Rumus lim θ → 0 sin ( θ ) θ \lim_{\theta\to 0} \frac{\sin(\theta)}{\theta} lim θ → 0 θ s i n ( θ )
Menggunakan Intuisi
lim θ → 0 sin ( θ ) θ \lim_{\theta\to 0} \frac{\sin(\theta)}{\theta}
θ → 0 lim θ sin ( θ )
Untuk menganalisanya kita membutuhkan sebuah unit lingkaran. Kita akan menggunakan sin definisi unit
lingkaran.
sin ( θ ) = y \sin(\theta) = y
sin ( θ ) = y
Panjang busur di depan sudut θ \theta θ θ ⋅ r = θ \theta \cdot r = \theta θ ⋅ r = θ θ \theta θ y y y θ \theta θ θ = y \theta = y θ = y y θ = 1 \dfrac{y}{\theta} = 1 θ y = 1
lim θ → 0 sin ( θ ) θ = y θ = 1 \lim_{\theta\to 0} \frac{\sin(\theta)}{\theta} = \frac{y}{\theta} = 1
θ → 0 lim θ sin ( θ ) = θ y = 1
Menggunakan Teorema Apit
Kita dapat menggunakan definisi fungsi trigonometri menurut unit lingkaran.
Kita membuat segitiga siku-siku dengan jari-jari yang berada di sumbu x positif sebagai alas dan
perpanjangan jari-jari yang lain sebagai sisi-miring.
Kita bisa memecah bangun segitiga siku-siku paling besar menjadi 3 bangun.
Bangun 1
Pada bangun ke 1 berlaku:
sin ( θ ) = t 1 t = sin ( θ ) \begin{align*}
\sin(\theta) &= \frac{t}{1} \\
t &= \sin(\theta)
\end{align*}
sin ( θ ) t = 1 t = sin ( θ )
Luas dari bangun 1 adalah:
L = 1 2 × 1 × ∣ t ∣ = ∣ sin ( θ ) ∣ 2 \begin{align*}
L &= \frac{1}{2}\times 1\times |t|\\
&= \frac{|\sin(\theta)|}{2}
\end{align*}
L = 2 1 × 1 × ∣ t ∣ = 2 ∣ sin ( θ ) ∣
Nilai mutlak digunakan karena nilai θ \theta θ θ \theta θ
Bangun 2
Luas dari bangun 2 yang berbentuk juring adalah:
L = ∣ θ ∣ 2 π × L Lingkaran = ∣ θ ∣ 2 π × π r 2 = ∣ θ ∣ 2 π × π = ∣ θ ∣ 2 \begin{align*}
L &= \frac{|\theta|}{2\pi}\times L_{\text{Lingkaran}}\\
&= \frac{|\theta|}{2\pi}\times\pi r^2\\
&= \frac{|\theta|}{2\pi}\times\pi\\
&= \frac{|\theta|}{2}
\end{align*}
L = 2 π ∣ θ ∣ × L Lingkaran = 2 π ∣ θ ∣ × π r 2 = 2 π ∣ θ ∣ × π = 2 ∣ θ ∣
Bangun 3
Pada bangun 3 berlaku:
tan ( θ ) = t 1 t = tan ( θ ) \begin{align*}
\tan(\theta) &= \frac{t}{1} \\
t &= \tan(\theta)
\end{align*}
tan ( θ ) t = 1 t = tan ( θ )
Luas bangun ke 3:
L = 1 2 × 1 × ∣ t ∣ = ∣ tan ( θ ) ∣ 2 \begin{align*}
L &= \frac{1}{2}\times 1\times |t|\\
&= \frac{|\tan(\theta)|}{2}
\end{align*}
L = 2 1 × 1 × ∣ t ∣ = 2 ∣ tan ( θ ) ∣
Dari gambar di atas kita tahu bahwa:
Luas Bangun 1 ≤ Luas Bangun 2 ≤ Luas Bangun 3 ∣ sin ( θ ) ∣ 2 ≤ ∣ θ ∣ 2 ≤ ∣ tan ( θ ) ∣ 2 ∣ sin ( θ ) ∣ ≤ ∣ θ ∣ ≤ ∣ tan ( θ ) ∣ \begin{array}{ccc}
\text{Luas Bangun 1} &\le \text{Luas Bangun 2} &\le \text{Luas Bangun 3} \\
\\
\dfrac{|\sin(\theta)|}{2} &\le \dfrac{|\theta|}{2} &\le \dfrac{|\tan(\theta)|}{2} \\
