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Selamat datang di halaman contoh soal dan pembahasan "Limit". Di halaman ini akan membahas tentang contoh soal dan pembahasan lengkap mengenai Limit yang berupa pecahan.
Untuk lebih jelasnya mari kita lihat contoh soal dan pembahasan dibawah ini!
1. Tentukan hasil dari soal limit berikut \(\lim\limits_{x\rightarrow 0}\frac{3x}{\sin 4x}\)=
a. \(\frac{1}{8}\)
b. \(\frac{3}{4}\)
c. \(\frac{2}{9}\)
d. \(\frac{3}{15}\)
e. \(\frac{1}{2}\)
Pembahasan :
\( \lim\limits_{x\rightarrow 0}\frac{3}{4\cos 4 (0)}\)
=\(\lim\limits_{x\rightarrow 0}\frac{3}{4\cos 0}\)
=\(\frac{3}{4(1)}\)
=\(\frac{3}{4}\)
2. Tentukan hasil soal limit berikut \(\lim\limits_{x\rightarrow 0}\left ( \frac{\sin 3x-\sin 2x}{4x} \right )\)=
a. \(\frac{1}{3}\)
b. \(\frac{1}{5}\)
c. \(\frac{2}{8}\)
d. \(\frac{3}{7}\)
e. \(\frac{1}{4}\)
Pembahasan :
\(\lim\limits_{x\rightarrow 0}\left ( \frac{\sin 3x-\sin 2x}{4x} \right )\)
=\(\lim\limits_{x\rightarrow 0}\left ( \frac{\sin 3x}{4x}- \frac{\sin 2x}{4x} \right )\)
=\(\frac{3}{4}- \frac{2}{4}\)
=\(\frac{1}{4}\)
3. Tentukan nilai dari \(\lim\limits_{x\rightarrow 0}\frac{2\sin 2x}{3\tan 3x}\)=
a. \(\frac{1}{12}\)
b. \(\frac{4}{9}\)
c. \(\frac{2}{3}\)
d. \(\frac{3}{15}\)
e. \(\frac{5}{6}\)
Pembahasan :
perhatikan rumus limit berikut :
\(\lim\limits_{x\rightarrow 0}\frac{\tan ax}{\tan bx}=
\frac{a}{b}\)
\(\lim\limits_{x\rightarrow 0}\frac{\sin ax}{\sin bx}=
\frac{a}{b}\)
\(\lim\limits_{x\rightarrow 0}\frac{\sin ax}{\tan bx}=
\frac{a}{b}\)
\(\lim\limits_{x\rightarrow 0}\frac{\tan ax}{\sin bx}=
\frac{a}{b}\)
diperoleh
\(\lim\limits_{x\rightarrow 0}\frac{2\sin 2x}{3\tan 3x}\)
=\(\frac{2.2}{3.3}\)
=\(\frac{4}{9}\)
4. Tenetukan hasil dari soal limit berikut \(\lim\limits_{x\rightarrow 0}\frac{1-\cos 2x}{2x\sin 2x}\)=
a. \(\frac{1}{2}\)
b. \(\frac{3}{2}\)
c. \(\frac{7}{6}\)
d. \(\frac{3}{4}\)
e. \(\frac{1}{3}\)
Pembahasan :
identitas trigonometri
\({\cos 2x}=
{1-2\sin^2x}\)
atau
\({1-\cos 2x}=
{2\sin^2x}\)
\(\lim\limits_{x\rightarrow 0}\frac{1 -\cos 2x}{2x\sin 2x}\)
=\(\lim\limits_{x\rightarrow 0}\frac{2\sin^2x}{2x\sin 2x}\)
=\(\lim\limits_{x\rightarrow 0}\frac{sin\times sinx}{x\sin 2x}\)
=\(\lim\limits_{x\rightarrow 0}\left ( \frac{sin x}{x} \right )\left ( \frac{\sin x}{\sin 2x} \right )\)
=\({1}. \frac{1}{2}\)
=\(\frac{1}{2}\)
5. Nilai dari \(\lim\limits_{x\rightarrow 0}\frac{1-\cos 2x}{x \tan 2x}\)=
a. 3
b. 5
c. 1
d. 2
e. 0
Pembahasan :
ubah \({1 -\cos 2x}\) menjadi \({2\sin^2x}\)
\(\lim\limits_{x\rightarrow 0}\frac{1-\cos 2x}{x\tan 2x}\)
=\(\lim\limits_{x\rightarrow 0}\frac{2\sin^2x}{x\tan 2x}\)
=\(\left ( \frac{2\sin x}{x} \right )\left ( \frac{\sin x}{\tan 2x} \right )\)
=\(\frac{2.1.1}{1.2}=1\)
