Basic trig limits are used to evaluate limits involving sine, cosine, tangent, and angle expressions as the input approaches zero. The most important idea is that \(\displaystyle \lim_{\theta\to0}\frac{\sin\theta}{\theta}=1\), along with related reciprocal and rewritten forms. These problems include direct trig limit forms, coefficient adjustments, shifted inputs, reciprocal trig functions, and common algebraic rewrites.
Notes
Questions
Find the limit
\(\textbf{1)}\) \(\displaystyle \lim_{\theta\to0} \frac{\sin 5\theta}{\theta} \) The answer is \(5\)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin 5\theta}{\theta}\)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin 5\theta}{\theta}\cdot \frac{5}{5}\)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin 5\theta}{5\theta} \cdot 5\)
\(\,\,\,\,\,\,\displaystyle (1) \cdot 5\)
\(\,\,\,\,\,\,\)The answer is \(5\)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin 5\theta}{\theta}\)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin 5\theta}{\theta}\cdot \frac{5}{5}\)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin 5\theta}{5\theta} \cdot 5\)
\(\,\,\,\,\,\,\displaystyle (1) \cdot 5\)
\(\,\,\,\,\,\,\)The answer is \(5\)
\(\textbf{2)}\) \(\displaystyle \lim_{\theta\to0} \frac{\sin \theta}{5\theta} \) The answer is \(\frac{1}{5}\)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin \theta}{5\theta} \)
\(\,\,\,\,\,\,\displaystyle \frac{1}{5} \lim_{\theta\to0} \frac{\sin \theta}{\theta} \)
\(\,\,\,\,\,\,\displaystyle \frac{1}{5} (1) \)
\(\,\,\,\,\,\,\)The answer is \(\frac{1}{5}\)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin \theta}{5\theta} \)
\(\,\,\,\,\,\,\displaystyle \frac{1}{5} \lim_{\theta\to0} \frac{\sin \theta}{\theta} \)
\(\,\,\,\,\,\,\displaystyle \frac{1}{5} (1) \)
\(\,\,\,\,\,\,\)The answer is \(\frac{1}{5}\)
\(\textbf{3)}\) \(\displaystyle \lim_{\theta\to0} \frac{\theta}{\sin 5\theta}\) The answer is \(\displaystyle \frac{1}{5}\)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\theta}{\sin 5\theta}\)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\theta}{\sin 5\theta}\cdot \frac{5}{5}\)
\(\,\,\,\,\,\,\displaystyle \frac{1}{5}\lim_{\theta\to0}\frac{5\theta}{\sin 5\theta}\)
\(\,\,\,\,\,\,\displaystyle \frac{1}{5}(1)\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle \frac{1}{5}\)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\theta}{\sin 5\theta}\)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\theta}{\sin 5\theta}\cdot \frac{5}{5}\)
\(\,\,\,\,\,\,\displaystyle \frac{1}{5}\lim_{\theta\to0}\frac{5\theta}{\sin 5\theta}\)
\(\,\,\,\,\,\,\displaystyle \frac{1}{5}(1)\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle \frac{1}{5}\)
\(\textbf{4)}\) \(\displaystyle \lim_{\theta\to0} \frac{5\theta}{\sin \theta}\) The answer is \(\displaystyle 5\)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{5\theta}{\sin \theta}\)
\(\,\,\,\,\,\,\displaystyle 5\lim_{\theta\to0}\frac{\theta}{\sin \theta}\)
\(\,\,\,\,\,\,\displaystyle 5(1)\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle 5\)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{5\theta}{\sin \theta}\)
\(\,\,\,\,\,\,\displaystyle 5\lim_{\theta\to0}\frac{\theta}{\sin \theta}\)
\(\,\,\,\,\,\,\displaystyle 5(1)\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle 5\)
\(\textbf{5)}\) \(\displaystyle \lim_{\theta\to0} \frac{\sin 3\theta}{\sin 4\theta} \) The answer is \(\displaystyle\frac{3}{4}\)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin 3\theta}{\sin 4\theta} \)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin 3\theta}{3\theta}\cdot \frac{4\theta}{\sin 4\theta}\cdot \frac{3}{4}\)