\\
|\sin(\theta)| &\le |\theta| &\le |\tan(\theta)|
\end{array}
Luas Bangun 1 2 ∣ sin ( θ ) ∣ ∣ sin ( θ ) ∣ ≤ Luas Bangun 2 ≤ 2 ∣ θ ∣ ≤ ∣ θ ∣ ≤ Luas Bangun 3 ≤ 2 ∣ tan ( θ ) ∣ ≤ ∣ tan ( θ ) ∣
Semua ruas dibagi dengan ∣ sin ( θ ) ∣ |\sin(\theta)| ∣ sin ( θ ) ∣
∣ sin ( θ ) ∣ ∣ sin ( θ ) ∣ ≤ ∣ θ ∣ ∣ sin ( θ ) ∣ ≤ ∣ tan ( θ ) ∣ ∣ sin ( θ ) ∣ 1 ≤ ∣ θ sin ( θ ) ∣ ≤ ∣ sin ( θ ) cos ( θ ) ∣ ∣ sin ( θ ) ∣ 1 ≤ ∣ θ sin ( θ ) ∣ ≤ ∣ sin ( θ ) ∣ ∣ cos ( θ ) ∣ ∣ sin ( θ ) ∣ 1 ≤ ∣ θ sin ( θ ) ∣ ≤ 1 ∣ cos ( θ ) ∣ \begin{array}{rcl}
\dfrac{|\sin(\theta)|}{|\sin(\theta)|} &\le \dfrac{|\theta|}{|\sin(\theta)|} &\le \dfrac{|\tan(\theta)|}{|\sin(\theta)|} \\
\\
1 &\le \left|\dfrac{\theta}{\sin(\theta)}\right| &\le \dfrac{\left|\dfrac{\sin(\theta)}{\cos(\theta)}\right|}{|\sin(\theta)|} \\
\\
1 &\le \left|\dfrac{\theta}{\sin(\theta)}\right| &\le \dfrac{|\sin(\theta)|}{|\cos(\theta)||\sin(\theta)|} \\
\\
1 &\le \left|\dfrac{\theta}{\sin(\theta)}\right| &\le \dfrac{1}{|\cos(\theta)|} \\
\end{array}
∣ sin ( θ ) ∣ ∣ sin ( θ ) ∣ 1 1 1 ≤ ∣ sin ( θ ) ∣ ∣ θ ∣ ≤ sin ( θ ) θ ≤ sin ( θ ) θ ≤ sin ( θ ) θ ≤ ∣ sin ( θ ) ∣ ∣ tan ( θ ) ∣ ≤ ∣ sin ( θ ) ∣ cos ( θ ) sin ( θ ) ≤ ∣ cos ( θ ) ∣∣ sin ( θ ) ∣ ∣ sin ( θ ) ∣ ≤ ∣ cos ( θ ) ∣ 1
lim θ → 0 \lim_{\theta \to 0} lim θ → 0 θ \theta θ θ \theta θ cos ( θ ) \cos(\theta) cos ( θ )
Pada kuadran 1, nilai θ \theta θ θ sin ( θ ) \frac{\theta}{\sin(\theta)} s i n ( θ ) θ θ \theta θ θ sin ( θ ) \frac{\theta}{\sin(\theta)} s i n ( θ ) θ sin ( θ ) θ \frac{\sin(\theta)}{\theta} θ s i n ( θ )
Pertidaksamaan kita tuliskan lagi menjadi:
1 ≤ θ sin ( θ ) ≤ 1 cos ( θ ) 1 \le \frac{\theta}{\sin(\theta)} \le \frac{1}{\cos(\theta)}
1 ≤ sin ( θ ) θ ≤ cos ( θ ) 1
Kita ambil limit dari ketiga ruas:
lim θ → 0 1 ≤ lim θ → 0 θ sin ( θ ) ≤ lim θ → 0 1 cos ( θ ) 1 ≤ lim θ → 0 θ sin ( θ ) ≤ 1 cos ( 0 ) 1 ≤ lim θ → 0 θ sin ( θ ) ≤ 1 \begin{array}{rcl}
\lim_{\theta \to 0}1 &\le \lim_{\theta \to 0}\dfrac{\theta}{\sin(\theta)} &\le \lim_{\theta \to 0}\dfrac{1}{\cos(\theta)} \\
\\
1 &\le \lim_{\theta \to 0}\dfrac{\theta}{\sin(\theta)} &\le \dfrac{1}{\cos(0)} \\
\\
1 &\le \lim_{\theta \to 0}\dfrac{\theta}{\sin(\theta)} &\le 1 \\
\end{array}
lim θ → 0 1 1 1 ≤ lim θ → 0 sin ( θ ) θ ≤ lim θ → 0 sin ( θ ) θ ≤ lim θ → 0 sin ( θ ) θ ≤ lim θ → 0 cos ( θ ) 1 ≤ cos ( 0 ) 1 ≤ 1
Menrut teorema apit berlaku:
lim θ → 0 θ sin ( θ ) = 1 \lim_{\theta \to 0}\dfrac{\theta}{\sin(\theta)} = 1 \\
θ → 0 lim sin ( θ ) θ = 1
Persamaan tersebut bisa kita balik.
