Nilai \( \displaystyle \lim_{x \to 0} \ \sqrt{\frac{x \tan x}{\sin^2 x - \cos 2x + 1}} = \cdots \)
- \( 3 \)
- \( \sqrt{3} \)
- \( \frac{\sqrt{3}}{3} \)
- \( \frac{1}{3} \)
- \( \frac{\sqrt{3}}{2} \)
\begin{aligned} \lim_{x \to 0} \ \sqrt{\frac{x \tan x}{\sin^2 x - \cos 2x + 1}} &= \lim_{x \to 0} \ \sqrt{\frac{x \tan x}{\sin^2 x + 1 - \cos 2x}} \\[8pt] &= \lim_{x \to 0} \ \sqrt{\frac{x \tan x}{\sin^2 x + 2 \sin^2 x}} \\[8pt] &= \lim_{x \to 0} \ \sqrt{\frac{x \tan x}{3 \sin^2 x}} = \sqrt{\lim_{x \to 0} \ \frac{x \tan x}{3 \sin^2 x}} \\[8pt] &= \sqrt{\frac{1}{3} \cdot \lim_{x \to 0} \ \frac{x}{\sin x} \cdot \lim_{x \to 0} \ \frac{\tan x}{\sin x}} \\[8pt] &= \sqrt{\frac{1}{3} \cdot 1 \cdot 1} = \sqrt{\frac{1}{3}} = \frac{1}{3} \sqrt{3} \end{aligned}
Jawaban C.
Contoh 6: UM UGM 2019
Nilai \( \displaystyle \lim_{x \to 0} \ \frac{\sqrt{1-\cos 4x^2}}{1-\cos 2x} = \cdots \)
- 1
- \( \sqrt{2} \)
- 2
- \( 2 \sqrt{2} \)
- 4
\begin{aligned} \lim_{x \to 0} \ \frac{\sqrt{1-\cos 4x^2}}{1-\cos 2x} &= \lim_{x \to 0} \ \frac{\sqrt{1-\cos (2 \cdot 2x^2)}}{2\sin^2 x} \\[8pt] &= \lim_{x \to 0} \ \frac{\sqrt{2\sin^2 2x^2}}{2\sin^2 x} \\[8pt] &= \lim_{x \to 0} \ \frac{\sqrt{2} \ \sin 2x^2}{2\sin^2 x} \cdot \frac{x^2}{x^2} \\[8pt] &= \frac{\sqrt{2}}{2} \cdot \lim_{x \to 0} \ \frac{\sin 2x^2}{x^2} \cdot \lim_{x \to 0} \ \frac{x^2}{\sin^2 x} \\[8pt] &= \frac{\sqrt{2}}{2} \cdot 2 \cdot (1)^2 = \sqrt{2} \end{aligned}
Jawaban C.
Contoh 7:
Nilai \( \displaystyle \lim_{x \to 0} \ \frac{x}{2 \csc x \ (1-\sqrt{\cos x})} = \cdots \)
- -2
- -1
- 0
- 1
- 2
\begin{aligned} \lim_{x \to 0} \ \frac{x}{2 \csc x \ (1-\sqrt{\cos x})} &= \lim_{x \to 0} \ \frac{x}{2 \csc x \ (1-\sqrt{\cos x})} \cdot \frac{1+\sqrt{\cos x}}{1+\sqrt{\cos x}} \\[8pt] &= \lim_{x \to 0} \ \frac{x \ (1+\sqrt{\cos x})}{\frac{2}{\sin x} \ (1-\cos x)} \\[8pt] &= \lim_{x \to 0} \ \frac{x \sin x \ (1+\sqrt{\cos x})}{2 \ (2 \sin^2 \frac{1}{2}x)} \\[8pt] &= \lim_{x \to 0} \ \frac{x}{4 \sin \frac{1}{2}x} \cdot \lim_{x \to 0} \ \frac{\sin x}{\sin \frac{1}{2}x} \cdot \lim_{x \to 0} \ (1+\sqrt{\cos x}) \\[8pt] &= \frac{1}{4 \cdot \frac{1}{2}} \cdot \frac{1}{\frac{1}{2}} \cdot (1+\sqrt{\cos 0}) \\[8pt] &= \frac{1}{2} \cdot 2 \cdot (1+1) = 2 \end{aligned}
Jawaban E.
