Soal, Materi Limit di Tak Hingga Fungsi Trigonometri Mirip Soal UTBK SBMPTN

Jibang P.Hutagaol,M.Pd October 03, 2020

Salam Para Bintang

Kali ini saya akan coba berbagi tentang materi matematika yaitu Limit Trigonometri di Tak Hingga. Sebelum kalian mempelajari materi ini, baiknya kalian harus menguasai materi Limit Trigonometri karena materi ini meskipun dari judulnya adalah Limit Trigonometri di Tak Hingga tetapi pelajaran ini menitikberatkan kepada materi Limit Trigonometri. Jadi, kita wajib paham apa itu dan bagaimana penggunaan Rumus-rumus, konsep Limit Trigonometri.

Baca:
Pengertian, Rumus Dasar , Contoh Soal Limit Fungsi Trigonometri pada Matematika Minat

Kita ketahui bahwa untuk bentuk sin x, cos x, tan x, sec x, cosec x dan cotan x, jika x bernilai tak hingga maka akan diperoleh bentuk:

$sin \, \, \infty ,cos\, \, \infty ,tan\, \, \infty ,sec\, \, \infty ,cosec\, \, \infty \, \, dan \, \, cotan\, \, \infty$

Jika sudut daripada sin x, cos x, tan x, sec x, cosec x dan cotan x adalah tak hingga, maka nilainya tidak dapat kita tentukan sama sekali karena nilai dari sin x, cos x, tan x, sec x, cosec x dan cotan x, memiliki batasan tertentu yaitu:

$-1\leq sin\, x\leq 1$

$-1\leq cos\, x\leq 1$

$-\infty \leq tan\, x\leq \infty$

Untuk menyelesaikan masalah Limit Trigonometri tak hingga, dapat diselesaikan dengan cara yaitu dengan menggunakan konsep:

$\frac{1}{x}\, \, untuk \, x\, \, mendekati \, \, \infty \, \,, maka\, \, \frac{1}{x}\, \, \, \, \, \,menjadi\, \, \, \frac{1}{\infty }=0$

Dari konsep di atas, limit trigonometri di tak hingga mudah diselesaikan yaitu menjadi:

$sin\, \, \frac{1}{\infty }=sin\, \, 0=0;cos\, \, \frac{1}{\infty }=cos\, \, 0=1\, \, dan\, \, tan\, \frac{1}{\infty }=0$

Dari konsep di atas, kita dapat menentukan hasil limit trigonometri di tak hingga. Ups...sebelum kita lanjut jangan lupa ingat materi ini ya, biar dapat diselesaikan soal-soalnya ya:

$\lim_{x\to 0 }\frac{sin (x)}{x} =1; \lim_{x\to 0 }\frac{x}{sin (x)} =1$

$\lim_{x\to 0 }\frac{tan (x)}{x} =1; \lim_{x\to 0 }\frac{x}{tan (x)} =1$

$\lim_{x\to 0 }\frac{sin (ax)}{x} =a; \lim_{x\to 0 }\frac{x}{sin (ax)} =\frac{1}{a}$
$\lim_{x\to 0 }\frac{tan (ax)}{x} =a; \lim_{x\to 0 }\frac{x}{tan (ax)} =\frac{1}{a}$

$\lim_{x\to 0 }\frac{sin (ax)}{bx} =\frac{a}{b}; \lim_{x\to 0 }\frac{bx}{sin (ax)} =\frac{b}{a}$

$\lim_{x\to 0 }\frac{sin (ax)}{tan (bx)} =\frac{a}{b}; \lim_{x\to 0 }\frac {tan (bx)}{sin (ax)} =\frac{b}{a}$

$\lim_{x\rightarrow 0}\frac{sin (x-a)}{(x-a)}=\lim_{x\rightarrow 0}\frac{(x-a)}{sin(x-a)}=1$

$\lim_{x\rightarrow 0}\frac{tan (x-a)}{(x-a)}=\lim_{x\rightarrow 0}\frac{(x-a)}{tan(x-a)}=1$