lim θ → 0 sin ( θ ) θ = lim θ → 0 1 ( θ sin ( θ ) ) = lim θ → 0 1 lim θ → 0 θ sin ( θ ) = 1 1 = 1 \begin{align*}
\lim_{\theta\to0}\frac{\sin(\theta)}{\theta} &= \lim_{\theta\to0}\frac{1}{\left(\dfrac{\theta}{\sin(\theta)}\right)}\\
& =\dfrac{\lim_{\theta\to0}1}{\lim_{\theta\to0}\dfrac{\theta}{\sin(\theta)}}\\
& =\frac{1}{1}\\
& =1
\end{align*}
θ → 0 lim θ sin ( θ ) = θ → 0 lim ( sin ( θ ) θ ) 1 = lim θ → 0 sin ( θ ) θ lim θ → 0 1 = 1 1 = 1
Perluasan
Penambahan Koefisien
Rumus tadi dapat diperluas dengan menambahkan koefisien pada θ \theta θ
lim θ → 0 sin ( a θ ) b θ = lim θ → 0 ( sin ( a θ ) b θ × a θ a θ ) = lim θ → 0 ( sin ( a θ ) a θ × a θ b θ ) = lim θ → 0 ( sin ( a θ ) a θ ) × lim θ → 0 ( a θ b θ ) \begin{align*}
\lim_{\theta\to0}\frac{\sin(a\theta)}{b\theta} & =\lim_{\theta\to0}\left(\frac{\sin(a\theta)}{b\theta}\times\frac{a\theta}{a\theta}\right)\\
& =\lim_{\theta\to0}\left(\frac{\sin(a\theta)}{a\theta}\times\frac{a\theta}{b\theta}\right)\\
& =\lim_{\theta\to0}\left(\frac{\sin(a\theta)}{a\theta}\right)\times\lim_{\theta\to0}\left(\frac{a\theta}{b\theta}\right)
\end{align*}
θ → 0 lim b θ sin ( a θ ) = θ → 0 lim ( b θ sin ( a θ ) × a θ a θ ) = θ → 0 lim ( a θ sin ( a θ ) × b θ a θ ) = θ → 0 lim ( a θ sin ( a θ ) ) × θ → 0 lim ( b θ a θ )
Kita misalkan x = a θ x=a\theta x = a θ θ → 0 \theta\to0 θ → 0 x → a × 0 x\to a\times 0 x → a × 0 x → 0 x\to0 x → 0
lim θ → 0 sin ( a θ ) b θ = lim x → 0 ( sin ( x ) x ) × lim θ → 0 ( a θ b θ ) = 1 × lim θ → 0 ( a b ) = a b \begin{align*}
\lim_{\theta\to0}\frac{\sin(a\theta)}{b\theta} & =\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)\times\lim_{\theta\to0}\left(\frac{a\theta}{b\theta}\right)\\
& =1\times\lim_{\theta\to0}\left(\frac{a}{b}\right)\\
& =\frac{a}{b}
\end{align*}
θ → 0 lim b θ sin ( a θ ) = x → 0 lim ( x sin ( x ) ) × θ → 0 lim ( b θ a θ ) = 1 × θ → 0 lim ( b a ) = b a
Rumus ini juga berlaku versi terbaliknya.
lim θ → 0 a θ sin ( b θ ) = lim θ → 0 ( a θ sin ( b θ ) × b θ b θ ) = lim θ → 0 ( b θ sin ( b θ ) × a θ b θ ) = lim θ → 0 ( b θ sin ( b θ ) ) × lim θ → 0 ( a θ b θ ) \begin{align*}
\lim_{\theta\to0}\frac{a\theta}{\sin(b\theta)} & =\lim_{\theta\to0}\left(\frac{a\theta}{\sin(b\theta)}\times\frac{b\theta}{b\theta}\right)\\
& =\lim_{\theta\to0}\left(\frac{b\theta}{\sin(b\theta)}\times\frac{a\theta}{b\theta}\right)\\
& =\lim_{\theta\to0}\left(\frac{b\theta}{\sin(b\theta)}\right)\times\lim_{\theta\to0}\left(\frac{a\theta}{b\theta}\right)
\end{align*}
θ → 0 lim sin ( b θ ) a θ = θ → 0 lim ( sin ( b θ ) a θ × b θ b θ ) = θ → 0 lim ( sin ( b θ ) b θ × b θ a θ ) = θ → 0 lim ( sin ( b θ ) b θ ) × θ → 0 lim ( b θ a θ )
Kita misalkan x = a θ x=a\theta x = a θ θ → 0 \theta\to0 θ → 0 x → a × 0 x\to a\times 0 x → a × 0 x → 0 x\to0 x → 0
lim θ → 0 a θ sin ( b θ ) = lim x → 0 ( x sin ( x ) ) × lim θ → 0 ( a θ b θ ) = 1 × lim θ → 0 ( a b ) = a b \begin{align*}
\lim_{\theta\to0}\frac{a\theta}{\sin(b\theta)} & =\lim_{x\to0}\left(\frac{x}{\sin(x)}\right)\times\lim_{\theta\to0}\left(\frac{a\theta}{b\theta}\right)\\
& =1\times\lim_{\theta\to0}\left(\frac{a}{b}\right)\\
& =\frac{a}{b}
\end{align*}
θ → 0 lim sin ( b θ ) a θ = x → 0 lim ( sin ( x ) x ) × θ → 0 lim ( b θ a θ ) = 1 × θ → 0 lim ( b a ) = b a
Perbandingan Sin dengan Sin
Perluasan lain dari rumus ini adalah perbandingan sin dengan sin.
