Question

bantu saya mengerjakan soal berikut​

Answer 1

[tex]\begin{aligned}\sf1.\ \ &\lim_{x\to\infty}\left(x-3-\sqrt{x^2-3x-2}\right)=\boxed{\,\bf-\frac{3}{2}\,}\\\sf2.\ \ &\lim_{x\to\infty}\left(\sqrt{x^2-x-3}-\sqrt{x^2-3x-2}\right)=\boxed{\,\bf1\,}\\\sf3.\ \ &\lim_{x\to\infty}x^2\left[1-\cos\left(\frac{4}{x}\right)\right]=\boxed{\,\bf8\,}\\\sf4.\ \ &\lim_{x\to\infty}\frac{x\sin\left(\dfrac{1}{x}\right)}{1-\sin\left(\dfrac{1}{x}\right)}=\boxed{\,\bf1\,}\end{aligned}[/tex]
 

Penjelasan dengan langkah-langkah:

Limit

Untuk soal nomor 1 dan 2, kita dapat menggunakan rumus cepat:

[tex]\begin{aligned}&\lim_{x\to\infty}\left(\sqrt{ax^2+bx+c}-\sqrt{px^2+qx+r}\right)\\&=\begin{cases}\infty\,,&{\sf jika\ }a > p\\\vphantom{\Bigg|}\dfrac{b-q}{2\sqrt{a}}\,,&{\sf jika\ }a = p\\-\infty\,,&{\sf jika\ }a < p\\\end{cases}\end{aligned}[/tex]
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Nomor 1

[tex]\begin{aligned}&\lim_{x\to\infty}\left(x-3-\sqrt{x^2-3x-2}\right)\\&{=\ }\lim_{x\to\infty}\left(\sqrt{(x-3)^2}-\sqrt{x^2-3x-2}\right)\\&{=\ }\lim_{x\to\infty}\left(\sqrt{x^2-6x+9}-\sqrt{x^2-3x-2}\right)\\&\quad...\ (a,b,c)=(1,-6,9)\,,\ (p,q,r)=(1,-3,-2)\\&\qquad\Rightarrow a=p\\&{=\ }\frac{-6-(-3)}{2\sqrt{1}}\\&{=\ }\boxed{\,\bf-\frac{3}{2}\,}\end{aligned}[/tex]
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Nomor 2

[tex]\begin{aligned}&\lim_{x\to\infty}\left(\sqrt{x^2-x-3}-\sqrt{x^2-3x-2}\right)\\&\quad...\ (a,b,c)=(1,-1,-3)\,,\ (p,q,r)=(1,-3,-2)\\&\qquad\Rightarrow a=p\\&{=\ }\frac{-1-(-3)}{2\sqrt{1}}=\frac{2}{2}\\&{=\ }\boxed{\,\bf1\,}\end{aligned}[/tex]
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Untuk nomor 3 dan 4, kita gunakan:

[tex]\begin{aligned}\boxed{\,\lim_{x\to\infty}x\sin\left(a\cdot\frac{1}{x}\right)=a\,}\end{aligned}[/tex]

BUKTI:

[tex]\begin{aligned}&\lim_{x\to\infty}x\sin\left(a\cdot\frac{1}{x}\right)=\lim_{x\to\infty}\frac{\sin\left(a\cdot\dfrac{1}{x}\right)}{\dfrac{1}{x}}\\&\quad...\ \textsf{Aturan L'H\^opital}\\&{=\ }\lim_{x\to\infty}\frac{\left[\sin\left(a\cdot\dfrac{1}{x}\right)\right]'}{\left(\dfrac{1}{x}\right)'}\\&{=\ }\lim_{x\to\infty}\frac{\cos\left(a\cdot\dfrac{1}{x}\right)\cdot\left(a\cdot\dfrac{1}{x}\right)'}{\left(\dfrac{1}{x}\right)'}\end{aligned}[/tex]
[tex]\begin{aligned}&{=\ }\lim_{x\to\infty}\frac{\cos\left(a\cdot\dfrac{1}{x}\right)\cdot a\cdot\cancel{\dfrac{-1}{x^2}}}{\cancel{\dfrac{-1}{x^2}}}\\&{=\ }\lim_{x\to\infty}a\cos\left(a\cdot\dfrac{1}{x}\right)\\&{=\ }\boxed{\,a\,}\end{aligned}[/tex]
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Nomor 3

[tex]\begin{aligned}&\lim_{x\to\infty}x^2\left[1-\cos\left(\frac{4}{x}\right)\right]\\&{=\ }\lim_{x\to\infty}\frac{1-\cos\left(\dfrac{4}{x}\right)}{\dfrac{1}{x^2}}\\&\quad...\ \textsf{Aturan L'H\^opital}\\&{=\ }\lim_{x\to\infty}\frac{\left(1-\cos\left(\dfrac{4}{x}\right)\right)'}{\left(\dfrac{1}{x^2}\right)'}\\&{=\ }\lim_{x\to\infty}\frac{\sin\left(\dfrac{4}{x}\right)\cdot\left(\dfrac{4}{x}\right)'}{\dfrac{-2}{x^3}}\end{aligned}[/tex]
[tex]\begin{aligned}&{=\ }\lim_{x\to\infty}\frac{\sin\left(\dfrac{4}{x}\right)\cdot\dfrac{-4}{x^2}}{\dfrac{-2}{x^3}}\\&{=\ }\lim_{x\to\infty}\sin\left(\dfrac{4}{x}\right)\cdot\dfrac{-4}{x^2}\cdot\frac{x^3}{-2}\\&{=\ }\lim_{x\to\infty}2x\sin\left(\dfrac{4}{x}\right)\\&{=\ }2\cdot\lim_{x\to\infty}x\sin\left(4\cdot\frac{1}{x}\right)\\&\quad... \lim_{x\to\infty}x\sin\left(a\cdot\frac{1}{x}\right)=a\\&{=\ }2\cdot4\\&{=\ }\boxed{\,\bf8\,}\end{aligned}[/tex]
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Nomor 4

[tex]\begin{aligned}&\lim_{x\to\infty}\frac{x\sin\left(\dfrac{1}{x}\right)}{1-\sin\left(\dfrac{1}{x}\right)}\\&{=\ }\frac{\lim\limits_{x\to\infty}x\sin\left(\dfrac{1}{x}\right)}{\lim\limits_{x\to\infty}\left[1-\sin\left(\dfrac{1}{x}\right)\right]}\\&{=\ }\frac{\lim\limits_{x\to\infty}x\sin\left(\dfrac{1}{x}\right)}{\lim\limits_{x\to\infty}1-\lim\limits_{x\to\infty}\sin\left(\dfrac{1}{x}\right)}\\&{=\ }\frac{\lim\limits_{x\to\infty}x\sin\left(\dfrac{1}{x}\right)}{1}\end{aligned}[/tex]
[tex]\begin{aligned}&{=\ }\lim_{x\to\infty}x\sin\left(\frac{1}{x}\right)\\&{=\ }\lim_{x\to\infty}x\sin\left(1\cdot\frac{1}{x}\right)\\&\quad... \lim_{x\to\infty}x\sin\left(a\cdot\frac{1}{x}\right)=a\\&{=\ }\boxed{\,\bf1\,}\end{aligned}[/tex]