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

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Friday, October 2, 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:

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

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