lim θ → 0 sin ( a θ ) sin ( b θ ) = lim θ → 0 ( sin ( a θ ) sin ( b θ ) × a θ a θ × b θ b θ ) = lim θ → 0 ( sin ( a θ ) a θ × a θ b θ × b θ sin ( b θ ) ) = lim θ → 0 ( sin ( a θ ) a θ ) lim θ → 0 ( a θ b θ ) lim θ → 0 ( b θ sin ( b θ ) ) \begin{align*}
\lim_{\theta\to0}\frac{\sin(a\theta)}{\sin(b\theta)} & =\lim_{\theta\to0}\left(\frac{\sin(a\theta)}{\sin(b\theta)}\times\frac{a\theta}{a\theta}\times\frac{b\theta}{b\theta}\right)\\
& =\lim_{\theta\to0}\left(\frac{\sin(a\theta)}{a\theta}\times\frac{a\theta}{b\theta}\times\frac{b\theta}{\sin(b\theta)}\right)\\
& =\lim_{\theta\to0}\left(\frac{\sin(a\theta)}{a\theta}\right)\lim_{\theta\to0}\left(\frac{a\theta}{b\theta}\right)\lim_{\theta\to0}\left(\frac{b\theta}{\sin(b\theta)}\right)
\end{align*}
θ → 0 lim sin ( b θ ) sin ( a θ ) = θ → 0 lim ( sin ( b θ ) sin ( a θ ) × a θ a θ × b θ b θ ) = θ → 0 lim ( a θ sin ( a θ ) × b θ a θ × sin ( b θ ) b θ ) = θ → 0 lim ( a θ sin ( a θ ) ) θ → 0 lim ( b θ a θ ) θ → 0 lim ( sin ( b θ ) b θ )
Sekarang kita bisa menggunakan perluasan rumus ini yang kita temukan sebelumnya.
lim θ → 0 sin ( a θ ) sin ( b θ ) = lim θ → 0 ( sin ( a θ ) a θ ) lim θ → 0 ( a θ b θ ) lim θ → 0 ( b θ sin ( b θ ) ) = a a × lim θ → 0 ( a b ) × b b = 1 × a b × 1 = a b \begin{align*}
\lim_{\theta\to0}\frac{\sin(a\theta)}{\sin(b\theta)} & =\lim_{\theta\to0}\left(\frac{\sin(a\theta)}{a\theta}\right)\lim_{\theta\to0}\left(\frac{a\theta}{b\theta}\right)\lim_{\theta\to0}\left(\frac{b\theta}{\sin(b\theta)}\right)\\
& =\frac{a}{a}\times\lim_{\theta\to0}\left(\frac{a}{b}\right)\times\frac{b}{b}\\
& =1\times\frac{a}{b}\times1\\
& =\frac{a}{b}
\end{align*}
θ → 0 lim sin ( b θ ) sin ( a θ ) = θ → 0 lim ( a θ sin ( a θ ) ) θ → 0 lim ( b θ a θ ) θ → 0 lim ( sin ( b θ ) b θ ) = a a × θ → 0 lim ( b a ) × b b = 1 × b a × 1 = b a
Rumus lim θ → 0 tan ( θ ) θ \lim_{\theta\to 0} \frac{\tan(\theta)}{\theta} lim θ → 0 θ t a n ( θ )
Rumus ini dapat kita selesaikan dengan hanya mengubah tan menjadi perbandingan sin dan cos.
lim θ → 0 tan ( θ ) θ = lim θ → 0 ( sin ( θ ) cos ( θ ) ) θ = lim θ → 0 sin ( θ ) cos ( θ ) θ \begin{align*}
\lim_{\theta\to0}\frac{\tan(\theta)}{\theta} & =\lim_{\theta\to0}\frac{\left(\dfrac{\sin(\theta)}{\cos(\theta)}\right)}{\theta}\\
& =\lim_{\theta\to0}\frac{\sin(\theta)}{\cos(\theta)\theta}
\end{align*}
θ → 0 lim θ tan ( θ ) = θ → 0 lim θ ( cos ( θ ) sin ( θ ) ) = θ → 0 lim cos ( θ ) θ sin ( θ )
Kita bisa menggunakan rumus lim θ → 0 sin ( θ ) θ \lim_{\theta\to 0} \frac{\sin(\theta)}{\theta} lim θ → 0 θ s i n ( θ )
lim θ → 0 tan ( θ ) θ = lim θ → 0 sin ( θ ) cos ( θ ) θ = lim θ → 0 ( ( sin ( θ ) θ ) ( 1 cos ( θ ) ) ) = lim θ → 0 ( sin ( θ ) θ ) lim θ → 0 ( 1 cos ( θ ) ) = 1 × 1 cos ( 0 ) = 1 × 1 1 = 1 \begin{align*}
\lim_{\theta\to0}\frac{\tan(\theta)}{\theta} & =\lim_{\theta\to0}\frac{\sin(\theta)}{\cos(\theta)\theta}\\
& =\lim_{\theta\to0}\left(\left(\frac{\sin(\theta)}{\theta}\right)\left(\frac{1}{\cos(\theta)}\right)\right)\\
& =\lim_{\theta\to0}\left(\frac{\sin(\theta)}{\theta}\right)\lim_{\theta\to0}\left(\frac{1}{\cos(\theta)}\right)\\
& =1\times\frac{1}{\cos(0)}\\
& =1\times\frac{1}{1}\\
& =1
\end{align*}
θ → 0 lim θ tan ( θ ) = θ → 0 lim cos ( θ ) θ sin ( θ ) = θ → 0 lim ( ( θ sin ( θ ) ) ( cos ( θ ) 1 ) ) = θ → 0 lim ( θ sin ( θ ) ) θ → 0 lim ( cos ( θ ) 1 ) = 1 × cos ( 0 ) 1 = 1 × 1 1 = 1
Rumus ini bisa kita balik.
lim θ → 0 θ tan ( θ ) = lim θ → 0 1 ( tan ( θ ) θ ) = lim θ → 0 1 lim θ → 0 tan ( θ ) θ = 1 1 = 1 \begin{align*}\lim_{\theta\to0}\frac{\theta}{\tan(\theta)} & =\lim_{\theta\to0}\frac{1}{\left(\dfrac{\tan(\theta)}{\theta}\right)}\\
& =\dfrac{\lim_{\theta\to0}1}{\lim_{\theta\to0}\dfrac{\tan(\theta)}{\theta}}\\
& =\frac{1}{1}\\
& =1
\end{align*}
θ → 0 lim tan ( θ ) θ = θ → 0 lim ( θ tan ( θ ) ) 1 = lim θ → 0 θ tan ( θ ) lim θ → 0 1 = 1 1 = 1
Perluasan
Perluasan rumus ini juga mirip perluasan rumus sebelumnya.