\(\,\,\,\,\,\,\displaystyle (1)(1)\cdot \frac{3}{4}\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle \frac{3}{4}\)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin 3\theta}{\sin 4\theta} \)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin 3\theta}{3\theta}\cdot \frac{4\theta}{\sin 4\theta}\cdot \frac{3}{4}\)
\(\,\,\,\,\,\,\displaystyle (1)(1)\cdot \frac{3}{4}\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle \frac{3}{4}\)
\(\textbf{6)}\) \(\displaystyle \lim_{x\to5} \frac{\sin (x-5)}{x-5}\) The answer is \(1\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to5} \frac{\sin (x-5)}{x-5}\)
\(\,\,\,\,\,\,\text{Let }u=x-5.\)
\(\,\,\,\,\,\,\text{As }x\to5,\text{ }u\to0.\)
\(\,\,\,\,\,\,\displaystyle \lim_{u\to0}\frac{\sin u}{u}\)
\(\,\,\,\,\,\,\)The answer is \(1\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to5} \frac{\sin (x-5)}{x-5}\)
\(\,\,\,\,\,\,\text{Let }u=x-5.\)
\(\,\,\,\,\,\,\text{As }x\to5,\text{ }u\to0.\)
\(\,\,\,\,\,\,\displaystyle \lim_{u\to0}\frac{\sin u}{u}\)
\(\,\,\,\,\,\,\)The answer is \(1\)
\(\textbf{7)}\) \(\displaystyle \lim_{x\to0} \frac{1}{x^2 \cot{x} \csc{3x}}\) The answer is \(3\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{1}{x^2 \cot{x} \csc{3x}}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\tan{x} \sin{3x}}{x^2}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin{x} \sin{3x}}{x^2 \cos{x}}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin{x}}{x} \cdot \frac{\sin{3x}}{x} \cdot \frac{1}{\cos{x}}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin{x}}{x} \cdot 3\frac{\sin{3x}}{3x} \cdot \frac{1}{\cos{x}}\)
\(\,\,\,\,\,\displaystyle (1)(3)(1)\)
\(\,\,\,\,\,\,\)The answer is \(3\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{1}{x^2 \cot{x} \csc{3x}}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\tan{x} \sin{3x}}{x^2}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin{x} \sin{3x}}{x^2 \cos{x}}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin{x}}{x} \cdot \frac{\sin{3x}}{x} \cdot \frac{1}{\cos{x}}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin{x}}{x} \cdot 3\frac{\sin{3x}}{3x} \cdot \frac{1}{\cos{x}}\)
\(\,\,\,\,\,\displaystyle (1)(3)(1)\)
\(\,\,\,\,\,\,\)The answer is \(3\)
\(\textbf{8)}\) \(\displaystyle \lim_{x\to\pi/6} \frac{\sin(6x)}{6x}\) The answer is \(0\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to\pi/6} \frac{\sin(6x)}{6x}\)
\(\,\,\,\,\,\,\displaystyle \frac{\sin\left(6\cdot\frac{\pi}{6}\right)}{6\cdot\frac{\pi}{6}}\)
\(\,\,\,\,\,\,\displaystyle \frac{\sin(\pi)}{\pi}\)
\(\,\,\,\,\,\,\displaystyle \frac{0}{\pi}\)
\(\,\,\,\,\,\,\)The answer is \(0\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to\pi/6} \frac{\sin(6x)}{6x}\)
\(\,\,\,\,\,\,\displaystyle \frac{\sin\left(6\cdot\frac{\pi}{6}\right)}{6\cdot\frac{\pi}{6}}\)
\(\,\,\,\,\,\,\displaystyle \frac{\sin(\pi)}{\pi}\)
\(\,\,\,\,\,\,\displaystyle \frac{0}{\pi}\)
\(\,\,\,\,\,\,\)The answer is \(0\)
\(\textbf{9)}\) \(\displaystyle \lim_{x\to0} \frac{1 – \cos x}{\sin x} \) The answer is \(\displaystyle 0\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{1 – \cos x}{\sin x} \)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{1 – \cos x}{\sin x} \cdot \frac{1+\cos x}{1+\cos x} \)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{1-\cos^2 x}{\sin x(1+\cos x)} \)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin^2 x}{\sin x(1+\cos x)} \)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin x}{1+\cos x} \)