$\lim_{x\rightarrow 0}\frac{sin\, k(x-a)}{(x-a)}=\lim_{x\rightarrow 0}\frac{tan\, k(x-a)}{(x-a)}=k$

$\lim_{x\rightarrow 0}\frac{sin\, (x-a)}{k(x-a)}=\lim_{x\rightarrow 0}\frac{tan\, (x-a)}{k(x-a)}=\frac{1}{k}$

$\lim_{x\rightarrow 0}\frac{sin\, \, l(x-a)}{k(x-a)}=\lim_{x\rightarrow 0}\frac{tan\, l(x-a)}{k(x-a)}=\frac{l}{k}$

$\lim_{x\rightarrow 0}\frac{sin\, \, l(x-a)}{tan\, k(x-a)}=\lim_{x\rightarrow 0}\frac{tan\,l(x-a)}{sin\, k(x-a)}=\frac{l}{k}$

Baca Juga:
Penertian Notasi Sigma dan Induksi Matematis beserta contoh soal

Untuk memahami penggunaan cara di atas maka, sekarang kita coba selesaikan permasalahan-permasalahan soal-soal yang ada di bawah ini.

1. Soal SPMB Tahun 2006

$\lim_{x\rightarrow \infty }\, \, 3x\, sin\, \frac{1}{x}=........$

Penyelesaian:

$Misalkan\, \, y=\frac{1}{x} , ekivalen\, \, dengan\, \, x=\frac{1}{y}$

$Jika \, \, x\rightarrow \infty \, \, ,maka\, \, y\rightarrow 0\, \, dapat \, \, dituliskan:$

$\lim_{x\rightarrow \infty }\, \, 3x\, sin\, \frac{1}{x}=\lim_{y\rightarrow 0}\, \, 3\left ( \frac{1}{y} \right )sin\, y$

$\lim_{x\rightarrow \infty }\, \, 3x\, sin\, \frac{1}{x}=3\, \, \lim_{y\rightarrow \, 0}\, \, \frac{sin\, y}{y}$

$\lim_{x\rightarrow \infty }\, \, 3x\, sin\, \frac{1}{x}=3.1=3$

2. SOAL LATIHAN

$Tentukan \, \, hasil\, \, \lim_{x\rightarrow \infty }\, \, \frac{2x\, cot\, \frac{2}{x}-3\, cot\, \frac{2}{x}}{5x^{2}-2x}.$

Penyelesaian:

$Misalkan\, \, y=\frac{1}{x} , ekivalen\, \, dengan\, \, x=\frac{1}{y}$

$Jika \, \, x\rightarrow \infty \, \, ,maka\, \, y\rightarrow 0\, \, dapat \, \, dituliskan:$

$\lim_{x\rightarrow \infty }\, \, \frac{2x\, cot\, \frac{2}{x}-3\, cot\, \frac{2}{x}}{5x^{2}-2x}=\lim_{x\rightarrow \infty }\, \, \frac{\left ( 2x-3 \right )cot\, \frac{2}{x}}{x\left ( 5x-2 \right )}$

$\lim_{x\rightarrow \infty }\, \, \frac{2x\, cot\, \frac{2}{x}-3\, cot\, \frac{2}{x}}{5x^{2}-2x}=\lim_{x\rightarrow \infty }\, \, \frac{\left ( 2x-3 \right )}{\left ( 5x-2 \right )}.\lim_{x\rightarrow \infty }\, \, \frac{1}{x}\, \, cot\, \, \frac{2}{x}$

$\lim_{x\rightarrow \infty }\, \, \frac{2x\, cot\, \frac{2}{x}-3\, cot\, \frac{2}{x}}{5x^{2}-2x}=\frac{2}{5}.\lim_{x\rightarrow \infty }\, \, y\, \, cot\, \, \2y$

$\lim_{x\rightarrow \infty }\, \, \frac{2x\, cot\, \frac{2}{x}-3\, cot\, \frac{2}{x}}{5x^{2}-2x}=\frac{2}{5}.\lim_{x\rightarrow 0 }\, \, y\, \, .\frac{1}{tan\, 2y}$