Penambahan Koefisien
Rumus tadi dapat diperluas dengan menambahkan koefisien pada θ \theta θ
lim θ → 0 tan ( a θ ) b θ = lim θ → 0 ( tan ( a θ ) b θ × a θ a θ ) = lim θ → 0 ( tan ( a θ ) a θ × a θ b θ ) = lim θ → 0 ( tan ( a θ ) a θ ) × lim θ → 0 ( a θ b θ ) \begin{align*}
\lim_{\theta\to0}\frac{\tan(a\theta)}{b\theta} & =\lim_{\theta\to0}\left(\frac{\tan(a\theta)}{b\theta}\times\frac{a\theta}{a\theta}\right)\\
& =\lim_{\theta\to0}\left(\frac{\tan(a\theta)}{a\theta}\times\frac{a\theta}{b\theta}\right)\\
& =\lim_{\theta\to0}\left(\frac{\tan(a\theta)}{a\theta}\right)\times\lim_{\theta\to0}\left(\frac{a\theta}{b\theta}\right)
\end{align*}
θ → 0 lim b θ tan ( a θ ) = θ → 0 lim ( b θ tan ( a θ ) × a θ a θ ) = θ → 0 lim ( a θ tan ( a θ ) × b θ a θ ) = θ → 0 lim ( a θ tan ( a θ ) ) × θ → 0 lim ( b θ a θ )
Kita misalkan x = a θ x=a\theta x = a θ θ → 0 \theta\to0 θ → 0 x → a × 0 x\to a\times 0 x → a × 0 x → 0 x\to0 x → 0
lim θ → 0 tan ( a θ ) b θ = lim x → 0 ( tan ( x ) x ) × lim θ → 0 ( a θ b θ ) = 1 × lim θ → 0 ( a b ) = a b \begin{align*}
\lim_{\theta\to0}\frac{\tan(a\theta)}{b\theta} & =\lim_{x\to0}\left(\frac{\tan(x)}{x}\right)\times\lim_{\theta\to0}\left(\frac{a\theta}{b\theta}\right)\\
& =1\times\lim_{\theta\to0}\left(\frac{a}{b}\right)\\
& =\frac{a}{b}
\end{align*}
θ → 0 lim b θ tan ( a θ ) = x → 0 lim ( x tan ( x ) ) × θ → 0 lim ( b θ a θ ) = 1 × θ → 0 lim ( b a ) = b a
Rumus ini juga berlaku versi terbaliknya.
lim θ → 0 a θ tan ( b θ ) = lim θ → 0 ( a θ tan ( b θ ) × b θ b θ ) = lim θ → 0 ( b θ tan ( b θ ) × a θ b θ ) = lim θ → 0 ( b θ tan ( b θ ) ) × lim θ → 0 ( a θ b θ ) \begin{align*}
\lim_{\theta\to0}\frac{a\theta}{\tan(b\theta)} & =\lim_{\theta\to0}\left(\frac{a\theta}{\tan(b\theta)}\times\frac{b\theta}{b\theta}\right)\\
& =\lim_{\theta\to0}\left(\frac{b\theta}{\tan(b\theta)}\times\frac{a\theta}{b\theta}\right)\\
& =\lim_{\theta\to0}\left(\frac{b\theta}{\tan(b\theta)}\right)\times\lim_{\theta\to0}\left(\frac{a\theta}{b\theta}\right)
\end{align*}
θ → 0 lim tan ( b θ ) a θ = θ → 0 lim ( tan ( b θ ) a θ × b θ b θ ) = θ → 0 lim ( tan ( b θ ) b θ × b θ a θ ) = θ → 0 lim ( tan ( b θ ) b θ ) × θ → 0 lim ( b θ a θ )
Kita misalkan x = a θ x=a\theta x = a θ θ → 0 \theta\to0 θ → 0 x → a × 0 x\to a\times 0 x → a × 0 x → 0 x\to0 x → 0
lim θ → 0 a θ tan ( b θ ) = lim x → 0 ( x tan ( x ) ) × lim θ → 0 ( a θ b θ ) = 1 × lim θ → 0 ( a b ) = a b \begin{align*}
\lim_{\theta\to0}\frac{a\theta}{\tan(b\theta)} & =\lim_{x\to0}\left(\frac{x}{\tan(x)}\right)\times\lim_{\theta\to0}\left(\frac{a\theta}{b\theta}\right)\\
& =1\times\lim_{\theta\to0}\left(\frac{a}{b}\right)\\
& =\frac{a}{b}
\end{align*}
θ → 0 lim tan ( b θ ) a θ = x → 0 lim ( tan ( x ) x ) × θ → 0 lim ( b θ a θ ) = 1 × θ → 0 lim ( b a ) = b a
Perbandingan Tan dengan Tan
Perluasan lain dari rumus ini adalah perbandingan tan dengan tan.