\(\,\,\,\,\,\,\displaystyle \frac{0}{1+1} \)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle 0\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{1 – \cos x}{\sin x} \)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{1 – \cos x}{\sin x} \cdot \frac{1+\cos x}{1+\cos x} \)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{1-\cos^2 x}{\sin x(1+\cos x)} \)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin^2 x}{\sin x(1+\cos x)} \)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin x}{1+\cos x} \)
\(\,\,\,\,\,\,\displaystyle \frac{0}{1+1} \)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle 0\)
\(\textbf{10)}\) \(\displaystyle \lim_{x\to0} \frac{\sin^2(6x)}{3x^2}\) The answer is \(12\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin^2(6x)}{3x^2}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin(6x)}{x}\cdot\frac{\sin(6x)}{3x}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} 6\frac{\sin(6x)}{6x}\cdot 2\frac{\sin(6x)}{6x}\)
\(\,\,\,\,\,\,\displaystyle 6(1)\cdot 2(1)\)
\(\,\,\,\,\,\,\)The answer is \(12\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin^2(6x)}{3x^2}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin(6x)}{x}\cdot\frac{\sin(6x)}{3x}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} 6\frac{\sin(6x)}{6x}\cdot 2\frac{\sin(6x)}{6x}\)
\(\,\,\,\,\,\,\displaystyle 6(1)\cdot 2(1)\)
\(\,\,\,\,\,\,\)The answer is \(12\)
\(\textbf{11)}\) \(\displaystyle \lim_{\theta\to0} \frac{\sin 7\theta}{\theta}\) The answer is \(7\)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin 7\theta}{\theta}\)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin 7\theta}{7\theta}\cdot 7\)
\(\,\,\,\,\,\,\displaystyle (1)(7)\)
\(\,\,\,\,\,\,\)The answer is \(7\)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin 7\theta}{\theta}\)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin 7\theta}{7\theta}\cdot 7\)
\(\,\,\,\,\,\,\displaystyle (1)(7)\)
\(\,\,\,\,\,\,\)The answer is \(7\)
\(\textbf{12)}\) \(\displaystyle \lim_{\theta\to0} \frac{\sin 2\theta}{\sin 9\theta}\) The answer is \(\displaystyle\frac{2}{9}\)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin 2\theta}{\sin 9\theta}\)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin 2\theta}{2\theta}\cdot \frac{9\theta}{\sin 9\theta}\cdot \frac{2}{9}\)
\(\,\,\,\,\,\,\displaystyle (1)(1)\cdot \frac{2}{9}\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle\frac{2}{9}\)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin 2\theta}{\sin 9\theta}\)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin 2\theta}{2\theta}\cdot \frac{9\theta}{\sin 9\theta}\cdot \frac{2}{9}\)
\(\,\,\,\,\,\,\displaystyle (1)(1)\cdot \frac{2}{9}\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle\frac{2}{9}\)
\(\textbf{13)}\) \(\displaystyle \lim_{x\to0} \frac{\tan 4x}{x}\) The answer is \(4\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\tan 4x}{x}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin 4x}{x\cos 4x}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} 4\frac{\sin 4x}{4x}\cdot \frac{1}{\cos 4x}\)
\(\,\,\,\,\,\,\displaystyle 4(1)\cdot \frac{1}{1}\)
\(\,\,\,\,\,\,\)The answer is \(4\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\tan 4x}{x}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin 4x}{x\cos 4x}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} 4\frac{\sin 4x}{4x}\cdot \frac{1}{\cos 4x}\)
\(\,\,\,\,\,\,\displaystyle 4(1)\cdot \frac{1}{1}\)
\(\,\,\,\,\,\,\)The answer is \(4\)