$\lim_{x\rightarrow \infty }\, \, \frac{2x\, cot\, \frac{2}{x}-3\, cot\, \frac{2}{x}}{5x^{2}-2x}=\frac{2}{5}.\frac{1}{2}=\frac{1}{5}$

3. SOAL SBMPTN 2017
$Hitunglah\, \, \lim_{x\rightarrow \infty }\frac{sin\, \frac{3}{x}}{\left ( 1-cos\, \frac{2}{x} \right ).x^{2}.sin\, \frac{1}{x}}.$

Penyelesaian:

$Misalkan\, \, y=\frac{1}{x} , ekivalen\, \, dengan\, \, x=\frac{1}{y}$

$Jika \, \, x\rightarrow \infty \, \, ,maka\, \, y\rightarrow 0\, \, dapat \, \, dituliskan:$

$\lim_{x\rightarrow \infty }\frac{sin\, \frac{3}{x}}{\left ( 1-cos\, \frac{2}{x} \right ).x^{2}.sin\, \frac{1}{x}}=\lim_{x\rightarrow 0}\, \, \frac{sin\, 3y}{\left ( 1-cos\, 2y \right ).\left ( \frac{1}{y} \right )^{2}.sin\, y}$

$\lim_{x\rightarrow \infty }\frac{sin\, \frac{3}{x}}{\left ( 1-cos\, \frac{2}{x} \right ).x^{2}.sin\, \frac{1}{x}}=\lim_{x\rightarrow0 }\, \, \frac{y^{2}.sin\, 3y}{\left ( 1-\left ( 1-2sin^{2}y \right ) \right ).siny}$

$\lim_{x\rightarrow \infty }\frac{sin\, \frac{3}{x}}{\left ( 1-cos\, \frac{2}{x} \right ).x^{2}.sin\, \frac{1}{x}}=\lim_{x\rightarrow0 }\, \, \frac{y^{2}.sin\, 3y}{2sin\, ^{2}y.siny}$

$\lim_{x\rightarrow \infty }\, \, \frac{sin\, \frac{3}{x}}{(1-cos\, \frac{2}{x}).x^{2}.sin\, \frac{1}{x}}=\frac{1^{2}.3}{2.1^{2}.1}=\frac{3}{2}$

4. Latihan Soal

$Hasil \, \, dari \, \, \lim_{x\rightarrow \infty }\, \frac{cos\, \frac{4}{x}\, +\, cos\, \frac{2}{x}.sin\frac{3}{\sqrt{x}}-cos\frac{4}{x}.sin\, \frac{3}{\sqrt{x}}-cos\, \frac{2}{x}}{sin^{2}\frac{1}{x}-cos\, \frac{2}{x}+1}.$

Penyelesaian:

$Misalkan \, \, y=\frac{1}{\sqrt{x}}, maka\, \frac{1}{x}=y^{2}, sehingga diperoleh:$

$\lim_{x\rightarrow \infty }\, \frac{cos\, \frac{4}{x}\, +\, cos\, \frac{2}{x}.sin\frac{3}{\sqrt{x}}-cos\frac{4}{x}.sin\, \frac{3}{\sqrt{x}}-cos\, \frac{2}{x}}{sin^{2}\frac{1}{x}-cos\, \frac{2}{x}+1}= \lim_{x\rightarrow \infty }\, \, \frac{cos\, 4y^{2}+cos\, 2y^{2}.sin\, 3y-cos\, 4y^{2}.sin\, 3y-cos\, 2y^{2}}{sin^{2}y^{2}-cos\, 2y^{2}+1}$

$\lim_{x\rightarrow \infty }\, \frac{cos\, \frac{4}{x}\, +\, cos\, \frac{2}{x}.sin\frac{3}{\sqrt{x}}-cos\frac{4}{x}.sin\, \frac{3}{\sqrt{x}}-cos\, \frac{2}{x}}{sin^{2}\frac{1}{x}-cos\, \frac{2}{x}+1}=\lim_{x\rightarrow \infty } \frac{cos\, 4y^{2}+sin\, 3y\left ( cos\, 2y^{2}-cos\, 4y^{2} \right )-cos\, 2y^{2}}{sin^{2}y^{2}-cos\, 2y^{2}+1}$