lim θ → 0 tan ( a θ ) tan ( b θ ) = lim θ → 0 ( tan ( a θ ) tan ( b θ ) × a θ a θ × b θ b θ ) = lim θ → 0 ( tan ( a θ ) a θ × a θ b θ × b θ tan ( b θ ) ) = lim θ → 0 ( tan ( a θ ) a θ ) lim θ → 0 ( a θ b θ ) lim θ → 0 ( b θ tan ( b θ ) ) \begin{align*}
\lim_{\theta\to0}\frac{\tan(a\theta)}{\tan(b\theta)} & =\lim_{\theta\to0}\left(\frac{\tan(a\theta)}{\tan(b\theta)}\times\frac{a\theta}{a\theta}\times\frac{b\theta}{b\theta}\right)\\
& =\lim_{\theta\to0}\left(\frac{\tan(a\theta)}{a\theta}\times\frac{a\theta}{b\theta}\times\frac{b\theta}{\tan(b\theta)}\right)\\
& =\lim_{\theta\to0}\left(\frac{\tan(a\theta)}{a\theta}\right)\lim_{\theta\to0}\left(\frac{a\theta}{b\theta}\right)\lim_{\theta\to0}\left(\frac{b\theta}{\tan(b\theta)}\right)
\end{align*}
θ → 0 lim tan ( b θ ) tan ( a θ ) = θ → 0 lim ( tan ( b θ ) tan ( a θ ) × a θ a θ × b θ b θ ) = θ → 0 lim ( a θ tan ( a θ ) × b θ a θ × tan ( b θ ) b θ ) = θ → 0 lim ( a θ tan ( a θ ) ) θ → 0 lim ( b θ a θ ) θ → 0 lim ( tan ( b θ ) b θ )
Sekarang kita bisa menggunakan perluasan rumus ini yang kita temukan sebelumnya.
lim θ → 0 tan ( a θ ) tan ( b θ ) = lim θ → 0 ( tan ( a θ ) a θ ) lim θ → 0 ( a θ b θ ) lim θ → 0 ( b θ tan ( b θ ) ) = a a × lim θ → 0 ( a b ) × b b = 1 × a b × 1 = a b \begin{align*}
\lim_{\theta\to0}\frac{\tan(a\theta)}{\tan(b\theta)} & =\lim_{\theta\to0}\left(\frac{\tan(a\theta)}{a\theta}\right)\lim_{\theta\to0}\left(\frac{a\theta}{b\theta}\right)\lim_{\theta\to0}\left(\frac{b\theta}{\tan(b\theta)}\right)\\
& =\frac{a}{a}\times\lim_{\theta\to0}\left(\frac{a}{b}\right)\times\frac{b}{b}\\
& =1\times\frac{a}{b}\times1\\
& =\frac{a}{b}
\end{align*}
θ → 0 lim tan ( b θ ) tan ( a θ ) = θ → 0 lim ( a θ tan ( a θ ) ) θ → 0 lim ( b θ a θ ) θ → 0 lim ( tan ( b θ ) b θ ) = a a × θ → 0 lim ( b a ) × b b = 1 × b a × 1 = b a
Rumus Perbandingan Sin dan Tan
Dua rumus yang kita temukan sebelumnya bisa kita gabungkan ubtuk membuat rumus perbandingan sin
dengan tan.
lim θ → 0 sin ( a θ ) tan ( b θ ) = lim θ → 0 ( sin ( a θ ) tan ( b θ ) × a θ a θ × b θ b θ ) = lim θ → 0 ( sin ( a θ ) a θ × a θ b θ × b θ tan ( b θ ) ) = lim θ → 0 ( sin ( a θ ) a θ ) lim θ → 0 ( a θ b θ ) lim θ → 0 ( b θ tan ( b θ ) ) = a a × lim θ → 0 ( a θ b θ ) × b b = 1 × a b × 1 = a b \begin{align*}
\lim_{\theta\to0}\frac{\sin(a\theta)}{\tan(b\theta)} & =\lim_{\theta\to0}\left(\frac{\sin(a\theta)}{\tan(b\theta)}\times\frac{a\theta}{a\theta}\times\frac{b\theta}{b\theta}\right)\\
& =\lim_{\theta\to0}\left(\frac{\sin(a\theta)}{a\theta}\times\frac{a\theta}{b\theta}\times\frac{b\theta}{\tan(b\theta)}\right)\\
& =\lim_{\theta\to0}\left(\frac{\sin(a\theta)}{a\theta}\right)\lim_{\theta\to0}\left(\frac{a\theta}{b\theta}\right)\lim_{\theta\to0}\left(\frac{b\theta}{\tan(b\theta)}\right)\\
& =\frac{a}{a}\times\lim_{\theta\to0}\left(\frac{a\theta}{b\theta}\right)\times\frac{b}{b}\\
& =1\times\frac{a}{b}\times1\\
& =\frac{a}{b}
\end{align*}
θ → 0 lim tan ( b θ ) sin ( a θ ) = θ → 0 lim ( tan ( b θ ) sin ( a θ ) × a θ a θ × b θ b θ ) = θ → 0 lim ( a θ sin ( a θ ) × b θ a θ × tan ( b θ ) b θ ) = θ → 0 lim ( a θ sin ( a θ ) ) θ → 0 lim ( b θ a θ ) θ → 0 lim ( tan ( b θ ) b θ ) = a a × θ → 0 lim ( b θ a θ ) × b b = 1 × b a × 1 = b a
Terdapat juga rumus perbandingan tan dengan sin.