6. \(\lim\limits_{x\rightarrow 0}\frac{6x\tan 2x}{1 -\cos 6x}\)=
a. \(\frac{1}{2}\)
b. \(\frac{2}{3}\)
c. \(\frac{2}{5}\)
d. \(\frac{5}{3}\)
e. \(\frac{3}{4}\)
Pembahasan :
\(\lim\limits_{x\rightarrow 0}\frac{6x\tan 2x}{1-\cos 6x}\)
=\(\lim\limits_{x\rightarrow 0}\frac{6x\tan 2x}{2\sin^23x}\)
=\(\lim\limits_{x\rightarrow 0}\left ( \frac{6x}{2\sin 3x} \right )\left ( \frac{\tan 2x}{\sin 3x} \right )\)
=\(\frac{6.2}{2.3.3}\)
=\(\frac{2}{3}\)
7. Nilai dari \(\lim\limits_{x\rightarrow 2}\frac{(x - 2)\cos (\pi x-2\pi )}{\tan2\pi x - 4\pi }\)=
a. \(2\pi\)
b. \(\pi\)
c. 0
d. \(\frac{1}{\pi}\)
e. \(\frac{1}{2\pi}\)
Pembahasan :
misalkan :
\({x-2}={y}\)
\(\lim\limits_{x\rightarrow 2}\frac{(x-2)\cos (\pi x-2\pi )}{\tan(2\pi x-4\pi )}\)
=\(\lim\limits_{(x-2)\rightarrow 0}\frac{(x-2)\cos \pi (x-2)}{\tan 2\pi(x-2) }\)
=\(\lim\limits_{y\rightarrow 0}\frac{y\cos \pi y}{\tan 2\pi y}\)
=\(\lim\limits_{y\rightarrow 0}\frac{y\sin \pi y}{\tan\pi y.\tan 2\pi y}\)
=\(\frac{y}{2\pi y}\)
=\(\frac{1}{2\pi }\)
8. Tentukan nilai dari \(\lim\limits_{x\rightarrow \frac{\pi }{4}}\frac{\cos 2x}{\sin x-\cos x}\)=
a. \({-\sqrt 2}\)
b. \({\sqrt 16}\)
c. \({\sqrt 76}\)
d. \({\sqrt 13}\)
e. \({\sqrt 81}\)
Pembahasan :
\(\lim\limits_{x\rightarrow \frac{\pi }{4}}\frac{\cos 2x}{\sin x - \cos x}\)
=\(\lim\limits_{x\rightarrow \frac{\pi }{4}}\frac{\cos^2x-\sin^2x}{\sin x-\cos x}\)
=\(\lim\limits_{x\rightarrow \frac{\pi }{4}}\frac{(\cos x-\sin x)(\cos x+\sin x)}{(\sin x-\cos x)}\)
=\(\lim\limits_{x\rightarrow \frac{\pi }{4}}-(\cos x+\sin x)\)
=\(-(\cos \frac{\pi }{4}+\sin \frac{\pi}{4})\)
=\(-(\frac{1}{2}\sqrt 2+\frac{1}{2}\sqrt 2)\)
=\(-{\sqrt 2}\)
9. Nilai dari \(\lim\limits_{x\rightarrow 1}\frac{(x^2-1)\tan (2x-2)}{\sin^2 (x-1)}\)=
a. 6
b. 5
c. 4
d. 2
e. 0
Pembahasan :
\(\lim\limits_{x\rightarrow 1}\frac{(x^2-1).\tan (2x-2)}{\sin^2(x+1)}\)
=\(\lim\limits_{x\rightarrow 1}\frac{(x-1)(x+1).\tan 2(x-1)}{\sin (x-1)\sin (x-1)}\)
=\(\lim\limits_{x\rightarrow 1}(x+1).2=(1+1).2=(2.2)=4\)
10. \(\lim\limits_{x\rightarrow 0}\frac{\sin^2 + \tan^2 x}{\frac{1}{2}x^2}\)=
a. 1
b. 3
c. 4
d. 5
e. 8
Pembahasan :
\(\lim\limits_{x\rightarrow 0}\frac{\sin^2x+\tan^2x}{\frac{1}{2}x^2}\)
=\(\lim\limits_{x\rightarrow 0}\frac{\sin^2x}{\frac{1}{2}x^2}+\lim\limits_{x\rightarrow 0} \frac{\tan^2x}{\frac{1}{2}x^2}\)
=\(\frac{1}{\frac{1}{2}}\lim\limits_{x\rightarrow 0}\frac{\sin x}{x}.\frac{\sin x}{x}+\frac{1}{\frac{1}{2}}\lim\limits_{x\rightarrow 0}\frac{\tan x}{x}.\frac{\tan x}{x}\)
=\({2}.\lim\limits_{x\rightarrow 0}\frac{\sin x}{x}.\lim\limits_{x\rightarrow 0}\frac{\sin x}{x}+2.\lim\limits_{x\rightarrow 0}\frac{\tan x}{x}.\lim\limits_{x\rightarrow 0}\frac{\tan x}{x}\)
=\({2.1.1 + 2.1.1}={4}\)