$\lim_{x\rightarrow \infty }\, \frac{cos\, \frac{4}{x}\, +\, cos\, \frac{2}{x}.sin\frac{3}{\sqrt{x}}-cos\frac{4}{x}.sin\, \frac{3}{\sqrt{x}}-cos\, \frac{2}{x}}{sin^{2}\frac{1}{x}-cos\, \frac{2}{x}+1}=\lim_{x\rightarrow \infty } \frac{cos\, 4y^{2}-cos\, 2y^{2}+sin\, 3y\left ( cos\, 2y^{2}-cos\, 4y^{2} \right )}{sin^{2}y^{2}-cos\, 2y^{2}+1}$

$\lim_{x\rightarrow \infty }\, \frac{cos\, \frac{4}{x}\, +\, cos\, \frac{2}{x}.sin\frac{3}{\sqrt{x}}-cos\frac{4}{x}.sin\, \frac{3}{\sqrt{x}}-cos\, \frac{2}{x}}{sin^{2}\frac{1}{x}-cos\, \frac{2}{x}+1}=\lim_{x\rightarrow \infty } \frac{cos\, 4y^{2}-cos\, 2y^{2}-sin\, 3y\left ( cos\, 4y^{2}-cos\, 2y^{2} \right )}{sin^{2}y^{2}-cos\, 2y^{2}+1}$

$\lim_{x\rightarrow \infty }\, \frac{cos\, \frac{4}{x}\, +\, cos\, \frac{2}{x}.sin\frac{3}{\sqrt{x}}-cos\frac{4}{x}.sin\, \frac{3}{\sqrt{x}}-cos\, \frac{2}{x}}{sin^{2}\frac{1}{x}-cos\, \frac{2}{x}+1}=\lim_{x\rightarrow \infty }\frac{\left ( cos\, 4y^{2}-cos\, 2y^{2}\right )\left ( 1-sin\, 3y \right )}{sin^{2}y^{2}-cos\, 2y^{2}+1}$

$\lim_{x\rightarrow \infty }\, \frac{cos\, \frac{4}{x}\, +\, cos\, \frac{2}{x}.sin\frac{3}{\sqrt{x}}-cos\frac{4}{x}.sin\, \frac{3}{\sqrt{x}}-cos\, \frac{2}{x}}{sin^{2}\frac{1}{x}-cos\, \frac{2}{x}+1}=\lim_{x\rightarrow \infty }\frac{\left ( -2\, sin\, 3y^{2}.sin\left ( -y^{2} \right )\right )\left ( 1-sin\, 3y \right )}{sin^{2}y^{2}-\left ( 1-2\, sin^{2}y^{2} \right )+1}$

$\lim_{x\rightarrow \infty }\, \frac{cos\, \frac{4}{x}\, +\, cos\, \frac{2}{x}.sin\frac{3}{\sqrt{x}}-cos\frac{4}{x}.sin\, \frac{3}{\sqrt{x}}-cos\, \frac{2}{x}}{sin^{2}\frac{1}{x}-cos\, \frac{2}{x}+1}= \lim_{x\rightarrow \infty } \frac{2\left ( \, sin\, 3y^{2}.sin\left ( y^{2} \right )\right )}{3sin^{2}y^{2}}.\lim_{x\rightarrow \infty }\, \, \left ( 1-sin\, 3y \right )$

$\lim_{x\rightarrow \infty }\, \frac{cos\, \frac{4}{x}\, +\, cos\, \frac{2}{x}.sin\frac{3}{\sqrt{x}}-cos\frac{4}{x}.sin\, \frac{3}{\sqrt{x}}-cos\, \frac{2}{x}}{sin^{2}\frac{1}{x}-cos\, \frac{2}{x}+1}=\frac{2}{3}\lim_{x\rightarrow \infty }\, \, \frac{sin\, \, 3y^{2}.sin\, y^{2}}{sin^{2}\, y^{2}}.\lim_{x\rightarrow \infty}\, \, \left ( 1-sin\, 3y \right )$

