Latihan Limit Trigonometri 01

\(\text{Note:}\) \(\color{red} \cos 2x = 1\:-\:\sin^2 x\) \(\color{red} \cos \frac{3}{2} x = 1\:-\:\sin^2 \frac{3}{4} x\) \(\lim\limits_{x \rightarrow 0} \dfrac{1\:-\:(1 \:-\: 2\sin^2 \frac{3}{4}x)}{4x^2}\) \(\lim\limits_{x \rightarrow 0} \dfrac{2\sin^2 \frac{3}{4}x}{4x^2}\) \(\lim\limits_{x \rightarrow 0} \dfrac{2}{4}\times \dfrac{\sin^2 \frac{3}{4}x}{x^2}\) \(\dfrac{2}{4}\times \left(\dfrac{3}{4}\right)^2\) \(\dfrac{9}{32}\)

\(\lim\limits_{x \rightarrow 0} \dfrac{\sin(\pi + 0) + \tan 0}{0 + \sin \left(\dfrac{\pi}{2} + 0 \right)}\) \(\dfrac{0}{1}\) \(0\)

\(\lim\limits_{x \rightarrow 0} \dfrac{6\sin x }{\tan 2x + 2\tan x}\) Bagi pembilang dan penyebut dengan \(x\) \(\lim\limits_{x \rightarrow 0} \dfrac{\dfrac{6\sin x}{x} }{\dfrac{\tan 2x}{x} + \dfrac{2\tan x}{x}}\) \(\dfrac{6}{2 + 2}\) \(\dfrac{3}{2}\)

\(\text{Note:}\) \(\color{red}\cos x \:-\: \cos y = -2\sin \frac{1}{2}(x + y)\sin \frac{1}{2}(x \:-\: y)\) \(\lim\limits_{x \rightarrow 0} \dfrac{2(1\:-\:\cos 4x)}{-2\sin \frac{1}{2}(3x + 5x)\sin \frac{1}{2}(3x \:-\: 5x)}\) \(\lim\limits_{x \rightarrow 0} \dfrac{2[1\:-\:(1\:-\:2\sin^2 2x)]}{-2\sin \frac{1}{2}(8x)\sin \frac{1}{2}(-2x)}\) \(\lim\limits_{x \rightarrow 0} \dfrac{4\sin^2 2x}{-2\sin 4x\sin (-x)}\) \(\lim\limits_{x \rightarrow 0} \dfrac{\cancelto{2}{4}\sin^2 2x}{\cancel{2}\sin 4x\sin x}\) \(2\lim\limits_{x \rightarrow 0} \dfrac{\sin 2x \cdot \sin 2x}{\sin 4x \cdot \sin x}\) \(2\cdot \dfrac{2}{4} \cdot \dfrac{2}{1}\) \(2\)

\(\lim\limits_{x \rightarrow -2} \dfrac{\sin (11\:-\:2x)(2 + x)}{(x + 5)\tan 5(2 + x)}\) Bagi pembilang dan penyebut dengan \((2 + x)\) \(\lim\limits_{x \rightarrow -2} \dfrac{\dfrac {\sin (11\:-\:2x)(2 + x)}{(2 + x)}}{\dfrac{(x + 5)\tan 5(2 + x)}{(2 + x)}}\) \(\lim\limits_{x \rightarrow -2} \dfrac{11\:-\: 2x}{5(x + 5)}\) \(\dfrac{11\:-\: 2(-2)}{5(-2 + 5)}\) \(\dfrac{11 + 4}{15}\) \(\dfrac{15}{15}\) \(1\)

\(\lim\limits_{a \rightarrow b} \dfrac{a\:-\:b}{\sin[a^2\:-\:b^2 + 2(a\:-\:b)] + \tan (4b\:-\:4a)}\) \(\lim\limits_{a \rightarrow b} \dfrac{a\:-\:b}{\sin[(a +b)(a\:-\:b) + 2(a\:-\:b)]\:-\:\tan 4(a \:-\:b)}\) \(\lim\limits_{a \rightarrow b} \dfrac{a\:-\:b}{\sin(a \:-\: b)(a + b + 2)\:-\:\tan 4(a \:-\:b)}\) Bagi pembilang dan penyebut dengan \((a\:-\:b)\) \(\lim\limits_{a \rightarrow b} \dfrac{\dfrac{a\:-\:b}{a\:-\:b}}{\dfrac{\sin (a\:-\:b)(a + b \:-\:2)}{a\:-\:b} \:-\:\dfrac{\tan 4(a \:-\:b)}{a\:-\:b}}\) \(\lim\limits_{a \rightarrow b} \dfrac{1}{a + b + 2\:-\:4}\) \(\dfrac{1}{2b\:-\:2}\)

