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 \(5\)
\(\textbf{2)}\) \(\displaystyle \lim_{\theta\to0} \frac{\sin \theta}{5\theta} \)
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 \(\frac{1}{5}\)
\(\textbf{3)}\) \(\displaystyle \lim_{\theta\to0} \frac{\theta}{\sin 5\theta}\)
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 \(\displaystyle \frac{1}{5}\)
\(\textbf{4)}\) \(\displaystyle \lim_{\theta\to0} \frac{5\theta}{\sin \theta}\)
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 \(\displaystyle 5\)
\(\textbf{5)}\) \(\displaystyle \lim_{\theta\to0} \frac{\sin 3\theta}{\sin 4\theta} \)
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 \(\displaystyle \frac{3}{4}\)
\(\textbf{6)}\) \(\displaystyle \lim_{x\to5} \frac{\sin (x-5)}{x-5}\)
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 \(1\)
\(\textbf{7)}\) \(\displaystyle \lim_{x\to0} \frac{1}{x^2 \cot{x} \csc{3x}}\)
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 \(3\)
\(\textbf{8)}\) \(\displaystyle \lim_{x\to\pi/6} \frac{\sin(6x)}{6x}\)
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 \(0\)
\(\textbf{9)}\) \(\displaystyle \lim_{x\to0} \frac{1 – \cos x}{\sin x} \)
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 \(\displaystyle 0\)
\(\textbf{10)}\) \(\displaystyle \lim_{x\to0} \frac{\sin^2(6x)}{3x^2}\)
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 \(12\)
\(\textbf{11)}\) \(\displaystyle \lim_{\theta\to0} \frac{\sin 7\theta}{\theta}\)
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 \(7\)
\(\textbf{12)}\) \(\displaystyle \lim_{\theta\to0} \frac{\sin 2\theta}{\sin 9\theta}\)
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 \(\displaystyle\frac{2}{9}\)
\(\textbf{13)}\) \(\displaystyle \lim_{x\to0} \frac{\tan 4x}{x}\)
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 \(4\)
\(\textbf{14)}\) \(\displaystyle \lim_{x\to0} \frac{x}{\tan 6x}\)
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 \(\displaystyle\frac{1}{6}\)
\(\textbf{15)}\) \(\displaystyle \lim_{x\to0} \frac{1-\cos(3x)}{x^2}\)
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 \(\displaystyle\frac{9}{2}\)
\(\textbf{16)}\) \(\displaystyle \lim_{x\to0} \frac{\sin(8x)}{\tan(2x)}\)
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 \(4\)
\(\textbf{17)}\) \(\displaystyle \lim_{x\to0} \frac{x\sin(5x)}{1-\cos x}\)
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 \(10\)
\(\textbf{18)}\) \(\displaystyle \lim_{x\to0} \frac{\sin(3x)\sin(4x)}{x^2}\)
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 \(12\)
\(\textbf{19)}\) \(\displaystyle \lim_{x\to0} \frac{\sin(2x)+\sin(5x)}{x}\)
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 \(7\)
\(\textbf{20)}\) \(\displaystyle \lim_{x\to0} \frac{\tan(3x)}{\sin(9x)}\)
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 \(\displaystyle\frac{1}{3}\)
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