5. Latihan Soal

$Nilai\, \, \lim_{x\rightarrow \infty }\, \, \frac{1-cos\, \frac{4}{x}}{\frac{2}{x}\, \, tan\, \frac{3}{x}}\, \, adalah....$

Penyelesaian:

$\lim_{x\rightarrow \infty }\, \, \frac{1-cos\, \frac{4}{x}}{\frac{2}{x}\, \, tan\, \frac{3}{x}}\, \, =\lim_{x\rightarrow 0 }\, \, \frac{1-cos\, 4y}{2y.tan\, 3y}$

$\lim_{x\rightarrow \infty }\, \, \frac{1-cos\, \frac{4}{x}}{\frac{2}{x}\, \, tan\, \frac{3}{x}}\, \, =\lim_{x\rightarrow 0 }\, \, \frac{1-(1-2sin^{2}2y)}{2y.tan\, 3y}$

$\lim_{x\rightarrow \infty }\, \, \frac{1-cos\, \frac{4}{x}}{\frac{2}{x}\, \, tan\, \frac{3}{x}}\, \, =\lim_{x\rightarrow 0 }\, \, \frac{2\, sin^{2}2y}{2y.tan\, 3y}$

$\lim_{x\rightarrow \infty }\, \, \frac{1-cos\, \frac{4}{x}}{\frac{2}{x}\, \, tan\, \frac{3}{x}}\, \, =\lim_{x\rightarrow 0 }\, \, \frac{\, sin\, 2y.sin\, 2y}{y.tan\, 3y}$

$\lim_{x\rightarrow \infty }\, \, \frac{1-cos\, \frac{4}{x}}{\frac{2}{x}\, \, tan\, \frac{3}{x}}\, \, =\frac{2.2}{1.3}=\frac{4}{3}$

6. SOAL UGM 2005

$Nilai\, \, dari\, \, \lim_{x\rightarrow \infty }\, \,x^{2} \left ( sec\frac{2}{x}-1 \right )adalah....$

Penyelesaian:

$\lim_{x\rightarrow \infty }\, \,x^{2} \left ( sec\frac{2}{x}-1 \right )=\lim_{x\rightarrow 0}\, \, \frac{1}{y^{2}}.\left ( sec\, 2y-1 \right )$

$\lim_{x\rightarrow \infty }\, \,x^{2} \left ( sec\frac{2}{x}-1 \right )=\lim_{x\rightarrow 0}\, \, \frac{1}{y^{2}}.\left ( \frac{1}{cos\, 2y}-1 \right )$

$\lim_{x\rightarrow \infty }\, \,x^{2} \left ( sec\frac{2}{x}-1 \right )=\lim_{x\rightarrow 0}\, \, \frac{1}{y^{2}}.\left ( \frac{1-cos\, 2y}{cos\, 2y}\right )$

$\lim_{x\rightarrow \infty }\, \,x^{2} \left ( sec\frac{2}{x}-1 \right )=\lim_{x\rightarrow 0}\, \, \frac{1}{y^{2}}.\left ( \frac{1-(1-2\, sin^{2}y)}{cos\, 2y}\right )$

$\lim_{x\rightarrow \infty }\, \,x^{2} \left ( sec\frac{2}{x}-1 \right )=\lim_{x\rightarrow 0}\, \, \frac{1}{y^{2}}.\left ( \frac{2\, sin^{2}y)}{cos\, 2y}\right )$

$\lim_{x\rightarrow \infty }\, \,x^{2} \left ( sec\frac{2}{x}-1 \right )=2\, \, \lim_{x\rightarrow 0}\, \, \frac{sin^{2}y}{y^{2}}.\lim_{x\rightarrow 0}\, \, \frac{1}{cos\, \, 2y}$

$\lim_{x\rightarrow \infty }\, \,x^{2} \left ( sec\frac{2}{x}-1 \right )=2.\frac{1^{2}}{1^{2}}.\frac{1}{cos\, 0}=2$

7. Soal SPMB 2005

$Nilai\, \, \lim_{x\rightarrow \infty }\, \,x^{2}.sin\, \frac{1}{x}.tan\, \frac{1}{x}\, \, adalah....$

