Selamat datang di Pembahasan Soal Matematika Pokok Bahasan " Soal Limit Trigonometri Part 14" untuk SMA/SMK. Kami akan membahas contoh soal dan pembahasan lengkap mengenai "Limit Trigonometri". Untuk lebih jelasnya mari kita lihat pembahasan soal dibawah ini!
1. Tentukan hasil dari \(\lim\limits_{x\rightarrow 0}\frac{tan\ 5x}{3x}\) = ...
a. \(\frac{5}{3}\)
b. \(\frac{5}{2}\)
c. \(\frac{3}{5}\)
d. \(\frac{2}{3}\)
e. \(\frac{6}{3}\)
Pembahasan :
\(\lim\limits_{x\rightarrow 0}\frac{tan\ 5x}{3x}\)
= \(\frac{tan\ 5x}{3x}\times\frac{5x}{3x}\)
= \(\frac{tan\ 5x}{3x}\times\frac{5}{3}\)
= \(1\times\frac{5}{3}\)
= \(\frac{5}{3}\)
2. Nilai dari \(\lim\limits_{x\rightarrow 0}\frac{sin\ 8x + tan\ 4x - sin\ 2x}{2x}\) = ...
a. 6
b. 7
c. 5
d. 4
e. 3
Pembahasan :
\(\lim\limits_{x\rightarrow 0}\frac{sin\ 8x}{2x} + \lim\limits_{x\rightarrow 0}\frac{sin\ 4x}{2x}-\frac{sin\ 2x}{2x}\)
= \({4\times(\lim\limits_{x\rightarrow 0}\frac{sin\ 8x}{8x})+ 2\times(\lim\limits_{x\rightarrow 0}\frac{sin\ 4x}{4x})-1\times(\lim\limits_{x\rightarrow 0}\frac{sin2x}{2x})}\)
= \((4\times 1)+(2\times 1)-(1\times 1)\)
= \(5\)
3.\(\lim\limits_{x\rightarrow 0}\frac{sin\ 24x + tan\ 12x + sin\ 6x}{6x}\) = ...
a. 5
b. 6
c. 9
d. 10
e. 7
Pembahasan :
\(\lim\limits_{x\rightarrow 0}\frac{sin\ 24x}{6x} + \lim\limits_{x\rightarrow 0}\frac{tan\ 12x}{6x}+\frac{sin\ 6x}{6x}\)
=\({4\times(\lim\limits_{x\rightarrow 0}\frac{sin\ 24x}{24x})+ 2\times(\lim\limits_{x\rightarrow 0}\frac{sin\ 12x}{12x})+1\times(\lim\limits_{x\rightarrow 0}\frac{sin\ 6x}{6x})}\)
=\((4\times 1)+(2\times 1)+(1\times 1)\)
=\(7\)
4. \(\lim\limits_{x\rightarrow 0}\frac{cot\ 6x}{cot\ 56x}\) = ...
a. 9
b. 6
c. 4
d. 2
e. 3
Pembahasan :
\(\lim\limits_{x\rightarrow 0}\frac{cot\ 6x}{cot\ 56x}\)
= \(\frac{1}{tan 6x}\times \frac{tan 56}{1}\)
= \(\frac{6x}{tan 6x}\times \frac{tan 56}{56}\times \frac{56}{6}\)
= \(9\times 1\times 1= 9\)
5. \(\lim\limits_{x\rightarrow 0}\frac{x\ sin\ x}{1-(1-2sin^2\frac{3}{2}x)}\) = ...
a. \(\frac{2}{10}\)
b. \(\frac{2}{9}\)
c. \(\frac{2}{11}\)
d. \(\frac{2}{19}\)
e. \(\frac{1}{4}\)
Pembahasan :
\(\lim\limits_{x\rightarrow 0}\frac{x\ sinx}{1-(1-2sin^2\frac{3}{2}x)}\)
=\( \frac{sin 3x}{4x}-\frac{sin 2x}{4x}\)
=\(\frac{3}{4}-\frac{2}{4}=\frac{1}{4}\)
6. \(\lim\limits_{x\rightarrow 1}\frac{(x^2-1)\times tan\ (2x-2)}{sin^2\ (x-1)}\) = ...
a. 4
b. 6
c. 7
d. 5
e. 2
Pembahasan :
\(\lim\limits_{x\rightarrow 1}\frac{(x^2-1)\times tan\ (2x-2)}{sin^2\ (x-1)}\)
= \(\frac{\left ( x-1 \right )\left ( x+1 \right )tan 2\left ( x-1 \right )}{sin\left ( x-1 \right )sin\left ( x-1 \right )}\)
= \(\left ( x+1 \right )\times 2=4\)
7. \(\lim\limits_{x\rightarrow 0}\frac{(cos\ 2\pi )-1}{x\ tan\ 2x}\) = ...
a. 4
b.-4
c.-6
d.-2
e. 6
Pembahasan :
\(\lim\limits_{x\rightarrow 0}\frac{(cos\ 2\pi )-1}{x\ tan\ 2x}\)
=\(\frac{-2sin^{2}x}{x tan 2x}\)
=\(\frac{-2sin 2x}{x}\times \frac{sin 2x}{tan 2x}=\frac{-2\times 2\times 2}{2}\)
=\(-4\)
8. \(\lim\limits_{x\rightarrow 2}\frac{(x-2)\ cos(\pi x-2x)}{tan (2\pi x-4\pi)}\) = ...
a. \(\frac{1}{3\pi}\)
b. \(\frac{1}{6\pi}\)
c. \(\frac{1}{5\pi}\)
d. \(\frac{1}{2\pi}\)
e. \(\frac{1}{4\pi}\)
Pembahasan :
misal \(x-2\)=\(y\)
\(\lim\limits_{x\rightarrow 2}\frac{(x-2)\ cos(\pi x-2x)}{tan (2\pi x-4\pi)}\)
=\(\frac{y cos \Pi y}{tan 2\Pi y}\)
=\(\frac{y sin \Pi y}{tan \Pi y\times tan 2\Pi y}\)
=\(\frac{y}{2\Pi y}=\frac{1}{2\Pi }\)
9. \(\lim\limits_{x\rightarrow 0}\frac{x\ tan\ 3x}{sin^{2}6x}\) = ...
a. \(\frac{1}{12}\)
b. \(\frac{1}{2}\)
c. \(\frac{1}{10}\)
d. \(\frac{1}{14}\)
e. \(\frac{1}{16}\)
Pembahasan :
\(\lim\limits_{x\rightarrow 0}\frac{x\ tan\ 3x}{sin^{2}6x}\)
=\(\frac{x tan 3x}{sin 6x\times sin 6x}\)
=\(\frac{x}{sin x}\times \frac{tan 3x}{sin 6x}\)
=\(\frac{1\times 3}{6\times 6}=\frac{1}{12}\)
10. \(\lim\limits_{x\rightarrow \frac{1}{8\pi}}\frac{sin^{2}2x-cos^{2}2x}{sin\ 2x-cos\ 2x}\) = ...
a. \(\sqrt{3}\)
b. \(\sqrt{\frac{1}{2}}\)
c. \(\sqrt{2}\)
d. \(\frac{\sqrt{1}}{2}\)
e. \(sqrt{4}\)
Pembahasan :
\(\lim\limits_{x\rightarrow \frac{1}{8\pi}}\frac{sin^{2}2x-cos^{2}2x}{sin\ 2x-cos\ 2x}\)
= \(sin 45^{\circ}+cos 45 ^{\circ}\)
= \(\frac{1}{2\sqrt2}+ \frac{1}{2\sqrt2}\)
= \(\sqrt2\)
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