Contoh 8: SBMPTN 2016
Nilai \( \displaystyle \lim_{x \to 0} \ \frac{\sqrt{x^2+1}-1}{\sqrt{3x^5+4\sin^4 x}} = \cdots \)
- \( 0 \)
- \( \frac{1}{4} \)
- \( \frac{1}{\sqrt{7}} \)
- \( \frac{1}{2} \)
- \( \frac{1}{\sqrt{3}} \)
\begin{aligned} \lim_{x \to 0} \ \frac{\sqrt{x^2+1}-1}{\sqrt{3x^5+4\sin^4 x}} &= \lim_{x \to 0} \ \frac{\sqrt{x^2+1}-1}{\sqrt{3x^5+4\sin^4 x}} \cdot \frac{\sqrt{x^2+1}+1}{\sqrt{x^2+1}+1} \\[8pt] &= \lim_{x \to 0} \ \frac{(x^2+1)-1}{(\sqrt{3x^5+4\sin^4 x})(\sqrt{x^2+1}+1)} \cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}} \\[8pt] &= \lim_{x \to 0} \ \frac{x^2 \cdot \frac{1}{x^2}}{\left(\sqrt{ \frac{3x^5}{x^4} + 4 \ \frac{\sin^4 x}{x^4} }\right)(\sqrt{x^2+1}+1)} \\[8pt] &= \frac{1}{\left( \sqrt{3(0) + 4 \cdot (1)^4}\right)(\sqrt{0^2+1}+1)} \\[8pt] &= \frac{1}{(\sqrt{4})(\sqrt{1}+1)} = \frac{1}{(2)(2)} = \frac{1}{4} \end{aligned}
Jawaban B.
Contoh 9: SIMAK UI 2013
Nilai \( \displaystyle \lim_{x \to 0} \ \frac{\sqrt{1+\tan x} - \sqrt{1+\sin x}}{x^3} = \cdots \)
- -1
- \( -\frac{1}{4} \)
- 0
- 1/4
- 1
\begin{aligned} \lim_{x \to 0} \ \frac{\sqrt{1+\tan x} - \sqrt{1+\sin x}}{x^3} &= \lim_{x \to 0} \ \frac{\sqrt{1+\tan x} - \sqrt{1+\sin x}}{x^3} \cdot \frac{\sqrt{1+\tan x} + \sqrt{1+\sin x}}{\sqrt{1+\tan x} + \sqrt{1+\sin x}} \\[8pt] &= \lim_{x \to 0} \ \frac{(1+\tan x) - (1+\sin x)}{x^3 \ (\sqrt{1+\tan x} + \sqrt{1+\sin x})} \\[8pt] &= \lim_{x \to 0} \ \frac{\tan x - \sin x}{x^3 \ (\sqrt{1+\tan x} + \sqrt{1+\sin x})} \\[8pt] &= \lim_{x \to 0} \ \frac{\tan x - \sin x}{x^3} \cdot \lim_{x \to 0} \ \frac{1}{(\sqrt{1+\tan x} + \sqrt{1+\sin x})} \\[8pt] &= \lim_{x \to 0} \ \frac{\frac{\sin x}{\cos x} - \sin x}{x^3} \cdot \frac{1}{(\sqrt{1+\tan 0} + \sqrt{1+\sin 0})} \\[8pt] &= \lim_{x \to 0} \ \frac{\sin x-\sin x \cos x}{x^3 \cos x} \cdot \frac{1}{\sqrt{1} + \sqrt{1}} \\[8pt] &= \lim_{x \to 0} \ \frac{\sin x (1-\cos x)}{x^3 \cos x} \cdot \frac{1}{2} \\[8pt] &= \frac{1}{2} \cdot \lim_{x \to 0} \ \frac{\sin x \ (2 \sin^2 \frac{1}{2}x)}{x^3 \cos x} \\[8pt] &= \frac{1}{2} \cdot \lim_{x \to 0} \ \frac{2\sin x}{x} \cdot \lim_{x \to 0} \ \frac{\sin^2 \frac{1}{2}x}{x^2} \cdot \lim_{x \to 0} \ \frac{1}{\cos x} \\[8pt] &= \frac{1}{2} \cdot 2 \cdot \left(\frac{1}{2}\right)^2 \cdot \frac{1}{\cos 0} = \frac{1}{4} \end{aligned}
Jawaban D.
