Question

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[tex]1. \: \displaystyle\lim_{x \to 0} \frac{ \sin(x)}{3x} = \dots[/tex]
[tex]2. \: \displaystyle\lim_{x \to 0} \frac{ \sin(3x) }{ \tan(3x) } = \dots[/tex]
[tex]3. \: \displaystyle\lim_{x \to 0} \frac{ \sin(x) + x }{ \tan(2x) } = \dots[/tex]
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Selamat Berjuang​

Answer 1

Penjelasan dengan langkah-langkah:

Limit FungSi

L'Hopital Metod

1)

[tex]\displaystyle\rm\lim_{x\to0}\frac{\sin(x)}{3x}\\ \\ \begin{aligned}&=\displaystyle\rm\lim_{x\to0} \frac{ \frac{d}{dx} \sin(x) }{ \frac{d}{dx}3x } \\ \\&=\displaystyle\rm\lim_{x\to0} \frac{ \cos(x) }{3} \\ \\& = \rm \frac{ \cos(0) }{3} \\ \\& = \rm \frac{1}{3} \end{aligned}[/tex]

2)

[tex]\displaystyle\rm\lim_{x\to0}\frac{\sin(x)}{\tan(x)}\\ \\ \begin{aligned}&=\displaystyle\rm\lim_{x\to0} \frac{ \frac{d}{dx} \sin(x) }{ \frac{d}{dx}\tan(x) } \\ \\&=\displaystyle\rm\lim_{x\to0} \frac{ \cos(x) }{\frac{1}{\cos(x)^2}} \\ \\& = \rm \cos(x)^3\\ \\& = \rm \cos(0)^3\\ \\ &=\rm 1 \end{aligned}[/tex]

3)

[tex]\displaystyle\rm\lim_{x\to0}\frac{\sin(x)+x}{\tan(2x)}\\ \\ \begin{aligned}&=\displaystyle\rm\lim_{x\to0} \frac{ \frac{d}{dx} \sin(x)+\frac{d}{dx}x }{ \frac{d}{dx}\tan(2x) } \\ \\&=\displaystyle\rm\lim_{x\to0} \frac{ \cos(x)+1 }{2\sec(x)^2} \\ \\& = \rm \frac{\cos(0)+1}{2\sec(0)^2}\\ \\& = \rm \frac{1+1}{2\times1}\\\\&=\rm 1 \end{aligned}[/tex]

Answer 2

Verified answer

[tex]\begin{aligned}\sf 1.\ \:&\lim_{x\to0}\frac{\sin x}{3x}=\boxed{\,\bf\frac{1}{3}\,}\\\vphantom{\Bigg|}\sf 2.\ \:&\lim_{x\to0}\frac{\sin(3x)}{\tan(3x)}=\boxed{\,\bf1\,}\\\sf 3.\ \:&\lim_{x\to0}\frac{\sin(x)+x}{\tan(2x)}=\boxed{\,\bf1\,}\end{aligned}[/tex]
 

Penjelasan

Limit Fungsi Trigonometri

Nomor 1

[tex]\begin{aligned}\lim_{x\to0}\frac{\sin x}{ax}&=\frac{1}{a}\left(\lim_{x\to0}\frac{\sin x}{x}\right)\\&=\frac{1}{a}\left(\lim_{x\to0}\frac{\sin x}{x}\right)\end{aligned}[/tex]

Misalkan terdapat sebuah lingkaran dengan panjang jari-jari = 1 satuan.

[tex]\sin x[/tex] adalah nilai tinggi segitiga sama kaki dari juring lingkaran dengan besar sudut pusat [tex]x[/tex]. Jarak dari garis tinggi segitiga tersebut ke pusat lingkaran adalah [tex]cos x[/tex], yang sama dengan [tex]\sin x[/tex].

Jika kita buat juring lingkaran yang lebih kecil dengan besar sudut pusat [tex]x[/tex] dan panjang jari-jari [tex]\cos x[/tex], panjang busurnya adalah:

[tex]\begin{aligned}\left(\frac{x}{\cancel{2\pi}}\right)\cancel{2\pi}\cdot\cos x=x\cos x\end{aligned}[/tex]
Perhatikan bahwa panjang busur ini pasti kurang dari [tex]\sin x[/tex].

