Soal, Materi Limit Tak Hingga Fungsi Trigonometri

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:

Memisalkan y = 1/x

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

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:

115 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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Unknown31 October 2021 at 07:28
Syawal suramana putra XII IA 4
Saya sudah membaca pak
ReplyDelete
Replies
Reza Ananda31 October 2021 at 09:54
saya sudah membaca pak
REZA ANANDA PANDIA
XII-IA4
ReplyDelete
Replies
Unknown3 November 2021 at 08:25
Mayang sari
XII-IA5 hdr pak
ReplyDelete
Replies
Unknown3 November 2021 at 08:26
Saya sudah baca pak
Dava syahputra
XII-IA5
ReplyDelete
Replies
Anonymous3 November 2021 at 08:29
Agnes XII IPA 5 hdr pak
ReplyDelete
Replies
Anonymous3 November 2021 at 08:30
Agnes XII IPA 5 hdr pak
ReplyDelete
Replies
Unknown3 November 2021 at 08:33
Gabriel Cristian Louis Lumban Gaol hadir pak Xll IA5
ReplyDelete
Replies
Unknown3 November 2021 at 09:01
Rio julianta Ginting Xll IA5 hadir pak
ReplyDelete
Replies
Anonymous6 November 2021 at 01:03
This comment has been removed by the author.
ReplyDelete
Replies
Anonymous6 November 2021 at 01:07
Anastasya Sijabat XII IPA 5 pak
ReplyDelete
Replies
Reyzauli4 December 2021 at 00:01
reyzauli sudah baca pak
ReplyDelete
Replies
Anonymous1 November 2023 at 15:33
Stanly 12 ipa-3 makasih pak
ReplyDelete
Replies
Anonymous1 November 2023 at 15:35
saya miftahul rizky anindia terimakasih pak dari 12 ipa 3 makasih pak
ReplyDelete
Replies
Anonymous1 November 2023 at 15:36
Saya Azrah Anggita XII IPA 3,terimakasih pak
ReplyDelete
Replies
Anonymous1 November 2023 at 15:51
Amanda Putri Natasya XII-IA2 , terimakasih pakk
ReplyDelete
Replies
Anonymous1 November 2023 at 16:08
M.syehan Al Fahri XII IA2
Terimakasih
ReplyDelete
Replies
Anonymous1 November 2023 at 16:21
M.Dava Caesar XII IA3
Terimakasih pak
ReplyDelete
Replies
Anonymous1 November 2023 at 16:22
Laksmi Dwi Oktavia XII IA3
Terimakasih pak
ReplyDelete
Replies
Anonymous1 November 2023 at 16:39
Jeremy Simatupang XII IPA 3
Makasih pak
ReplyDelete
Replies
Anonymous1 November 2023 at 16:39
Doni Zebua XII IPA 3
Makasih Pak
ReplyDelete
Replies
Anonymous1 November 2023 at 16:40
Arjen Robben XII IPA 3
Makasih Pak
ReplyDelete
Replies
Anonymous1 November 2023 at 16:57
Audyra Hidayat XII IPA 2
Terimakasih sir
ReplyDelete
Replies
Anonymous1 November 2023 at 17:03
Mayurika Kresensia Sembiring XII IPA 2
Terimakasih e
ReplyDelete
Replies
Maria1 November 2023 at 17:10
Maria Angelica XII IPA 2
Terima kasih pak
ReplyDelete
Replies
Anonymous1 November 2023 at 17:11
Fadila Naima Syamila
XII IPA 2
Terimakasih pak
ReplyDelete
Replies
Anonymous1 November 2023 at 17:30
Adzra Alfiyah
XII IA 2
Terimakasih pak
ReplyDelete
Replies
Anonymous1 November 2023 at 17:45
ALDO ALFONSO
XII A2
TERIMAH KASIH PAK
ReplyDelete
Replies
Anonymous1 November 2023 at 18:17
syatrio pradana XII IPA2
terimakasih pak
ReplyDelete
Replies
Anonymous1 November 2023 at 19:48
Deanisa Syafitri
XII IPA3

Terimakasih Banyak Pak
ReplyDelete
Replies
Anonymous1 November 2023 at 19:57
naurah zakyyah
XII IPA 3

