# Limit Trigonometri

###### tags: `math` `calculus` `limits` `trigonometry`

## ★ Bentuk-bentuk Dasar

### ★ Hasil Tes Limit

$$
\frac{0}{\text{positif}}=\frac{0}{\text{negatif}}=0
$$

$$
\frac{\text{positif}}{0} = \infty
$$

$$
\frac{\text{negatif}}{0} = -\infty
$$

### ★ *Seven Deadly Sins*

Semua bentuk di bawah ini tidak terdefinisi:
$$
\begin{aligned}
1 &.\; \frac{0}{0} \\
2 &.\; \frac{\infty}{\infty} \\
3 &.\; 0 \times \infty \\
4 &.\; \infty - \infty \\
5 &.\; 0^0 \\
6 &.\; 1^\infty \\
7 &.\; \infty^0
\end{aligned}
$$

### ★ Limit Trigonometri Dasar

$$
\lim_{x\to0}
\begin{Bmatrix}
\frac{ax}{bx} & \frac{\sin ax}{bx} & \frac{\tan ax}{bx} \\
\frac{ax}{\sin bx} & \frac{\sin ax}{\sin bx} & \frac{\tan ax}{\sin bx} \\
\frac{ax}{\tan bx} & \frac{\sin ax}{\tan bx} & \frac{\tan ax}{\tan bx}
\end{Bmatrix} = \frac{a}{b}
$$

## ★★ Dalil L'Hôpital

:::warning
**Dalil L'Hôpital**
Jika $f(a) / g(a)=0/0$, maka
$$
\lim_{x\to a}\frac{f(x)}{g(x)}=
\lim_{x\to a}\frac{f'(x)}{g'(x)}
$$ dengan $f'(x)$ dan $g'(x)$ adalah turunan pertama dari $f(x)$ dan $g(x)$.
:::

| $\mathbf{\boldsymbol{f}(\boldsymbol{x})}$ | $\mathbf{\boldsymbol{f}'(\boldsymbol{x})}$ |
|:--------------------- | ---------------------------------------- |
| konstanta | $0$ |
| $a\cdot g(x)$ | $a \cdot g'(x)$ |
| $ax^n$ | $anx^{n-1}$ |
| $f(x)\cdot g(x)$ | $f'(x)\cdot g(x)+f(x)\cdot g'(x)$ |
| $$\frac{f(x)}{g(x)}$$ | $$\frac{f'(x)g(x) - f(x)g'(x)}{g^2(x)}$$ |
| $f(g(x))$ | $g'(x) \cdot f'(g(x))$ |
| $\sin x$ | $\cos x$ |
| $\cos x$ | $-\sin x$ |
| $\tan x$ | $\sec^2 x$ |

**Kapan menggunakan dalil ini?**
- Bentuk $\text{polinom}/\text{polinom}$
- Bentuk $\cos ax$

## WIP
:::spoiler
$$
{1. \;} \lim_{x \to 5} \frac{\sin(3x-15^\circ)}{\tan(2x-10^\circ)} = \lim_{x \to 5} \frac{\sin(3(x-5^\circ))}{\tan(2(x-5^\circ))} =\frac{3}{2}
$$

$$
\begin{aligned}
{4. \;} \lim_{x \to 0} \frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{x} &= \lim_{x \to 0} \left[ \frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{x} \cdot \frac{\sqrt{1+\sin x}-\sqrt{1+\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}} \right] \\
&=\lim_{x \to 0} \frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{x}
\end{aligned}
$$
:::