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Guess:
\[ \lim_{x \to 0} \frac{B}{R} = 1 \] \[ \lim_{x \to 0} \frac{G}{B} = 0 \]
\[ \left. \begin{aligned} \lim_{x \to 0} \frac{\sin x}{x} &= 1 \\[1em] \lim_{x \to 0} \frac{\cos x - 1}{x} &= 0 \end{aligned} \right\} \quad \text{visually.} \] \[ \lim_{x \to 0} \frac{x}{\sin x} = 1 \] \[ \lim_{x \to 0} \frac{\tan x}{x} = \lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{1}{\cos x} = 1 \]
Eg: Evaluate \[\lim_{x \to 0} \frac{\sin(2x)}{x}\]
\[\lim_{x \to 0} \frac{\sin(2x)}{2x} = 1\] \[\lim_{x \to 0} \frac{\sin(mx)}{mx} = 1\] \[\lim_{x \to 0} \frac{\sin(2x)}{x} = \lim_{x \to 0} \frac{\sin(2x)}{2x} \cdot 2 = 1 \cdot 2 = 2\]
for \(\theta \to 0 \quad \sin\theta \approx \theta\)
As \( x \to 0 \):
\[ \sin(2x) \approx 2x \] \[ \sin(3x) \approx 3x \] \[ \lim_{x \to 0} \frac{\sin(2x)}{\sin(3x)} = \lim_{x \to 0} \frac{2x}{3x} = \frac{2}{3} \]
\[ \lim_{x \to 0} \frac{\sin(2x)}{\sin(3x)} = \lim_{x \to 0} \frac{\frac{\sin 2x}{x}}{\frac{\sin 3x}{x}} = \frac{2}{3} \]
\( N: \lim_{x \to 0} \frac{\sin 2x}{x} = 2 \)
\( D: \lim_{x \to 0} \frac{\sin 3x}{x} = 3 \)
\[ \frac{d}{dx}(\sin x) \]
\[ \frac{df}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
\[ \frac{d}{dx}(\sin x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin(x)}{h} \]
use:
\[ \sin(A+B) = \sin A \cos B + \cos A \sin B \]
\[ \sin(x+h) = \sin x \cos h + \cos x \sin h \]
\[ \frac{d}{dx}(\sin x) = \lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h} \]
\[ = \lim_{h \to 0} \sin x \left[ \frac{\cos h - 1}{h} \right]^{\to 0} + \cos x \left[ \frac{\sin h}{h} \right]^{\to 1} \]
\[ = \cos x \]
\[ \frac{d}{dx}(\cos x) = \lim_{h \to 0} \frac{\cos(x+h) - \cos x}{h} \]
\[ \cos(A+B) = \cos A \cos B - \sin A \sin B \]
\[ \cos(x+h) = \cos x \cos h - \sin x \sin h \]
\[ \frac{d}{dx}[\cos x] = \lim_{h \to 0} \frac{\cos x \cos h - \sin x \sin h - \cos x}{h} \]
\[ = \lim_{h \to 0} \cos x \left[ \frac{\cos h - 1}{h} \right] - \sin x \left[ \frac{\sin h}{h} \right] \]
\[ = -\sin x \]
\[ \frac{d}{dx}(\tan x) = \frac{d}{dx}\left[\frac{\sin x}{\cos x}\right] \]
\[ N = \sin x \rightsquigarrow \frac{dN}{dx} = \cos x \] \[ D = \cos x \rightsquigarrow \frac{dD}{dx} = -\sin x \]
Quotient Rule:
\[ \frac{d}{dx}\left(\frac{N}{D}\right) = \frac{N'D - ND'}{D^2} \]
\[ \frac{d}{dx}(\tan x) = \frac{d}{dx}\left(\frac{\sin x}{\cos x}\right) = \frac{\cos x \cos x - \sin x(-\sin x)}{\cos^2 x} \]
\[ = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x \]
\[ \frac{d}{dx}\tan x = \sec^2 x \]
\[ \frac{d}{dx}(\sec x) = \frac{d}{dx}\left[\frac{1}{\cos x}\right] \]
\( N = 1, \; N' = 0, \quad D = \cos x, \; D' = -\sin x \)
\[ = \frac{N'D - ND'}{D^2} = \frac{0(\cos x) - 1(-\sin x)}{\cos^2 x} \] \[ = \frac{\sin x}{\cos^2 x} = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} \] \[ = \tan x \sec x \] \[ \frac{d}{dx}(\sec x) = \tan x \sec x \]
\[ \frac{d}{dx} \cot x = \frac{d}{dx} \left[ \frac{1}{\tan x} \right] \] \[ = \frac{\left(\frac{d}{dx} 1\right)^{\to 0} \tan x - 1 \cdot \left(\frac{d}{dx} \tan x\right)^{\to \sec^2 x}}{(\tan x)^2} \] \[ = \frac{-\sec^2 x}{\tan^2 x} = \frac{-\left(\frac{1}{\cos^2 x}\right)}{\left(\frac{\sin^2 x}{\cos^2 x}\right)} = -\frac{1}{\sin^2 x} \] \[ = -\operatorname{cosec}^2 x \]
\[ \frac{d}{dx} [\operatorname{cosec} x] = \frac{d}{dx} \left[ \frac{1}{\sin x} \right] \]
\[ = \frac{\left(\frac{d}{dx} 1\right) \sin x - 1 \cdot \frac{d}{dx} \sin x}{\sin^2 x} \]
\[ = -\frac{\cos x}{\sin^2 x} = -\frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} \]
\[ = -\cot x \operatorname{cosec} x \]
\[ \frac{d}{dx}(\sin x) = \cos x \] \[ \frac{d}{dx}(\cos x) = -\sin x \] \[ \frac{d}{dx}(\tan x) = \sec^2 x \] \[ \frac{d}{dx}(\sec x) = \sec x \tan x \] \[ \frac{d}{dx}(\cot x) = -\operatorname{cosec}^2 x \] \[ \frac{d}{dx}(\operatorname{cosec} x) = -\operatorname{cosec} x \cot x \]
\[ \frac{d}{dx}\left[\frac{\tan x - 1}{\sec x}\right] \]
\[ \frac{d}{dx}\left[\frac{\frac{\sin x}{\cos x} - 1}{\frac{1}{\cos x}}\right] \] \[ = \frac{d}{dx}\left[\sin x - \cos x\right] \] \[ = \cos x + \sin x \]
Derivatives of the components:
\[ \frac{d}{dx}(\tan x - 1) = \sec^2 x \] \[ \frac{d}{dx}\sec x = \sec x \tan x \]
Use Quotient Rule:
\[ \frac{d}{dx}\left[\frac{\tan x - 1}{\sec x}\right] \] \[ = \frac{(\sec^2 x)\sec x - (\tan x - 1)\sec x \tan x}{\sec^2 x} \] \[ = \frac{\sec^2 x \sec x - \tan^2 x \sec x + \sec x \tan x}{\sec^2 x} \]
Using the identity \(\sec^2 x - \tan^2 x = 1\):
\[ = \frac{\sec x + \sec x \tan x}{\sec^2 x} \] \[ = \dots = \sin x + \cos x \]
eg:
\[ \frac{d}{dx} (\underbrace{\cos x \tan x}_{\sin x} - e^x \sec x) \] \[ = \frac{d}{dx} \sin x - \frac{d}{dx} [e^x \sec x] \] \[ = \cos x - \left[ \frac{d}{dx} e^x \sec x + e^x \frac{d}{dx} \sec x \right] \] \[ = \cos x - [e^x \sec x + e^x \sec x \tan x] \]