Latihan Limit Trigonometri 02

Gunakan identitas: \(\color{blue} \sin^2 x + \cos^2 x = 1\) \(\color{blue} \cos x \:-\: \cos y = -2\sin \frac{1}{2} (x + y) \sin \frac{1}{2} (x \:-\: y)\)   \(\lim\limits_{x \rightarrow 0} \dfrac{4(\cos 17x \:-\:\cos 7x)}{\sin^2 8x}\) \(\lim\limits_{x \rightarrow 0} \dfrac{4(-2\sin \frac{1}{2} (17x + 7x) \sin \frac{1}{2} (17x \:-\:7x)}{\sin^2 8x}\) \(\lim\limits_{x \rightarrow 0} \dfrac{4(-2\sin 12x \sin 5x)}{\sin^2 8x}\) \(\lim\limits_{x \rightarrow 0} \dfrac{-8 \cdot \sin 12x  \cdot \sin 5x}{\sin 8x \cdot \sin 8x}\) \(-\cancel{8} \times \dfrac{12}{\cancel{8}} \times \dfrac{5}{8}\) \(-7,5\)   Jadi, \(\lim\limits_{x \rightarrow 0} \dfrac{4\cos 17x \:-\:4\cos 7x}{1\:-\:\cos^2 8x} = -7,5\)

\(\lim\limits_{x \rightarrow 0} [1 + (-2\sin^2 3x)]^{\tfrac{\sin 4x \:-\: 4\sin x}{\tan^5 2x}}\)

\(\color{blue} \text{Note: }\)

\(\color{blue} \lim\limits_{x \rightarrow 0} [1 + f(x)]^{g(x)} = e^{\lim\limits_{x \rightarrow 0} f(x)\cdot g(x)}\)

\(\color{blue} \sin 2x = 2\sin x \cos x\)

\(\color{blue} \cos 2x = 1\:-\:2\sin^2 x\)

\(e^{\lim\limits_{x \rightarrow 0} -2\sin^2 3x \cdot \tfrac{\sin 4x \:-\: 4\sin x}{\tan^5 2x}}\) \(e^{\lim\limits_{x \rightarrow 0} -2\sin^2 3x \cdot \tfrac{\sin 4x \:-\: 4\sin x}{\tan^2 2x \cdot \tan^3 2x}}\) \(e^{\lim\limits_{x \rightarrow 0} \color{red}\tfrac{-2\sin^2 3x}{\tan^2 2x}\color{black}\cdot \tfrac{\sin 4x \:-\: 4\sin x}{\tan^3 2x}}\) \(e^{\color{red}-2\cdot (\tfrac{3}{2})^2 \color{black}\lim\limits_{x \rightarrow 0} \cdot \tfrac{2\sin 2x\cdot \cos 2x \:-\: 4\sin x}{\tan^3 2x}}\) \(e^{\color{red}-\tfrac{9}{2}\color{black}\lim\limits_{x \rightarrow 0} \cdot \tfrac{2\cdot 2\sin x \cos x \cdot \cos 2x \:-\: 4\sin x}{\tan^3 2x}}\) \(e^{\color{red}-\tfrac{9}{2}\color{black}\lim\limits_{x \rightarrow 0} \cdot \tfrac{4\sin x (\cos x \cdot \cos 2x \:-\: 1)}{\tan 2x \cdot \tan^2 2x}}\) \(e^{\color{red}-\tfrac{9}{2}\color{black}\lim\limits_{x \rightarrow 0} \cdot \tfrac{4\sin x}{\tan 2x}\cdot \tfrac{\cos x \cdot \cos 2x \:-\: 1}{\tan^2 2x}}\) \(e^{\color{red}-\tfrac{9}{2}\cdot \tfrac{4}{2} \color{black}\cdot \lim\limits_{x \rightarrow 0} \tfrac{\cos x \cdot \cos 2x \:-\: 1}{\tan^2 2x}}\) \(e^{\color{red}-9\color{black}\cdot \lim\limits_{x \rightarrow 0} \tfrac{\cos x \cdot (1\:-\:2\sin^2 x) \:-\: 1}{\tan^2 2x}}\) \(e^{\color{red}-9\color{black}\cdot \lim\limits_{x \rightarrow 0} \tfrac{\cos x\:-\:2\cdot \cos x \cdot \sin^2 x \:-\: 1}{\tan^2 2x}}\) \(e^{\color{red}-9\color{black}\cdot \lim\limits_{x \rightarrow 0} \tfrac{\cos x\:-\: 1 \: – \: 2\cdot \cos x \cdot \sin^2 x}{\tan^2 2x}}\) \(e^{\color{red}-9\color{black}\cdot \lim\limits_{x \rightarrow 0} \tfrac{1\:-\:2\sin^2 \frac{1}{2}x\:-\: 1 \: – \: 2\cdot \cos x \cdot \sin^2 x}{\tan^2 2x}}\) \(e^{\color{red}-9\color{black}\cdot \lim\limits_{x \rightarrow 0} \tfrac{-2\sin^2 \frac{1}{2}x\: – \: 2\cdot \cos x \cdot \sin^2 x}{\tan^2 2x}}\) \(e^{\color{red}-9\color{black}\cdot \lim\limits_{x \rightarrow 0} \tfrac{-2\sin^2 \frac{1}{2}x}{\tan^2 2x}\: – \: \tfrac{2\cdot \cos x \cdot \sin^2 x}{\tan^2 2x}}\) \(e^{-9[-2\cdot (\tfrac{1}{4})^2\: – \: 2\cdot (\tfrac{1}{2})^2]}\) \(e^{-9[-2\cdot \tfrac{1}{16}\: – \: 2\cdot \tfrac{1}{4}]}\) \(e^{-9[-\tfrac{1}{8}\: – \: \tfrac{1}{2}]}\) \(e^{-9[-\tfrac{1}{8}\: – \: \tfrac{4}{8}]}\) \(e^{-9(-\tfrac{5}{8})}\) \(e^{\tfrac{45}{8}}\)