Contoh 10: SBMPTN 2016
Nilai \( \displaystyle \lim_{x \to 0} \ \frac{x^3}{\sqrt{1+\sin x} - \sqrt{1+\tan x}} = \cdots \)
- -4
- -2
- 0
- 2
- 4
\begin{aligned} \lim_{x \to 0} \ \frac{x^3}{\sqrt{1+\sin x} - \sqrt{1+\tan x}} &= \lim_{x \to 0} \ \frac{x^3}{\sqrt{1+\sin x} - \sqrt{1+\tan x}} \cdot \frac{\sqrt{1+\sin x} + \sqrt{1+\tan x}}{\sqrt{1+\sin x} + \sqrt{1+\tan x}} \\[8pt] &= \lim_{x \to 0} \ \frac{x^3 \ (\sqrt{1+\sin x} + \sqrt{1+\tan x})}{(1+\sin x) - (1+\tan x)} \\[8pt] &= \lim_{x \to 0} \ \frac{x^3 \ (\sqrt{1+\sin x} + \sqrt{1+\tan x})}{\sin x - \tan x} \\[8pt] &= \lim_{x \to 0} \ \frac{x^3 \ (\sqrt{1+\sin x} + \sqrt{1+\tan x})}{\sin x \ (1-\frac{1}{\cos x})} \\[8pt] &= \lim_{x \to 0} \ \frac{x^3 \ (\sqrt{1+\sin x} + \sqrt{1+\tan x})}{\sin x \ (\frac{\cos x - 1}{\cos x})} \\[8pt] &= \lim_{x \to 0} \ \frac{x^3 \cos x \ (\sqrt{1+\sin x} + \sqrt{1+\tan x})}{\sin x \ (\cos x - 1)} \\[8pt] &= \lim_{x \to 0} \ \frac{x^3 \cos x \ (\sqrt{1+\sin x} + \sqrt{1+\tan x})}{\sin x \ (-2 \sin^2 \frac{1}{2}x)} \\[8pt] &= -\frac{1}{2} \cdot \lim_{x \to 0} \ \frac{x}{\sin x} \cdot \lim_{x \to 0} \ \frac{x^2}{\sin^2 \frac{1}{2}x} \cdot \lim_{x \to 0} \ \cos x \ (\sqrt{1+\sin x} + \sqrt{1+\tan x}) \\[8pt] &= -\frac{1}{2} \cdot 1 \cdot \left( \frac{1}{\frac{1}{2}} \right)^2 \cdot \left(\cos 0 \ (\sqrt{1+\sin 0} + \sqrt{1+\tan 0})\right) \\[8pt] &= -\frac{1}{2} \cdot 1 \cdot (2)^2 \cdot \left(1 \ (\sqrt{1+0} + \sqrt{1+0})\right) \\[8pt] &= -2 \cdot (\sqrt{1} + \sqrt{1}) = -4 \end{aligned}
Jawaban A.
Contoh 11: SBMPTN 2018
Nilai \( \displaystyle \lim_{x \to 0} \ \frac{\sin x \cos x}{\sqrt{\pi+2\sin x} - \sqrt{\pi}} = \cdots \)
- \( -2 \sqrt{\pi} \)
- \( -\sqrt{\pi} \)
- 0
- \( \sqrt{\pi} \)
- \( 2 \sqrt{\pi} \)
\begin{aligned} \lim_{x \to 0} \ \frac{\sin x \cos x}{\sqrt{\pi+2\sin x} - \sqrt{\pi}} &= \lim_{x \to 0} \ \frac{\sin x \cos x}{\sqrt{\pi+2\sin x} - \sqrt{\pi}} \times \frac{\sqrt{\pi+2\sin x} + \sqrt{\pi}}{\sqrt{\pi+2\sin x} + \sqrt{\pi}} \\[8pt] &= \lim_{x \to 0} \ \frac{(\sqrt{\pi+2\sin x} + \sqrt{\pi}) \sin x \cos x}{(\pi+2\sin x) - \pi} \\[8pt] &= \lim_{x \to 0} \ \frac{(\sqrt{\pi+2\sin x} + \sqrt{\pi}) \sin x \cos x}{2\sin x} \\[8pt] &= \frac{1}{2} \cdot \lim_{x \to 0} \ (\sqrt{\pi+2\sin x} + \sqrt{\pi}) \cos x \\[8pt] &= \frac{1}{2} \cdot (\sqrt{\pi+2\sin 0} + \sqrt{\pi}) \cos 0 \\[8pt] &= \frac{1}{2} \cdot (\sqrt{\pi} + \sqrt{\pi}) \cdot 1 \\[8pt] &= \sqrt{\pi} \end{aligned}
Jawaban D.