lim θ → 0 tan ( a θ ) sin ( b θ ) = lim θ → 0 ( tan ( a θ ) sin ( b θ ) × a θ a θ × b θ b θ ) = lim θ → 0 ( tan ( a θ ) a θ × a θ b θ × b θ sin ( b θ ) ) = lim θ → 0 ( tan ( a θ ) a θ ) lim θ → 0 ( a θ b θ ) lim θ → 0 ( b θ sin ( b θ ) ) = a a × lim θ → 0 ( a θ b θ ) × b b = 1 × a b × 1 = a b \begin{align*}\lim_{\theta\to0}\frac{\tan(a\theta)}{\sin(b\theta)} & =\lim_{\theta\to0}\left(\frac{\tan(a\theta)}{\sin(b\theta)}\times\frac{a\theta}{a\theta}\times\frac{b\theta}{b\theta}\right)\\
& =\lim_{\theta\to0}\left(\frac{\tan(a\theta)}{a\theta}\times\frac{a\theta}{b\theta}\times\frac{b\theta}{\sin(b\theta)}\right)\\
& =\lim_{\theta\to0}\left(\frac{\tan(a\theta)}{a\theta}\right)\lim_{\theta\to0}\left(\frac{a\theta}{b\theta}\right)\lim_{\theta\to0}\left(\frac{b\theta}{\sin(b\theta)}\right)\\
& =\frac{a}{a}\times\lim_{\theta\to0}\left(\frac{a\theta}{b\theta}\right)\times\frac{b}{b}\\
& =1\times\frac{a}{b}\times1\\
& =\frac{a}{b}
\end{align*}
θ → 0 lim sin ( b θ ) tan ( a θ ) = θ → 0 lim ( sin ( b θ ) tan ( a θ ) × a θ a θ × b θ b θ ) = θ → 0 lim ( a θ tan ( a θ ) × b θ a θ × sin ( b θ ) b θ ) = θ → 0 lim ( a θ tan ( a θ ) ) θ → 0 lim ( b θ a θ ) θ → 0 lim ( sin ( b θ ) b θ ) = a a × θ → 0 lim ( b θ a θ ) × b b = 1 × b a × 1 = b a
Rumus lim θ → 0 cos ( θ ) − 1 θ \lim_{\theta\to 0} \frac{\cos(\theta)-1}{\theta} lim θ → 0 θ c o s ( θ ) − 1
Pertama, kita bisa mengalikan sekawan pembilan.
lim θ → 0 cos ( θ ) − 1 θ = lim θ → 0 ( cos ( θ ) − 1 θ × cos ( θ ) + 1 cos ( θ ) + 1 ) = lim θ → 0 cos ( θ ) 2 − 1 θ ( cos ( θ ) + 1 ) \begin{align*}
\lim_{\theta\to0}\frac{\cos(\theta)-1}{\theta} & =\lim_{\theta\to0}\left(\frac{\cos(\theta)-1}{\theta}\times\frac{\cos(\theta)+1}{\cos(\theta)+1}\right)\\
& =\lim_{\theta\to0}\frac{\cos(\theta)^{2}-1}{\theta(\cos(\theta)+1)}
\end{align*}
θ → 0 lim θ cos ( θ ) − 1 = θ → 0 lim ( θ cos ( θ ) − 1 × cos ( θ ) + 1 cos ( θ ) + 1 ) = θ → 0 lim θ ( cos ( θ ) + 1 ) cos ( θ ) 2 − 1
Sekarang pembilang seperti mempunyai hubungan dengan identitas phytagoras.
sin ( θ ) 2 + cos ( θ ) 2 = 1 sin ( θ ) 2 = 1 − cos ( θ ) 2 − sin ( θ ) 2 = cos ( θ ) 2 − 1 \begin{align*}
\sin(\theta)^{2}+\cos(\theta)^{2} & =1\\
\sin(\theta)^{2} & =1-\cos(\theta)^{2}\\
-\sin(\theta)^{2} & =\cos(\theta)^{2}-1
\end{align*}
sin ( θ ) 2 + cos ( θ ) 2 sin ( θ ) 2 − sin ( θ ) 2 = 1 = 1 − cos ( θ ) 2 = cos ( θ ) 2 − 1
Kita masukkan kembali persamaan tadi.
