Materi: Matematika (Limit)
A. Aturan Dasar Pengerjaan
- Jika \( f(x) = k \), maka \[ \lim_{x \to a} f(x) = k, \quad \text{dengan } k \text{ konstanta},\ k, a \in \mathbb{R} \]
- Jika \( f(x) = x \), maka \[ \lim_{x \to a} f(x) = a \]
- \[ \lim_{x \to a} \{f(x) \pm g(x)\} = \lim_{x \to a} f(x) \pm \lim_{x \to a} g(x) \]
- Jika \( k \) konstanta, maka \[ \lim_{x \to a} k \cdot f(x) = k \cdot \lim_{x \to a} f(x) \]
- \[ \lim_{x \to a} \{f(x) \cdot g(x)\} = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x) \]
- \[ \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}, \quad \text{dengan } \lim_{x \to a} g(x) \ne 0 \]
- \[ \lim_{x \to a} \{f(x)\}^n = \left( \lim_{x \to a} f(x) \right)^n \]
A. Rumus-Rumus Penting
Model 1
Cara Cepat
Bentuk:
\[
\lim_{x \to c} \frac{f(x)}{g(x)} \quad \text{dan hasilnya } \frac{0}{0} \text{ atau } \frac{\infty}{\infty}
\]
Gunakan dalil L'Hospital jika:
\[
\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{f(c)}{g(c)} = \frac{0}{0}
\]
Penyelesaian:
\[
\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} = \frac{f'(c)}{g'(c)} \ne \frac{0}{0}
\]
(Turunkan atas dan bawah, lalu substitusikan nilai \( c \))
Model 2
Cara Cepat
\[ \lim_{x \to 0} \frac{a x}{\sqrt{n + b x} - \sqrt{n + c x}} = \frac{2 a \sqrt{n}}{b - c} \] dan \[ \lim_{x \to 0} \frac{\sqrt{n + b x} - \sqrt{n + c x}}{a x} = \frac{b - c}{2 a \sqrt{n}} \]Model 3
Cara Cepat
Bentuk: \[ \lim_{x \to \infty} \frac{a x^m + \cdots}{b x^n + \cdots} \] Keterangan: \( m \) = pangkat tertinggi pembilang, \( n \) = pangkat tertinggi penyebut. \[ \lim_{x \to \infty} \frac{a x^m + \cdots}{b x^n + \cdots} = \begin{cases} \infty, & \text{jika } m > n \\ \frac{a}{b}, & \text{jika } m = n \\ 0, & \text{jika } m < n \end{cases} \]Model 4
Cara Cepat
\[ \lim_{x \to \infty} \left( \sqrt{a x^2 + b x + c} - \sqrt{p x^2 + q x + r} \right) \] \[ = \begin{cases} +\infty, & \text{jika } a > p \\ \dfrac{b - q}{2 \sqrt{a}}, & \text{jika } a = p \\ -\infty, & \text{jika } a < p \end{cases} \]Variasinya:
\[ \lim_{x \to \infty} \frac{ \sqrt{a x^2 + b x + c} - \sqrt{p x^2 + q x + r} }{ \sqrt{m x^2 + n x + s + \dots} } \] Jika \( \sqrt{a} - \sqrt{p} - \sqrt{m} + \dots = 0 \), maka: \[ \lim = \frac{b}{2\sqrt{a}} - \frac{q}{2\sqrt{p}} - \frac{n}{2\sqrt{m}} + \dots \] ---Model 5
Cara Cepat
Bentuk: \[ \lim_{x \to \infty} \sqrt{a x^2 + b x + c} - p x + q, \quad \text{dengan } p^2 = a \] \[ \Rightarrow \lim = \frac{b}{2\sqrt{a}} + q \] ---Model 6
Cara Cepat
\[ \lim_{x \to \infty} \left( \sqrt[3]{a x^3 + b x^2 + c x + d} - \sqrt[3]{a x^3 + m x^2 + n x + s} \right) \] \[ = \frac{b - m}{3 \sqrt[3]{a^2}} \] ---Model 7
Rumus Praktis – Cara Cepat
\[ \lim_{x \to \infty} \left( \sqrt{a x \pm b} - \sqrt{c x \pm d} \right) \] \[ = \begin{cases} \infty, & \text{untuk } a > c \\ 0, & \text{untuk } a = c \\ -\infty, & \text{untuk } a < c \end{cases} \]Limit Fungsi Trigonometri
Perhatikan Polanya!
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1\]\[ \lim_{x \to 0} \frac{x}{\sin x} = 1 \] \[ \lim_{x \to 0} \frac{\tan x}{x} = 1\]\[ \lim_{x \to 0} \frac{x}{\tan x} = 1 \]
\[ \lim_{x \to 0} \frac{\sin(mx)}{nx} = \frac{m}{n}\]\[ \lim_{x \to 0} \frac{\tan(mx)}{nx} = \frac{m}{n} \] \[ \lim_{x \to 0} \frac{mx}{\sin(nx)} = \frac{m}{n}\]\[ \lim_{x \to 0} \frac{mx}{\tan(nx)} = \frac{m}{n} \]
\[ \lim_{x \to a} \frac{\sin(m(x-a))}{n(x-a)} = \frac{m}{n}\]\[ \lim_{x \to a} \frac{\tan(m(x-a))}{n(x-a)} = \frac{m}{n} \] \[ \lim_{x \to a} \frac{m(x-a)}{\sin(n(x-a))} = \frac{m}{n}\]\[ \lim_{x \to a} \frac{m(x-a)}{\tan(n(x-a))} = \frac{m}{n} \]
Rumus Bantu yang Sering Dipakai
- \(\sin^2 x + \cos^2 x = 1\)
- \(\sin 2x = 2 \sin x \cos x\)
- \(\cos 2x = \cos^2 x - \sin^2 x\)
- \(1 - \cos 2x = 2 \sin^2 x\)
- \(1 + \cos 2x = 2 \cos^2 x\)