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

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:

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

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

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

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

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