Penyelesaian:

$\lim_{x\rightarrow \infty }\, \,x^{2}.sin\, \frac{1}{x}.tan\, \frac{1}{x}\, \,=\lim_{x\rightarrow 0}\left ( \frac{1}{y} \right )^{2}.sin\, y.tan\, y$

$\lim_{x\rightarrow \infty }\, \,x^{2}.sin\, \frac{1}{x}.tan\, \frac{1}{x}\, \,=\lim_{y\rightarrow 0}\frac{sin\, y.tan\, y}{y^{2}}$

$\lim_{y\rightarrow \infty }\, x^{2}.sin\frac{1}{x}.tan\, \frac{1}{x}=\lim_{x\rightarrow 0}\, \frac{sin\, y}{y}.\lim_{y\rightarrow 0}\, \frac{sin\, y}{y}$

$\lim_{y\rightarrow \infty }\, x^{2}.sin\frac{1}{x}.tan\, \frac{1}{x}=1.1=1$

8.Soal SPMB 2017

$Nilai\, \, dari\, \, \lim_{x\rightarrow \infty }\, \, \frac{x^{4}.sin\frac{1}{x}+x^{2}}{1+x^{3}}\, \, adalah...$

Penyelesaian:

$\lim_{x\rightarrow \infty }\, \, \frac{x^{4}.sin\frac{1}{x}+x^{2}}{1+x^{3}}\, \,=\lim_{y\rightarrow 0}\, \, \frac{\left ( \frac{1}{y} \right )^{4}sin\, y+\left ( \frac{1}{y} \right )^{2}}{1+\left ( \frac{1}{y} \right )^{3}}$

$\lim_{x\rightarrow \infty }\, \, \frac{x^{4}.sin\frac{1}{x}+x^{2}}{1+x^{3}}\, \,=\lim_{y\rightarrow 0}\, \frac{\frac{1}{y^{2}}\left ( \frac{sin\, y}{y^{2}}+1 \right )}{1+\frac{1}{y^{3}}}$

$\lim_{x\rightarrow \infty }\, \, \frac{x^{4}.sin\frac{1}{x}+x^{2}}{1+x^{3}}\, \,=\lim_{y\rightarrow 0}\, \frac{\frac{1}{y^{2}}\left ( \frac{sin\, y}{y^{2}}+1 \right )}{\frac{1}{y^{3}}\left ( y^{3}+1 \right )}$

$\lim_{x\rightarrow \infty }\, \, \frac{x^{4}.sin\frac{1}{x}+x^{2}}{1+x^{3}}\, \,=\lim_{y\rightarrow 0}\, \frac{y\left ( \frac{sin\, y}{y^{2}}+1 \right )}{y^{3}+1}$

$\lim_{x\rightarrow \infty }\, \, \frac{x^{4}.sin\frac{1}{x}+x^{2}}{1+x^{3}}\, \,=\lim_{y\rightarrow 0}\, \frac{\frac{sin\, y}{y}+y}{y^{3}+1}$

$\lim_{x\rightarrow \infty }\, \, \frac{x^{4}.sin\frac{1}{x}+x^{2}}{1+x^{3}}\, \,=\frac{1+0}{0^{3}+1}=\frac{1}{1}=1$

9. Soal SBMPTN 2017

$Nilai\, \, dari\, \, \lim_{x\rightarrow \infty }\, \,cosec\, \frac{1}{x}-cotan\, \frac{1}{x}\, \, adalah...$

Penyelesaian:

$\lim_{x\rightarrow \infty }\, \,cosec\, \frac{1}{x}-cotan\, \frac{1}{x}\, \,=\lim_{y\rightarrow 0}\, \, cosec\, y-cotan\, y$

$\lim_{x\rightarrow \infty }\, \,cosec\, \frac{1}{x}-cotan\, \frac{1}{x}\, \,=\lim_{y\rightarrow 0}\, \, \frac{1}{sin\, y}-\frac{1}{tan\, y}$