\(\lim\limits_{x \rightarrow 0} \dfrac{\sqrt{1 + \sin 6x}\:-\:\sqrt{1\:-\:\sin 6x}}{\tan 3x}\times \color{red}\dfrac{\sqrt{1 + \sin 6x}+\sqrt{1\:-\:\sin 6x}}{\sqrt{1 + \sin 6x}+\sqrt{1\:-\:\sin 6x}}\) \(\lim\limits_{x \rightarrow 0} \dfrac{1 + \sin 6x \:-\: (1 \:-\: \sin 6x)}{\tan 3x(\sqrt{1 + \sin 6x}+\sqrt{1\:-\:\sin 6x})}\) \(\lim\limits_{x \rightarrow 0} \dfrac{2\sin 6x}{\tan 3x(\sqrt{1 + \sin 6x}+\sqrt{1\:-\:\sin 6x})}\) \(2\cdot \dfrac{6}{3}\cdot \dfrac{1}{(\sqrt{1} + \sqrt{1})}\) \(4 \cdot \dfrac{1}{2}\) \(2\)

\(\lim\limits_{x \rightarrow \frac{\pi}{2}} \dfrac{\dfrac {(\sin x \:-\: 1)}{\sin x}[\sin \frac{3}{2}x + \sin (\frac{\pi}{2}\:-\: \frac{3}{2}x)]}{1 + (2\cos^2 x \:-\: 1)}\) \(\lim\limits_{x \rightarrow \frac{\pi}{2}} \dfrac{(\sin x \:-\: 1)[2\sin \frac{1}{2}(\frac{\pi}{2})\cos \frac{1}{2}(3x\:-\:\frac{\pi}{2})]}{\sin x (2\cos^2 x)}\) \(\lim\limits_{x \rightarrow \frac{\pi}{2}} \dfrac{(\sin x \:-\: 1)[2\sin \frac{1}{2}(\frac{\pi}{2})\cos \frac{1}{2}(3x\:-\:\frac{\pi}{2})]}{\sin x (2\cos^2 x)}\) \(\lim\limits_{x \rightarrow \frac{\pi}{2}} \dfrac{(\sin x \:-\: \sin \dfrac{\pi}{2})[\sqrt{2}\cos (\frac{3}{2}x \:-\:\frac{\pi}{4})]}{2\sin x \cdot \cos x \cdot \cos x}\) \(\text{Note:}\) \(\color{red}\sin x \:-\: \sin y = 2\cos \frac{1}{2}(x + y)\sin \frac{1}{2} (x\:-\:y)\) \(\lim\limits_{x \rightarrow \frac{\pi}{2}} \dfrac{\sqrt{2}\cdot 2\cos \frac{1}{2}(x + \frac{\pi}{2})\cdot \cancelto{-\frac{1}{2}}{\sin \frac{1}{2}(x\:-\:\frac{\pi}{2})}\cdot\sin[\frac{\pi}{2}\:-\:(\frac{3}{2}x\:-\:\frac{\pi}{4})]}{2\sin \frac {\pi}{2}\cdot \cancel{\sin (\frac{\pi}{2}\:-\: x)}\cdot \sin (\frac{\pi}{2}\:-\:x)}\) \(\lim\limits_{x \rightarrow \frac{\pi}{2}} \dfrac{-\sqrt{2}\cdot \cos (\frac{1}{2}x + \frac{\pi}{4})\cdot \sin(\frac{3\pi}{4}\:-\:\frac{3}{2}x)}{2\sin(\frac{\pi}{2}\:-\:x)}\) \(\lim\limits_{x \rightarrow \frac{\pi}{2}} \dfrac{-\sqrt{2}\cdot\cos (\frac{1}{2}x + \frac{\pi}{4})\cdot \cancelto{\frac{3}{2}}{\sin \frac{3}{2}(\frac{\pi}{2}\:-\:x)}}{2\cancel{\sin(\frac{\pi}{2}\:-\:x)}}\) \(0\)