\(\textbf{14)}\) \(\displaystyle \lim_{x\to0} \frac{x}{\tan 6x}\) The answer is \(\displaystyle\frac{1}{6}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{x}{\tan 6x}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{x\cos 6x}{\sin 6x}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{1}{6}\cdot \frac{6x}{\sin 6x}\cdot \cos 6x\)
\(\,\,\,\,\,\,\displaystyle \frac{1}{6}(1)(1)\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle\frac{1}{6}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{x}{\tan 6x}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{x\cos 6x}{\sin 6x}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{1}{6}\cdot \frac{6x}{\sin 6x}\cdot \cos 6x\)
\(\,\,\,\,\,\,\displaystyle \frac{1}{6}(1)(1)\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle\frac{1}{6}\)
\(\textbf{15)}\) \(\displaystyle \lim_{x\to0} \frac{1-\cos(3x)}{x^2}\) The answer is \(\displaystyle\frac{9}{2}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{1-\cos(3x)}{x^2}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{1-\cos(3x)}{x^2}\cdot\frac{1+\cos(3x)}{1+\cos(3x)}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{1-\cos^2(3x)}{x^2(1+\cos(3x))}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin^2(3x)}{x^2(1+\cos(3x))}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \left(\frac{\sin(3x)}{x}\right)^2\cdot \frac{1}{1+\cos(3x)}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \left(3\frac{\sin(3x)}{3x}\right)^2\cdot \frac{1}{1+\cos(3x)}\)
\(\,\,\,\,\,\,\displaystyle 3^2(1)^2\cdot \frac{1}{1+1}\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle\frac{9}{2}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{1-\cos(3x)}{x^2}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{1-\cos(3x)}{x^2}\cdot\frac{1+\cos(3x)}{1+\cos(3x)}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{1-\cos^2(3x)}{x^2(1+\cos(3x))}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin^2(3x)}{x^2(1+\cos(3x))}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \left(\frac{\sin(3x)}{x}\right)^2\cdot \frac{1}{1+\cos(3x)}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \left(3\frac{\sin(3x)}{3x}\right)^2\cdot \frac{1}{1+\cos(3x)}\)
\(\,\,\,\,\,\,\displaystyle 3^2(1)^2\cdot \frac{1}{1+1}\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle\frac{9}{2}\)
\(\textbf{16)}\) \(\displaystyle \lim_{x\to0} \frac{\sin(8x)}{\tan(2x)}\) The answer is \(4\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin(8x)}{\tan(2x)}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin(8x)\cos(2x)}{\sin(2x)}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin(8x)}{8x}\cdot\frac{2x}{\sin(2x)}\cdot 4\cos(2x)\)
\(\,\,\,\,\,\,\displaystyle (1)(1)(4)(1)\)
\(\,\,\,\,\,\,\)The answer is \(4\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin(8x)}{\tan(2x)}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin(8x)\cos(2x)}{\sin(2x)}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin(8x)}{8x}\cdot\frac{2x}{\sin(2x)}\cdot 4\cos(2x)\)
\(\,\,\,\,\,\,\displaystyle (1)(1)(4)(1)\)
\(\,\,\,\,\,\,\)The answer is \(4\)
\(\textbf{17)}\) \(\displaystyle \lim_{x\to0} \frac{x\sin(5x)}{1-\cos x}\) The answer is \(10\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{x\sin(5x)}{1-\cos x}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{x\sin(5x)}{1-\cos x}\cdot\frac{1+\cos x}{1+\cos x}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{x\sin(5x)(1+\cos x)}{\sin^2 x}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin(5x)}{x}\cdot\left(\frac{x}{\sin x}\right)^2(1+\cos x)\)
\(\,\,\,\,\,\,\displaystyle 5(1)^2(1+1)\)