$\lim_{x\rightarrow \infty }\, \,cosec\, \frac{1}{x}-cotan\, \frac{1}{x}\, \,=\lim_{y\rightarrow 0}\, \, \frac{1}{sin\, y}-\frac{cos\, y}{sin\, y}$

$\lim_{x\rightarrow \infty }\, \,cosec\, \frac{1}{x}-cotan\, \frac{1}{x}\, \,=\lim_{y\rightarrow 0}\, \, \frac{1-cos\, y}{sin\, y}$

$\lim_{x\rightarrow \infty }\, \,cosec\, \frac{1}{x}-cotan\, \frac{1}{x}\, \,=\lim_{y\rightarrow 0}\, \, \frac{1-\left ( 1-2sin^{2}\frac{1}{2}y \right )}{sin\, y}$

$\lim_{x\rightarrow \infty }\, \,cosec\, \frac{1}{x}-cotan\, \frac{1}{x}\, \,=\lim_{y\rightarrow 0}\, \, \frac{2sin^{2}\frac{1}{2}y}{sin\, y}$

$\lim_{x\rightarrow \infty }\, \,cosec\, \frac{1}{x}-cotan\, \frac{1}{x}\, \,=2.\lim_{y\rightarrow 0}\, \,\frac{sin^{2}\frac{1}{2}y}{sin\, y}.\frac{\frac{1}{2}y}{\frac{1}{2}y}. \frac{\frac{1}{2}y}{\frac{1}{2}y}.\frac{y}{y}$

$\lim_{x\rightarrow \infty }\, \,cosec\, \frac{1}{x}-cotan\, \frac{1}{x}\, \,=2\lim_{y\rightarrow 0}\frac{sin\frac{1}{2}y}{\frac{1}{2}y}.\lim_{y\rightarrow 0}\frac{sin\frac{1}{2}y }{\frac{1}{2}y}.\lim_{y\rightarrow 0}\frac{y}{sin\, y}.\lim_{y\rightarrow 0}\frac{\frac{1}{4}y^{2}}{y}$

$\lim_{x\rightarrow \infty }\, \,cosec\, \frac{1}{x}-cotan\, \frac{1}{x}\, \,=2.1.1.1.\lim_{y\rightarrow 0}\frac{1}{4}y$

$\lim_{x\rightarrow \infty }\, \,cosec\, \frac{1}{x}-cotan\, \frac{1}{x}\, \,=2.0=0$

10. Soal SBMPTN 2017

$Nilai\, \, dari\, \, \lim_{x\rightarrow \infty }\, \,\frac{2x^{2}tan\frac{1}{x}-xsin\, \left ( \frac{1}{x} \right )+\frac{1}{x}}{x.cos\, \frac{2}{x}}\, \,adalah....$

Penyelesaian:

$\lim_{x\rightarrow \infty }\, \,\frac{2x^{2}tan\frac{1}{x}-xsin\, \left ( \frac{1}{x} \right )+\frac{1}{x}}{x.cos\, \frac{2}{x}}\, \, =\lim_{y\rightarrow 0}\frac{\frac{2}{y^{2}}.tan\, y-\frac{1}{y}.sin\, y+y}{\frac{1}{y}.cos\, 2y}$

$\lim_{x\rightarrow \infty }\, \,\frac{2x^{2}tan\frac{1}{x}-xsin\, \left ( \frac{1}{x} \right )+\frac{1}{x}}{x.cos\, \frac{2}{x}}\, \,=\lim_{y\rightarrow 0}\, \, \frac{\frac{2}{y^{2}}.tan\, y-\frac{1}{y}\, sin\, y+y}{\frac{1}{y}.cos\, 2y}.\left ( \frac{y}{y} \right )$

$\lim_{x\rightarrow \infty }\, \,\frac{2x^{2}tan\frac{1}{x}-xsin\, \left ( \frac{1}{x} \right )+\frac{1}{x}}{x.cos\, \frac{2}{x}}\, \,=\lim_{y\rightarrow 0}\, \, \frac{\frac{2.tan\, y}{y}}{cos\, y}+\lim_{y\rightarrow 0}\, \frac{y^{2}-sin\, y}{cos\, 2y}$