\(\color{blue} \text{Note: }\) \(\color{blue}\cos x \:-\:\cos y = -2\sin \tfrac{1}{2}(x + y) \sin \tfrac{1}{2}(x \:-\: y)\) \(\color{blue}\text{Jumlah deret geometri tak hingga } s_{\infty} = \dfrac{a}{1\:-\:r}\) \(\lim\limits_{x \rightarrow 0} \left(\dfrac{1}{1\:-\:x}\right)^{\tfrac{\sin 2x}{-2\sin \tfrac{1}{2} (5x + x) \sin \tfrac{1}{2} (5x\:-\:x)}}\) \(\lim\limits_{x \rightarrow 0} \left(\dfrac{1\:-\:x + x}{1\:-\:x}\right)^{\tfrac{\sin 2x}{-2\sin \tfrac{1}{2} (6x) \sin \tfrac{1}{2} (4x)}}\) \(\lim\limits_{x \rightarrow 0} \left(1 + \dfrac{x}{1\:-\:x}\right)^{\tfrac{\cancel{\sin 2x}}{-2\sin 3x\cdot \cancel{\sin 2x}}}\) \(e^{\lim\limits_{x \rightarrow 0} \tfrac{x}{1\:-\:x}\cdot \tfrac{1}{-2\sin 3x}}\) \(e^{\lim\limits_{x \rightarrow 0} \tfrac{x}{-2\sin 3x}\cdot \tfrac{1}{1\:-\:x}}\) \(e^{-\tfrac{1}{6}\cdot \lim\limits_{x \rightarrow 0}\tfrac{1}{1\:-\:x}}\) \(e^{-\tfrac{1}{6}\cdot\tfrac{1}{1\:-\:0}}\) \(e^{-\tfrac{1}{6}\cdot\tfrac{1}{1\:-\:0}}\) \(e^{-\tfrac{1}{6}}\)