Contoh 12: SBMPTN 2018
Nilai \( \displaystyle \lim_{x \to 0} \ \frac{\sin x}{\sqrt{\pi+\tan x} - \sqrt{\pi-\tan x}} = \cdots \)
- \( -2 \sqrt{\pi} \)
- \( -\sqrt{\pi} \)
- 0
- \( \sqrt{\pi} \)
- \( 2 \sqrt{\pi} \)
\begin{aligned} \lim_{x \to 0} \ \frac{\sin x}{\sqrt{\pi+\tan x} - \sqrt{\pi-\tan x}} &= \lim_{x \to 0} \ \frac{\sin x}{\sqrt{\pi+\tan x} - \sqrt{\pi-\tan x}} \times \frac{\sqrt{\pi+\tan x} + \sqrt{\pi-\tan x}}{\sqrt{\pi+\tan x} + \sqrt{\pi-\tan x}} \\[8pt] &= \lim_{x \to 0} \ \frac{(\sqrt{\pi+\tan x} + \sqrt{\pi-\tan x}) \ \sin x}{(\pi+\tan x) - (\pi-\tan x)} \\[8pt] &= \lim_{x \to 0} \ \frac{(\sqrt{\pi+\tan x} + \sqrt{\pi-\tan x}) \ \sin x}{2 \tan x} \\[8pt] &= \lim_{x \to 0} \ \frac{\sin x}{2 \tan x} \cdot \lim_{x \to 0} \ (\sqrt{\pi+\tan x} + \sqrt{\pi-\tan x}) \\[8pt] &= \frac{1}{2} \cdot (\sqrt{\pi+\tan 0} + \sqrt{\pi-\tan 0}) \\[8pt] &= \frac{1}{2} \ (\sqrt{\pi} + \sqrt{\pi}) = \sqrt{\pi} \end{aligned}
Jawaban D.
Contoh 13:
Nilai \( \displaystyle \lim_{x\to 0} \ \frac{\sqrt{x}-\sqrt{\sin x}}{x^2\sqrt{x}} = \cdots \)
- 12
- 6
- 1/3
- 1/12
- 2/3
\begin{aligned} &\lim_{x\to 0} \ \frac{x-\sin x}{x^3} = \frac{1}{6} \\[8pt] &\lim_{x\to 0} \ \frac{(2x)-\sin (2x)}{(2x)^3} = \frac{1}{6} \\[8pt] &\lim_{x\to 0} \ \frac{(3x)-\sin (3x)}{(3x)^3} = \frac{1}{6} \\[8pt] \end{aligned}
\begin{aligned} \lim_{x\to 0} \ \frac{\sqrt{x}-\sqrt{\sin x}}{x^2\sqrt{x}} &= \lim_{x\to 0} \ \frac{\sqrt{x}-\sqrt{\sin x}}{x^2\sqrt{x}} \times \frac{\sqrt{x}+\sqrt{\sin x}}{\sqrt{x}+\sqrt{\sin x}} \\[8pt] &= \lim_{x\to 0} \ \frac{x-\sin x}{x^2\sqrt{x} \ (\sqrt{x}+\sqrt{\sin x})} \\[8pt] &= \lim_{x\to 0} \ \frac{x-\sin x}{x^2\sqrt{x} \cdot \sqrt{x} \ \left(1+\sqrt{ \frac{\sin x}{x}}\right)} \\[8pt] &= \lim_{x\to 0} \ \frac{x-\sin x}{x^3 \ \left(1+\sqrt{ \frac{\sin x}{x}}\right)} \\[8pt] &= \lim_{x\to 0} \ \frac{x-\sin x}{x^3} \cdot \lim_{x\to 0} \ \frac{1}{\left(1+\sqrt{ \frac{\sin x}{x}}\right)} \\[8pt] &= \frac{1}{6} \cdot \frac{1}{\left(1+\sqrt{ \displaystyle \lim_{x\to 0} \ \frac{\sin x}{x}}\right)} \\[8pt] &= \frac{1}{6} \cdot \frac{1}{1+1} = \frac{1}{12} \end{aligned}
Jawaban D.
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