Terimakasih pak
ReplyDelete
Replies
Anonymous1 November 2023 at 20:09
Jonathan Guido Nambah
XII-IPA 3
Terimakasih pak
ReplyDelete
Replies
Anonymous1 November 2023 at 20:10
PAREL RAHMADSYAH
XII-IA³
TERIMAKASIH PAKK
ReplyDelete
Replies
Anonymous1 November 2023 at 20:10
Bima Arya
XII IPA 3
Makasih Pak
ReplyDelete
Replies
Apryuva Sormin XII IPA 21 November 2023 at 20:12
terima kasih Sir
ReplyDelete
Replies
Anonymous1 November 2023 at 20:51
Mutia Zahra
XII IPA 2
Terimakasih Bapakk
ReplyDelete
Replies
Anonymous1 November 2023 at 20:53
Syifa Azrina Tasya
XII A2
terimakasih pak
ReplyDelete
Replies
Anonymous3 November 2023 at 07:02
Rafael sinukaban
XII IA 2
Terimakasih pak
ReplyDelete
Replies
Anonymous3 November 2023 at 07:06
Nazwa Cahyani Firgianty
XII IPA 2
Terimakasih Pak
ReplyDelete
Replies
Anonymous3 November 2023 at 12:46
Chalista salsabila
XII IPA 2
Terimakasih pak
ReplyDelete
Replies
Anonymous3 November 2023 at 13:19
Santa mikhal br bangun
12 IPA 2
Terimakasih pakkk
ReplyDelete
Replies
Anonymous3 November 2023 at 13:27
Dita darma
12 ipa 2
terimaakasih pak
ReplyDelete
Replies
Anonymous3 November 2023 at 19:15
pricillia putri
XII IPA 2
ReplyDelete
Replies
Anonymous3 November 2023 at 21:28
lilis dwinta sijabat
XII IA 2
terimakasih pak
ReplyDelete
Replies
Anonymous3 November 2023 at 22:09
Andreas Alfredo
12 IPA 2
trimaksih pak
ReplyDelete
Replies
Anonymous5 November 2023 at 11:23
SYAFRIZA WASLAINI
12 IPA 3
TERIMAKASIH PAK
ReplyDelete
Replies
Anonymous5 November 2023 at 11:53
Zaskia Nurfika Priani
12 IPA 3
Terimakasih pak
ReplyDelete
Replies
Anonymous5 November 2023 at 12:08
STEVANY AMELIA S MUHAM
12 IPA 3
TERIMAKASIH PAK
ReplyDelete
Replies
Anonymous5 November 2023 at 14:14
aisyah daulay
XIIA3
Terimakasih pak
ReplyDelete
Replies
Anonymous5 November 2023 at 15:03
jalfa adinda
XII IA 3
terimakasih pak
ReplyDelete
Replies
Anonymous5 November 2023 at 17:46
Nesa Aulia Marpaung
XII IPA 3
Terimakasih pak
ReplyDelete
Replies
Anonymous5 November 2023 at 19:33
Marsya Harani Aulia Br Tambunan XII IPA 3
ReplyDelete
Replies
Anonymous5 November 2023 at 19:41
Jenny Angelica XII-A3
Terimakasih pak
ReplyDelete
Replies
Vina Grescia br. Manullang5 November 2023 at 21:31
Vina Grescia br. Manullang
XII IPA 3
Terimakasih, Pak.
ReplyDelete
Replies
Anonymous5 November 2023 at 22:23
clara yunika
XII-A3
trimakasih pak
ReplyDelete
Replies
Anonymous6 November 2023 at 04:29
sahwa amelia
XII IPA 3
terimakasih pak
ReplyDelete
Replies
Anonymous6 November 2023 at 05:02
siti humairah
Xll ipa 3
terima kasih pak
ReplyDelete
Replies
Anonymous26 October 2024 at 02:14
Wahyu Tri Asih Putri
XII IPA 3
Saya sudah membaca, terima kasih banyak Pak.
ReplyDelete
Replies
Anonymous26 October 2024 at 07:55
Lusiana simamora
XII IPA 2
saya sudah membaca, terima kasih banyak pak.
ReplyDelete
Replies
Anonymous26 October 2024 at 07:55
Saya sudah membaca pak (arrash bima faturrahman (XII-IA2))
ReplyDelete
Replies
Anonymous26 October 2024 at 07:55
Felicia Benadicta Ginting
saya sudah membaca
XII IPA 2
tahun 2024/2025
ReplyDelete
Replies
Anonymous26 October 2024 at 07:55
Widia Sabrina
XII IPA 2
terima kasih pak
ReplyDelete
Replies
tasya26 October 2024 at 08:27
Tasya Saragih
XII IA 2 (2024)
ReplyDelete
Replies
Anonymous26 October 2024 at 08:27
Sofiyana
XII IPA 2
Terimakasih Pak
ReplyDelete
Replies
Anonymous26 October 2024 at 08:27
Chrisanta Imanuel Sebayang
XII IPA 2
Saya sudah membaca, terima kasih pak
ReplyDelete
Replies
Anonymous28 October 2024 at 12:15
Dhini yunita sari
Xll-5
Terimakasih ser
ReplyDelete
Replies
Ramaaa28 October 2024 at 12:25
Rama Pratama
XII IA 5
ReplyDelete
Replies
Anonymous28 October 2024 at 16:33
Asela Tarigan XII IPA 4 Hadirr pakk
ReplyDelete
Replies
Anonymous29 October 2024 at 20:34
Rani Manurung
XII IA4 hadir pakkk
ReplyDelete
Replies
Anonymous3 November 2024 at 12:17
dameria simatupang XII IPA 5
ReplyDelete
Replies
Anonymous3 November 2024 at 18:55
Nesya Adiva Xll IA5
ReplyDelete
Replies
Anonymous3 November 2024 at 20:37
Olivia jois XII IA5
ReplyDelete
Replies
Anonymous8 November 2024 at 12:24
Jelita Sinarta Br Limbong
XII IPA 3
Terima kasih Pak
ReplyDelete
Replies
Anonymous8 November 2024 at 18:40
IRNA NARULITA GULTOM
XII IPA 3
ReplyDelete
Replies

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Soal, Materi Limit Tak Hingga Fungsi Trigonometri

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115 comments:

BIMBEL ONLINE MASUK SMA UNGGULAN TAHUN 2026 UNTUK KELAS IX SMP

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