Sedangkan panjang busur dari juring lingkaran yang lebih besar adalah:

[tex]\begin{aligned}\left(\frac{x}{\cancel{2\pi}}\right)\cancel{2\pi}\cdot1=x\end{aligned}[/tex]

Maka, hubungannya adalah:

[tex]\begin{aligned}x\cos x \le \sin x \le x\end{aligned}[/tex]

Ketiga ruas dibagi [tex]x[/tex], kita peroleh:

[tex]\begin{aligned}\cos x \le \frac{\sin x}{x} \le 1\end{aligned}[/tex]

Ketika [tex]x[/tex] mendekati 0:

[tex]\begin{aligned}&\lim_{x\to0}\cos x \le \lim_{x\to0}\frac{\sin x}{x} \le 1\\&\Leftrightarrow \cos 0 \le \lim_{x\to0}\frac{\sin x}{x} \le 1\\&\Leftrightarrow 1 \le \lim_{x\to0}\frac{\sin x}{x} \le 1\\&\Leftrightarrow \lim_{x\to0}\frac{\sin x}{x}=1\end{aligned}[/tex]

Oleh karena itu:

[tex]\begin{aligned}\lim_{x\to0}\frac{\sin x}{ax}&=\frac{1}{a}\left(\lim_{x\to0}\frac{\sin x}{x}\right)\\&=\frac{1}{a}\end{aligned}[/tex]

Dengan demikian, nilai limit yang kita cari adalah:

[tex]\begin{aligned}&\lim_{x\to0}\frac{\sin x}{3x}=\boxed{\,\bf\frac{1}{3}\,}\end{aligned}[/tex]
____________

Nomor 2

[tex]\begin{aligned}\lim_{x\to0}\frac{\sin(3x)}{\tan(3x)}&=\lim_{x\to0}\left(\cancel{\sin(3x)}\cdot\frac{\cos(3x)}{\cancel{\sin(3x)}}\right)\\&=\lim_{x\to0}\cos(3x)\\&=\cos0\\\lim_{x\to0}\frac{\sin(3x)}{\tan(3x)}&=\boxed{\,\bf1\,}\end{aligned}[/tex]
____________

Nomor 3

[tex]\begin{aligned}&\lim_{x\to0}\frac{\sin(x)+x}{\tan(2x)}\\&{=\ }\lim_{x\to0}\left(\frac{\sin(x)}{\tan(2x)}+\frac{x}{\tan(2x)}\right)\\&{=\ }\lim_{x\to0}\frac{\sin(x)}{\tan(2x)}+\lim_{x\to0}\frac{x}{\tan(2x)}\\&{=\ }\lim_{x\to0}\frac{\sin(x)\cos(2x)}{\sin(2x)}+\lim_{x\to0}\frac{x\cos(2x)}{\sin(2x)}\\&{=\ }\lim_{x\to0}\frac{\cancel{\sin(x)}\cos(2x)}{2\cancel{\sin(x)}\cos x}+\lim_{x\to0}\frac{x\cos(2x)}{2\sin x\cos x}\end{aligned}[/tex]
[tex]\begin{aligned}&{=\ }\frac{1}{2}\left[\lim_{x\to0}\frac{\cos(2x)}{\cos x}+\lim_{x\to0}\frac{x\cos(2x)}{\sin x\cos x}\right]\\&\quad\textsf{dengan pengecualian pada bentuk tak tentu}:\\&{=\ }\frac{1}{2}\left[1+\lim_{x\to0}\frac{x}{\sin x}\cdot\lim_{x\to0}\frac{\cos(2x)}{\cos x}\right]\\&{=\ }\frac{1}{2}\left[1+1\cdot1\right]\\&{=\ }\boxed{\,\bf1\,}\end{aligned}[/tex]
 
 
[tex]\overline{\begin{array}{l}\small\textsf{Duc In Altum}\\\small\text{bertolaklah\;ke\;tempat}\\\small\text{yang\;lebih\;dalam}\end{array}}[/tex]