Notes

 

Questions

Find the limit

\(\textbf{1)}\) \(\displaystyle \lim_{\theta\to0} \frac{\sin 5\theta}{\theta}⁡ \)

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The answer is \(5\)

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\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin 5\theta}{\theta}⁡\)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin 5\theta}{\theta}\cdot \frac{5}{5}⁡\)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin 5\theta}{5\theta} \cdot 5⁡\)
\(\,\,\,\,\,\,\displaystyle (1) \cdot 5⁡\)
\(\,\,\,\,\,\,\)The answer is \(5\)

 

\(\textbf{2)}\) \(\displaystyle \lim_{\theta\to0} \frac{\sin \theta}{5\theta}⁡ \)

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The answer is \(\frac{1}{5}\)

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\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin \theta}{5\theta}⁡ \)
\(\,\,\,\,\,\,\displaystyle \frac{1}{5} \lim_{\theta\to0} \frac{\sin \theta}{\theta}⁡ \)
\(\,\,\,\,\,\,\displaystyle \frac{1}{5} (1)⁡ \)
\(\,\,\,\,\,\,\)The answer is \(\frac{1}{5}\)

 

\(\textbf{3)}\) \(\displaystyle \lim_{\theta\to0} \frac{\theta}{\sin 5\theta}\)

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The answer is \(\displaystyle \frac{1}{5}\)

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\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\theta}{\sin 5\theta}\)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\theta}{\sin 5\theta}\cdot \frac{5}{5}\)
\(\,\,\,\,\,\,\displaystyle \frac{1}{5},\lim_{\theta\to0}\frac{5\theta}{\sin 5\theta}\)
\(\,\,\,\,\,\,\displaystyle \frac{1}{5},(1)\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle \frac{1}{5}\)

 

\(\textbf{4)}\) \(\displaystyle \lim_{\theta\to0} \frac{5\theta}{\sin \theta}\)

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The answer is \(\displaystyle 5\)

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\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{5\theta}{\sin \theta}\)
\(\,\,\,\,\,\,\displaystyle 5\lim_{\theta\to0}\frac{\theta}{\sin \theta}\)
\(\,\,\,\,\,\,\displaystyle 5,(1)\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle 5\)

 

\(\textbf{5)}\) \(\displaystyle \lim_{\theta\to0} \frac{\sin 3\theta}{\sin 4\theta} \)

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The answer is \(\displaystyle\frac{3}{4}\)

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\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin 3\theta}{\sin 4\theta} \)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin 3\theta}{\sin 4\theta} \cdot \frac{\theta}{\theta}\)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\theta \sin 3\theta}{\theta \sin 4\theta} \)
\(\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin 3\theta}{theta} \cdot \lim_{\theta\to0} \frac{\theta}{\sin 4\theta}\)
\(\,\,\,\,\,\,\displaystyle \frac{3}{3} \cdot \lim_{\theta\to0} \frac{\sin 3\theta}{\theta} \cdot \lim_{\theta\to0} \frac{\theta}{\sin 4\theta} \cdot \frac{4}{4}\)
\(\,\,\,\,\,\,\displaystyle \frac{3}{1} \cdot \lim_{\theta\to0} \frac{\sin 3\theta}{3\theta} \cdot \lim_{\theta\to0} \frac{4\theta}{\sin 4\theta} \cdot \frac{1}{4}\)
\(\,\,\,\,\,\,\displaystyle \frac{3}{1} \cdot (1) \cdot (1) \cdot \frac{1}{4}\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle \frac{3}{4}\)

 

\(\textbf{6)}\) \(\displaystyle \lim_{x\to5} \frac{\sin (x-5)}{x-5}\)

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The answer is \(1\)

 

\(\textbf{7)}\) \(\displaystyle \lim_{x\to0} \frac{1}{x^2 \cot{x} \csc{3x}}\)

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The answer is \(3\)