\(\,\,\,\,\,\,\)The answer is \(10\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{x\sin(5x)}{1-\cos x}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{x\sin(5x)}{1-\cos x}\cdot\frac{1+\cos x}{1+\cos x}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{x\sin(5x)(1+\cos x)}{\sin^2 x}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin(5x)}{x}\cdot\left(\frac{x}{\sin x}\right)^2(1+\cos x)\)
\(\,\,\,\,\,\,\displaystyle 5(1)^2(1+1)\)
\(\,\,\,\,\,\,\)The answer is \(10\)
\(\textbf{18)}\) \(\displaystyle \lim_{x\to0} \frac{\sin(3x)\sin(4x)}{x^2}\) The answer is \(12\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin(3x)\sin(4x)}{x^2}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin(3x)}{x}\cdot\frac{\sin(4x)}{x}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} 3\frac{\sin(3x)}{3x}\cdot4\frac{\sin(4x)}{4x}\)
\(\,\,\,\,\,\,\displaystyle 3(1)\cdot4(1)\)
\(\,\,\,\,\,\,\)The answer is \(12\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin(3x)\sin(4x)}{x^2}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin(3x)}{x}\cdot\frac{\sin(4x)}{x}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} 3\frac{\sin(3x)}{3x}\cdot4\frac{\sin(4x)}{4x}\)
\(\,\,\,\,\,\,\displaystyle 3(1)\cdot4(1)\)
\(\,\,\,\,\,\,\)The answer is \(12\)
\(\textbf{19)}\) \(\displaystyle \lim_{x\to0} \frac{\sin(2x)+\sin(5x)}{x}\) The answer is \(7\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin(2x)+\sin(5x)}{x}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0}\frac{\sin(2x)}{x}+\lim_{x\to0}\frac{\sin(5x)}{x}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0}2\frac{\sin(2x)}{2x}+\lim_{x\to0}5\frac{\sin(5x)}{5x}\)
\(\,\,\,\,\,\,\displaystyle 2(1)+5(1)\)
\(\,\,\,\,\,\,\)The answer is \(7\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin(2x)+\sin(5x)}{x}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0}\frac{\sin(2x)}{x}+\lim_{x\to0}\frac{\sin(5x)}{x}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0}2\frac{\sin(2x)}{2x}+\lim_{x\to0}5\frac{\sin(5x)}{5x}\)
\(\,\,\,\,\,\,\displaystyle 2(1)+5(1)\)
\(\,\,\,\,\,\,\)The answer is \(7\)
\(\textbf{20)}\) \(\displaystyle \lim_{x\to0} \frac{\tan(3x)}{\sin(9x)}\) The answer is \(\displaystyle\frac{1}{3}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\tan(3x)}{\sin(9x)}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin(3x)}{\cos(3x)\sin(9x)}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0}\frac{\sin(3x)}{3x}\cdot\frac{9x}{\sin(9x)}\cdot\frac{1}{3\cos(3x)}\)
\(\,\,\,\,\,\,\displaystyle (1)(1)\cdot\frac{1}{3(1)}\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle\frac{1}{3}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\tan(3x)}{\sin(9x)}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin(3x)}{\cos(3x)\sin(9x)}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0}\frac{\sin(3x)}{3x}\cdot\frac{9x}{\sin(9x)}\cdot\frac{1}{3\cos(3x)}\)
\(\,\,\,\,\,\,\displaystyle (1)(1)\cdot\frac{1}{3(1)}\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle\frac{1}{3}\)
See Related Pages\(\)
\(\bullet\text{ Limit Calculator}\)
\(\,\,\,\,\,\,\,\,\text{(Symbolab.com)}\)
\(\bullet\text{ Calculus Homepage}\)
\(\,\,\,\,\,\,\,\,\text{All the Best Topics…}\)
\(\bullet\text{ Limits on Graphs}\)
\(\,\,\,\,\,\,\,\,\)
\(…\)
\(\bullet\text{ Continuity on Graphs}\)
\(\,\,\,\,\,\,\,\,\)\(…\)
\(\bullet\text{ Piecewise Functions- Limits and Continuity}\)
\(\,\,\,\,\,\,\,\,\)
\(…\)
\(\bullet\text{ Infinite Limits}\)
\(\,\,\,\,\,\,\,\,\displaystyle \lim_{x\to 4^{+}} \frac{5}{x-4}…\)
\(\bullet\text{ Limits at Infinity}\)
\(\,\,\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{5x^2+2x-10}{3x^2+4x-5}…\)