$\lim_{x\rightarrow \infty }\, \,\frac{2x^{2}tan\frac{1}{x}-xsin\, \left ( \frac{1}{x} \right )+\frac{1}{x}}{x.cos\, \frac{2}{x}}\, \,=\frac{2}{cos\, 0}+\frac{0-sin\, 0}{cos\, 0}$

$\lim_{x\rightarrow \infty }\, \,\frac{2x^{2}tan\frac{1}{x}-xsin\, \left ( \frac{1}{x} \right )+\frac{1}{x}}{x.cos\, \frac{2}{x}}\, \,=\frac{2}{1}+\frac{0}{1}$

$\lim_{x\rightarrow \infty }\, \,\frac{2x^{2}tan\frac{1}{x}-xsin\, \left ( \frac{1}{x} \right )+\frac{1}{x}}{x.cos\, \frac{2}{x}}\, \,=2+0=2$

Silahkan lihat juga pembahasan soal-soalnya di sini:

31 comments:

jibangpardamean@gmail.com8 October 2020 at 09:03
Selamat membaca,semoga bermanfaat...
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Unknown8 October 2020 at 09:03
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Syawal suramana putra XII IA 4
Saya sudah membaca pak
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Reza Ananda31 October 2021 at 09:54
saya sudah membaca pak
REZA ANANDA PANDIA
XII-IA4
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Unknown3 November 2021 at 08:25
Mayang sari
XII-IA5 hdr pak
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Unknown3 November 2021 at 08:26
Saya sudah baca pak
Dava syahputra
XII-IA5
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Unknown3 November 2021 at 08:29
Agnes XII IPA 5 hdr pak
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Anonymous3 November 2021 at 08:30
Agnes XII IPA 5 hdr pak
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Unknown3 November 2021 at 08:33
Gabriel Cristian Louis Lumban Gaol hadir pak Xll IA5
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Unknown3 November 2021 at 09:01
Rio julianta Ginting Xll IA5 hadir pak
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daniel silalahi 66676 November 2021 at 01:03
This comment has been removed by the author.
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Anonymous6 November 2021 at 01:07
Anastasya Sijabat XII IPA 5 pak
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Reyzauli4 December 2021 at 00:01
reyzauli sudah baca pak
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Soal, Materi Limit di Tak Hingga Fungsi Trigonometri Mirip Soal UTBK SBMPTN

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BIMBEL ONLINE MASUK SMA PLUS YASOP DAN SMA UNGGUL DEL 2023

BIMBEL ONLINE FOKUS LULUS SIMAMA POLTEKKES 2022

BIMBEL ONLINE FOKUS LULUS UJIAN SBMPN (POLITEKNIK) 2022

AYO BERLANGGANAN DI RUANG PARA BINTANG

Fanpage Ruang Para Bintang

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ARTIKEL TERBAIK

Artikel Mingguan

KATEGORI PILIHAN

INFORMASI UTBK SBMPTN/SIMAMA POLTEKKES/UM-PTKIN/SBMPN POLITEKNIK

INFO KEDINASAN/TUTORIAL PENDIDIKAN

UNBK & INFO SMA

Soal, Materi Limit di Tak Hingga Fungsi Trigonometri Mirip Soal UTBK SBMPTN

Related Posts

Leave a Comment

31 comments:

BIMBEL ONLINE MASUK SMA PLUS YASOP DAN SMA UNGGUL DEL 2023

BIMBEL ONLINE FOKUS LULUS SIMAMA POLTEKKES 2022

BIMBEL ONLINE FOKUS LULUS UJIAN SBMPN (POLITEKNIK) 2022

AYO BERLANGGANAN DI RUANG PARA BINTANG

Fanpage Ruang Para Bintang

Anda Pengunjung ke-

ARTIKEL TERBAIK

Artikel Mingguan

KATEGORI PILIHAN

INFORMASI UTBK SBMPTN/SIMAMA POLTEKKES/UM-PTKIN/SBMPN POLITEKNIK

INFO KEDINASAN/TUTORIAL PENDIDIKAN

UNBK & INFO SMA