\(\lim\limits_{x \rightarrow \frac{\pi}{4}} \dfrac{\sin \frac{\pi}{2}\:-\:\sin 2x}{(4x\:-\:\pi)\sin 4x}\) \(\lim\limits_{x \rightarrow \frac{\pi}{4}} \dfrac{2\cos \frac{1}{2}(\frac{\pi}{2} + 2x)\sin \frac{1}{2}(\frac{\pi}{2} \:-\: 2x)}{(4x\:-\:\pi)\sin 4x}\) \(\lim\limits_{x \rightarrow \frac{\pi}{4}} \dfrac{2\cos (\frac{\pi}{4} + x)\sin (\frac{\pi}{4} \:-\: x)}{4(x\:-\:\frac{\pi}{4})\sin 4x}\) \(\lim\limits_{x \rightarrow \frac{\pi}{4}} \dfrac{2}{4}\cdot \dfrac{\cos (\frac{\pi}{4} + x)}{\sin 4x} \cdot \dfrac{\cancelto{-1}{\sin (\frac{\pi}{4} \:-\: x)}}{\cancel{(x\:-\:\frac{\pi}{4})}}\) \(-\dfrac{2}{4}\times \lim_{x \rightarrow \frac{\pi}{4}} \dfrac{\cos(\frac{\pi}{4} + x)}{\sin 4x}\) \(-\dfrac{2}{4}\times \lim_{x \rightarrow \frac{\pi}{4}} \dfrac{\sin [\frac{\pi}{2}\:-\:(\frac{\pi}{4} + x)]}{\sin(\pi \:-\:4x)}\) \(-\dfrac{2}{4}\times \lim_{x \rightarrow \frac{\pi}{4}} \dfrac{\sin (\frac{\pi}{4}\:-\:x)}{\sin 4(\frac{\pi}{4}\:-\:x)}\) \(-\dfrac{2}{4}\times \dfrac{1}{4}\) \(-\dfrac{2}{16}\) \(-\dfrac{1}{8}\)

\(\text{Note:}\) \(\color{red} \sin^2 x + \cos^2 x = 1\) \(\color{red} \cos 2x = 1\:-\:2\sin^2 x\) \(\lim\limits_{x \rightarrow 0} \dfrac{2\:-\:(1\:-\:\sin^2 4x) \:-\: (1\:-\:\sin^2 8x)}{2\:-\: (1\:-\:2\sin^2 x) \:-\: (1\:-\:\sin^2 2x)}\) \(\lim\limits_{x \rightarrow 0} \dfrac{\sin^2 4x + \sin^2 8x}{2\sin^2 x + 2\sin^2 2x}\) \(\lim\limits_{x \rightarrow 0} \dfrac{\dfrac{\sin^2 4x}{x^2} + \dfrac{\sin^2 8x}{x^2}}{\dfrac{2\sin^2 x}{x^2} + \dfrac{2\sin^2 2x}{x^2}}\) \(\dfrac{4^2 + 8^2}{2 + 2(2)^2}\) \(\dfrac{16 + 64}{2 + 8}\) \(\dfrac{80}{10}\) \(8\)

\(\text{Note:}\) \(\color{red} \cos 2x = 1\:-\:\sin^2 x\) \(\color{red} \cos 4x = 1\:-\:\sin^2 x\) \(\lim\limits_{x \rightarrow 0} \dfrac{1\:-\:\cos x(1\:-\:2\sin^2 x)}{1\:-\:\cos 2x (1\:-\:2\sin^2 2x)}\) \(\lim\limits_{x \rightarrow 0} \dfrac{1\:-\:\cos x + \cos x\cdot 2\sin^2 x}{1\:-\:\cos 2x + \cos 2x\cdot 2\sin^2 2x}\) Hilangkan angka 1 dengan menggunakan rumus sudut rangkap cosinus \(\lim\limits_{x \rightarrow 0} \dfrac{1\:-\:(1\:-\:2\sin^2 \frac{1}{2} x) + \cos x\cdot 2\sin^2 x}{1\:-\:(1\:-\:2\sin^2 x) + \cos 2x\cdot 2\sin^2 2x}\) \(\lim\limits_{x \rightarrow 0} \dfrac{2\sin^2 \frac{1}{2} x + \cos x\cdot 2\sin^2 x}{2\sin^2 x + \cos 2x\cdot 2\sin^2 2x}\) \(\text{Note:}\) Selanjutnya gunakan rumus sudut rangkap sinus \(\color{red} \sin x = 2\sin \frac{1}{2} x \cos \frac{1}{2} x\) \(\color{red} \sin 2x = 2\sin x \cos x\) \(\lim\limits_{x \rightarrow 0} \dfrac{2\sin^2 \frac{1}{2} x + \cos x\cdot 2\cdot (2\sin \frac{1}{2} x \cos \frac{1}{2} x)^2}{2\sin^2 x + \cos 2x\cdot 2 \cdot (2\sin x \cos x)^2}\) \(\lim\limits_{x \rightarrow 0} \dfrac{2\sin^2 \frac{1}{2} x + 2\cos x\cdot 4\sin^2 \frac{1}{2} x \cos^2 \frac{1}{2} x}{2\sin^2 x + 2\cos 2x \cdot 4\sin^2 x \cos^2 x }\) \(\lim\limits_{x \rightarrow 0} \dfrac{\cancel{2}\color{red}\sin^2 \frac{1}{2} x\color{black}(1 + 4\cos x \cos^2 \frac{1}{2} x)}{\cancel{2}\color{red}\sin^2 x\color{black}(1 + 4\cos 2x \cos^2 x)}\) \(\lim\limits_{x \rightarrow 0} \dfrac{\color{red}(\frac{1}{2})^2\color{black}(1 + 4\cos x \cos^2 \frac{1}{2} x)}{1 + 4\cos 2x \cos^2 x}\) \(\left(\dfrac{1}{2}\right)^2\cdot \dfrac{(1 + 4)}{(1 + 4)}\) \(\dfrac{5}{20}\) \(\dfrac{1}{4}\)