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\(\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{1}{x^2 \cot{x} \csc{3x}}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\tan{x} \sin{3x}}{x^2}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin{x} \sin{3x}}{x^2 \cos{x}}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin{x}}{x} \cdot \frac{\sin{3x}}{x} \cdot \frac{1}{\cos{x}}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin{x}}{x} \cdot 3\frac{\sin{3x}}{3x} \cdot \frac{1}{\cos{x}}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin{x}}{x} \cdot \frac{\sin{3x}}{3x} \cdot \frac{3}{\cos{x}}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin{x}}{x} \cdot \lim_{x\to0} \frac{\sin{3x}}{3x} \cdot \lim_{x\to0} \frac{3}{\cos{x}}\)
\(\,\,\,\,\,\displaystyle \lim_{\theta \to 0} \frac{\sin{\theta}}{\theta}=1 \text{ (From the notes)}\)
\(\,\,\,\,\,\displaystyle 1 \cdot 1 \cdot \frac{3}{\cos{0}}\)
\(\,\,\,\,\,\displaystyle 1 \cdot 1 \cdot \frac{3}{1}\)
\(\,\,\,\,\,3\)

 

\(\textbf{8)}\) \(\displaystyle \lim_{x\to\pi/6} \frac{\sin(6x)}{6x}\)

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The answer is \(0\)

 

\(\textbf{9)}\) \(\displaystyle \lim_{x\to0} \frac{1 – \cos x}{\sin x} \)

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The answer is \(\displaystyle 0\)

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\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{1 – \cos x}{\sin x} \)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{1 – \cos x}{\sin x} \cdot \frac{x}{x} \)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{1 – \cos x}{x} \cdot \frac{x}{\sin x} \)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{x(1 – \cos x)}{x \sin x} \)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \left( \frac{1 – \cos x}{x} \right) \cdot \lim_{x\to0} \left( \frac{x}{\sin x} \right) \)
\(\,\,\,\,\,\,\displaystyle 0 \cdot 1 = 0 \)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle 0\)

 

\(\textbf{10)}\) \(\displaystyle \lim_{x\to0} \frac{\sin^2(6x)}{3x^2}\)

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The answer is \(12\)

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\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin^2(6x)}{3x^2}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin(6x)\cdot \sin(6x)}{3(x)(x)}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin(6x)}{3x}\cdot \frac{\sin(6x)}{x}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to0} \frac{\sin(6x)}{3x}\cdot \frac{2}{2}\cdot \frac{\sin(6x)}{x}\cdot \frac{6}{6}\)
\(\,\,\,\,\,\,\displaystyle 2\cdot 6 \cdot \lim_{x\to0}\frac{\sin(6x)}{6x}\cdot \lim_{x\to0}\frac{\sin(6x)}{6x}\)
\(\,\,\,\,\,\,\displaystyle 12 \cdot (1)\cdot (1)\)
\(\,\,\,\,\,\,\displaystyle 12\)
\(\,\,\,\,\,\,\)The answer is \(12\)

 

 

See Related Pages\(\)

\(\bullet\text{ Limit Calculator}\)
\(\,\,\,\,\,\,\,\,\text{(Symbolab.com)}\)

\(\bullet\text{ Calculus Homepage}\)
\(\,\,\,\,\,\,\,\,\text{All the Best Topics…}\)

\(\bullet\text{ Limits on Graphs}\)
\(\,\,\,\,\,\,\,\,\)\(…\)

\(\bullet\text{ Continuity on Graphs}\)
\(\,\,\,\,\,\,\,\,\)\(…\)

\(\bullet\text{ Piecewise Functions- Limits and Continuity}\)
\(\,\,\,\,\,\,\,\,\)\(…\)

\(\bullet\text{ Infinite Limits}\)
\(\,\,\,\,\,\,\,\,\displaystyle \lim_{x\to 4^{+}} \frac{5}{x-4}…\)

\(\bullet\text{ Limits at Infinity}\)
\(\,\,\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{5x^2+2x-10}{3x^2+4x-5}…\)

 

In Summary

Trigonometry is a branch of mathematics that deals with the relationships between angles and the sides of triangles. Trig limits are typically covered in a calculus course. In most cases, trig limits are introduced after students have learned about limits of functions in general, and after they have learned about trigonometry and the basic properties of trigonometric functions.

Topics that use Trig Limits

Calculus: Trigonometric limits are an important concept in calculus, where they are used to evaluate limits involving trigonometric functions.

Differential equations: Trigonometric limits can be used to solve differential equations involving trigonometric functions.

Vector calculus: Trigonometric limits are used in vector calculus to evaluate limits involving vector fields.

Numerical analysis: Trigonometric limits are used in numerical analysis to approximate the value of certain functions.