lim θ → 0 cos ( θ ) − 1 θ = lim θ → 0 cos ( θ ) 2 − 1 θ ( cos ( θ ) + 1 ) = lim θ → 0 − sin ( θ ) 2 θ ( cos ( θ ) + 1 ) = lim θ → 0 − sin ( θ ) sin ( θ ) θ ( cos ( θ ) + 1 ) \begin{align*}
\lim_{\theta\to0}\frac{\cos(\theta)-1}{\theta} & =\lim_{\theta\to0}\frac{\cos(\theta)^{2}-1}{\theta(\cos(\theta)+1)}\\
& =\lim_{\theta\to0}\frac{-\sin(\theta)^{2}}{\theta(\cos(\theta)+1)}\\
& =\lim_{\theta\to0}\frac{-\sin(\theta)\sin(\theta)}{\theta(\cos(\theta)+1)}
\end{align*}
θ → 0 lim θ cos ( θ ) − 1 = θ → 0 lim θ ( cos ( θ ) + 1 ) cos ( θ ) 2 − 1 = θ → 0 lim θ ( cos ( θ ) + 1 ) − sin ( θ ) 2 = θ → 0 lim θ ( cos ( θ ) + 1 ) − sin ( θ ) sin ( θ )
Kita bisa menggunakan rumus lim θ → 0 sin ( θ ) θ \lim_{\theta\to 0} \frac{\sin(\theta)}{\theta} lim θ → 0 θ s i n ( θ )
lim θ → 0 cos ( θ ) − 1 θ = lim θ → 0 − sin ( θ ) sin ( θ ) θ ( cos ( θ ) + 1 ) = lim θ → 0 ( ( sin ( θ ) θ ) ( − sin ( θ ) cos ( θ ) + 1 ) ) = lim θ → 0 ( sin ( θ ) θ ) lim θ → 0 ( − sin ( θ ) cos ( θ ) + 1 ) = 1 × − sin ( 0 ) cos ( 0 ) + 1 = 1 × − 0 1 + 1 = 1 × 0 = 0 \begin{align*}
\lim_{\theta\to0}\frac{\cos(\theta)-1}{\theta} & =\lim_{\theta\to0}\frac{-\sin(\theta)\sin(\theta)}{\theta(\cos(\theta)+1)}\\
& =\lim_{\theta\to0}\left(\left(\frac{\sin(\theta)}{\theta}\right)\left(\frac{-\sin(\theta)}{\cos(\theta)+1}\right)\right)\\
& =\lim_{\theta\to0}\left(\frac{\sin(\theta)}{\theta}\right)\lim_{\theta\to0}\left(\frac{-\sin(\theta)}{\cos(\theta)+1}\right)\\
& =1\times\frac{-\sin(0)}{\cos(0)+1}\\
& =1\times\frac{-0}{1+1}\\
& =1\times0\\
& =0
\end{align*}
θ → 0 lim θ cos ( θ ) − 1 = θ → 0 lim θ ( cos ( θ ) + 1 ) − sin ( θ ) sin ( θ ) = θ → 0 lim ( ( θ sin ( θ ) ) ( cos ( θ ) + 1 − sin ( θ ) ) ) = θ → 0 lim ( θ sin ( θ ) ) θ → 0 lim ( cos ( θ ) + 1 − sin ( θ ) ) = 1 × cos ( 0 ) + 1 − sin ( 0 ) = 1 × 1 + 1 − 0 = 1 × 0 = 0
Kesimpulan
lim θ → 0 θ sin ( θ ) = 1 lim θ → 0 sin ( θ ) θ = 1 lim θ → 0 sin ( a θ ) b θ = a b lim θ → 0 a θ sin ( b θ ) = a b lim θ → 0 θ tan ( θ ) = 1 lim θ → 0 tan ( θ ) θ = 1 lim θ → 0 tan ( a θ ) b θ = a b lim θ → 0 a θ tan ( b θ ) = a b lim θ → 0 sin ( a θ ) tan ( b θ ) = a b lim θ → 0 tan ( a θ ) sin ( b θ ) = a b lim θ → 0 cos ( θ ) − 1 θ = 0 \begin{align*}
\lim_{\theta \to 0}\dfrac{\theta}{\sin(\theta)} &= 1 \\
\lim_{\theta\to0}\frac{\sin(\theta)}{\theta} &= 1\\
\lim_{\theta\to0}\frac{\sin(a\theta)}{b\theta} & = \frac{a}{b}\\
\lim_{\theta\to0}\frac{a\theta}{\sin(b\theta)} & =\frac{a}{b}\\
\lim_{\theta \to 0}\dfrac{\theta}{\tan(\theta)} &= 1 \\
\lim_{\theta\to0}\frac{\tan(\theta)}{\theta} &= 1\\
\lim_{\theta\to0}\frac{\tan(a\theta)}{b\theta} & = \frac{a}{b}\\
\lim_{\theta\to0}\frac{a\theta}{\tan(b\theta)} & =\frac{a}{b}\\
\lim_{\theta\to0}\frac{\sin(a\theta)}{\tan(b\theta)} &=\frac{a}{b}\\
\lim_{\theta\to0}\frac{\tan(a\theta)}{\sin(b\theta)} &=\frac{a}{b}\\
\lim_{\theta\to0}\frac{\cos(\theta)-1}{\theta} & =0\\
\end{align*}
θ → 0 lim sin ( θ ) θ θ → 0 lim θ sin ( θ ) θ → 0 lim b θ sin ( a θ ) θ → 0 lim sin ( b θ ) a θ θ → 0 lim tan ( θ ) θ θ → 0 lim θ tan ( θ ) θ → 0 lim b θ tan ( a θ ) θ → 0 lim tan ( b θ ) a θ θ → 0 lim tan ( b θ ) sin ( a θ ) θ → 0 lim sin ( b θ ) tan ( a θ ) θ → 0 lim θ cos ( θ ) − 1 = 1 = 1 = b a = b a = 1 = 1 = b a = b a = b a = b a = 0