\(\lim\limits_{x \rightarrow 0} \dfrac{2(\cos 2x \:-\: \frac{1}{2})}{\sin (x \:-\: \frac{\pi}{6})}\) \(\lim\limits_{x \rightarrow 0} \dfrac{2(\cos 2x \:-\: \cos \frac{\pi}{6})}{\sin (x \:-\: \frac{\pi}{6})}\) \(\text{Note:}\) \(\color{red} \cos x \:-\: \cos y = -2\sin \frac{1}{2}(x + y)\sin \frac{1}{2} (x \:-\: y)\) \(\lim\limits_{x \rightarrow 0} \dfrac{2[-2\sin \frac{1}{2}(2x + \frac{\pi}{3})\sin \frac{1}{2}(2x\:-\:\frac{\pi}{3})]}{\sin (x \:-\: \frac{\pi}{6})}\) \(\lim\limits_{x \rightarrow 0}\dfrac{-4\sin(x + \frac{\pi}{6})\cancel{\sin(x\:-\:\frac{\pi}{6})}}{\cancel{\sin(x\:-\:\frac{\pi}{6})}}\) \(\lim\limits_{x \rightarrow 0} -4\sin(x + \frac{\pi}{6})\) \(-4\sin(\frac{\pi}{6}+ \frac{\pi}{6})\) \(-4\sin(\frac{\pi}{3})\) \(-4\cdot \frac{1}{2}\sqrt{3}\) \(-2\sqrt{3}\)

\(\lim\limits_{x \rightarrow 0} \left(\dfrac{\color{red}1 + 6\sin x + 9\sin^2 x \color{black} + 2\sin x + 7\sin^2 x}{\color{red}1 + 6\sin x + 9\sin^2 x }\right)^{\dfrac{\sin 3x}{1\:-\:(1\:-\:2\sin^2 2x)}}\) \(\lim\limits_{x \rightarrow 0}\left(1 + \dfrac{2\sin x + 7\sin^2 x}{1 + 6\sin x + 9\sin^2 x}\right)^{\dfrac{\sin 3x}{2\sin^2 2x}}\) \(\text{Note:}\) \(\color{red}\lim_{x \rightarrow 0}[1 + \text{f(x)}]^{\text{g(x)}}\) \(e^{\lim\limits_{x \rightarrow 0}\left(\dfrac{2\sin x + 7\sin^2x}{1 + 6\sin x + 9\sin^2 x}\right)\left(\dfrac{\sin 3x}{2\sin 2x \sin 2x}\right)}\) \(e^{\lim_{x \rightarrow 0}\left(\dfrac{\color{red}\sin x\color{black}\cdot (2 + 7\sin x )\cdot \color{blue}\sin 3x}{2\color{red}\sin 2x \color{black}\cdot (1 + 3\sin x)^2 \cdot \color{blue}\sin 2x}\right)}\) \(e^{\lim\limits_{x \rightarrow 0} \dfrac{\color{red}\frac{1}{2}\cdot\color{black}(2 + 7\sin x)\cdot \color{blue}\frac{3}{2}}{2(1 + 3\sin x)^2}}\) \(e^{\lim\limits_{x \rightarrow 0} \dfrac{3}{8}\cdot \dfrac{(2 + 7\sin x)}{(1 + 3\sin x)^2}}\) \(e^{\frac{3}{8}\cdot \frac{(2 + 0)}{(1 + 0)^2}}\) \(e^{\frac{3}{4}}\)

Misalkan \(\dfrac{1}{x} = y\), sehingga untuk \(x \rightarrow \infty,\:\: y \rightarrow 0\) \(\lim\limits_{y \rightarrow 0} \dfrac{\sin 3y}{(1\:-\:\cos 2y)\cdot \left(\dfrac{1}{y}\right)^2 \cdot \sin y}\) \(\lim\limits_{y \rightarrow 0} \dfrac{\sin 3y}{[1\:-\:(1\:-\:2\sin^2 y)]\cdot \dfrac{1}{y^2} \cdot \sin y}\) \(\lim\limits_{y \rightarrow 0} \dfrac{\sin 3y}{2\sin^2 y\cdot \dfrac{1}{y^2}\cdot \sin y}\) \(\lim\limits_{y \rightarrow 0} \dfrac{\sin 3y}{\sin y}\cdot \dfrac{1}{2\cdot \dfrac{\sin^2 y}{y^2}}\) \(3 \cdot \dfrac{1}{2}\) \(\frac{3}{2}\)