Misal \(\frac{1}{x} = y\) Jika \(x \rightarrow \infty\), maka \(y \rightarrow 0\) \(\lim\limits_{y \rightarrow 0} \dfrac{\tan 5y\:-\:\tan 4y \:-\:\tan y}{\sin 10y\:-\:2\sin 5y}\) \(\lim\limits_{y \rightarrow 0} \dfrac{\tan 5y\:-\:(\tan 4y + \tan y)}{2\sin 5y\cdot \cos 5y\:-\:2\sin 5y}\) \(\lim\limits_{y \rightarrow 0} \dfrac{\tan 5y\:-\:[\tan 5y(1\:-\:\tan4y\cdot \tan y)]}{2\sin 5y(\cos 5y\:-\:1)}\) \(\lim\limits_{y \rightarrow 0} \dfrac{\tan 5y\:-\:[\tan 5y\:-\:\tan 5y\cdot \tan4y\cdot \tan y]}{2\sin 5y(1\:-\:2\sin^2 \frac{5}{2}y\:-\:1)}\) \(\lim\limits_{y \rightarrow 0} \dfrac{\tan 5y\cdot \tan4y\cdot \tan y}{-4\sin 5y \cdot \sin^2 \frac{5}{2}y}\) \(\lim\limits_{y \rightarrow 0} \dfrac{\tan 5y}{-4\sin 5y}\cdot \dfrac{\tan 4y}{\sin \frac{5}{2}y} \cdot \dfrac{\tan y}{\sin \frac{5}{2}y}\) \(-\frac{1}{\cancel{4}} \cdot \frac{\cancelto{2}{8}}{5} \cdot \frac{2}{5}\) \(-\frac{4}{25}\) \(-0,16\)

Misal \(\dfrac{1}{x} = y\) Jika \(x \rightarrow \infty\), maka \(y \rightarrow 0\)   \(\lim\limits_{y\rightarrow 0} \dfrac{3\:-\:\cos 6y \:-\:\cos 8y \:-\:\cos 10y}{\tan 4y \sin 8y}\) Gunakan rumus sudut rangkap \(\color{blue}\cos 2x = 1\:-\:2\sin^2x\)   \(\lim\limits_{y\rightarrow 0} \dfrac{3\:-\:(1\:-\:2\sin^2 3y) \:-\:(1\:-\:2\sin^2 4y) \:-\:(1\:-\:2\sin^2 5y)}{\tan 4y \sin 8y}\) \(\lim\limits_{y\rightarrow 0} \dfrac{2\sin^2 3y + 2\sin^2 4y + 2\sin^2 5y}{\tan 4y \sin 8y}\) Bagi pembilang dan penyebut dengan \(\color{blue} \sin^2y\) \(\dfrac{2(3)^2 + 2(4)^2 + 2(5)^2}{4 \cdot 8}\) \(\dfrac{18 + 32 + 50}{32}\) \(\dfrac{100}{32}\) \(\dfrac{25}{8}\)

Gunakan rumus: \(\color{blue}\lim\limits_{x \rightarrow 0} (1 + f(x))^{g(x)} = e^{\lim\limits_{x \rightarrow 0} f(x) \cdot g(x)}\)   \(\lim\limits_{x \rightarrow 0} \left(1 + 8x \right)^{\dfrac{1}{4} \cdot \dfrac{\sin 10x}{2\:-\:\cos^2 x \:-\:\cos^2 3x}}\) \(\lim\limits_{x \rightarrow 0} \left(1 + 8x \right)^{\dfrac{1}{4} \cdot \dfrac{\sin 10x}{2 \: – \: (1\:-\:\sin^2x) \:-\:(1\:-\:\sin^2 3x)}}\) \(\lim\limits_{x \rightarrow 0} \left(1 + 8x \right)^{\dfrac{1}{4} \cdot \dfrac{\sin 10x}{\sin^2x + \sin^2 3x}}\) \(\lim\limits_{x \rightarrow 0} \left(1 + 8x \right)^{\dfrac{1}{4} \cdot \dfrac{\sin 10x}{\sin^2x +\sin^2 3x}}\) \(e^{\lim\limits_{x \rightarrow 0} 8x \cdot \dfrac{1}{4} \cdot \dfrac{\sin 10x}{\sin^2x +\sin^2 3x}}\) \(e^{\lim\limits_{x \rightarrow 0} \dfrac{2x \cdot \sin 10x}{\sin^2x +\sin^2 3x}}\) Bagi bagian pembilang dan penyebut dengan \(x^2\) \(e^{\lim\limits_{x \rightarrow 0} \dfrac{\frac{2\sin 10x}{x}}{\frac{\sin^2x +\sin^2 3x}{x^2}}}\) \(e^{\frac{20}{1 + 9}}\) \(e^{2}\)