# Nakafa Learning Content > For AI agents: use [llms.txt](https://nakafa.com/llms.txt) for the site index. Markdown versions are available by appending `.md` to content URLs or sending `Accept: text/markdown`. URL: https://nakafa.com/en/subjects/mathematics/limit/limit-of-trigonometric-function Source: https://raw.githubusercontent.com/nakafaai/nakafa.com/refs/heads/main/packages/contents/material/lesson/mathematics/limit/limit-of-trigonometric-function/en.mdx Learn trigonometric limits using fundamental theorems. Learn sine ratio properties, identities, and advanced substitution techniques. --- ## Understanding Limits of Trigonometric Functions Imagine you observe a clock pendulum swinging very slowly approaching its equilibrium point. This motion is similar to the behavior of trigonometric functions when their variable approaches a certain value. **Limits of trigonometric functions** examine how the values of sine, cosine, and tangent functions behave when the input approaches critical points. Unlike limits of algebraic functions that can often be solved by direct substitution, trigonometric functions have special characteristics due to their periodic and oscillating nature. This makes us need to use **theorems and special properties** to solve trigonometric limits. ## Fundamental Theorem of Trigonometric Limits The most important foundation in trigonometric limits is the theorem stating that the sine function approaches its gradient when the angle approaches zero. ### Basic Sine Limit The most fundamental theorem is: ```math \lim_{x \to 0} \frac{\sin x}{x} = 1 ``` This theorem **cannot be proven** using direct substitution because it results in the form $$\frac{0}{0}$$. Its proof requires a geometric approach using the unit circle and the squeeze theorem. Visible text: This theorem **cannot be proven** using direct substitution because it results in the form . Its proof requires a geometric approach using the unit circle and the squeeze theorem. > In this theorem, $$x$$ must be in **radians**, not degrees. If using degrees, the result will be different. Visible text: > In this theorem, must be in **radians**, not degrees. If using degrees, the result will be different. ### Consequences of the Basic Limit From the fundamental limit above, we can derive several other important limits: Component: MathContainer Children: ```math \lim_{x \to 0} \frac{\tan x}{x} = 1 \quad \text{(because } \tan x = \frac{\sin x}{\cos x} \text{ and} \cos x \to 1\text{)} ``` ```math \lim_{x \to 0} \frac{1 - \cos x}{x} = 0 \quad \text{(use L'Hôpital or Taylor expansion)} ``` ```math \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \quad \text{(identity } 1 - \cos x = 2\sin^2\frac{x}{2}\text{)} ``` ## Properties of Trigonometric Limits Based on the fundamental theorem, we can build a series of very useful properties: ### Trigonometric Ratios For constants $$a \neq 0$$ and $$b \neq 0$$, and **all following limits when $$x \to 0$$**: Visible text: For constants and , and **all following limits when **: | Limit | Result | Notes | |-------|--------|-------| | $$\lim_{x \to 0} \frac{\sin ax}{bx}$$ | $$\frac{a}{b}$$ | Manipulation from basic theorem | | $$\lim_{x \to 0} \frac{\tan ax}{bx}$$ | $$\frac{a}{b}$$ | Because $$\tan ax = \frac{\sin ax}{\cos ax}$$ | | $$\lim_{x \to 0} \frac{\sin ax}{\sin bx}$$ | $$\frac{a}{b}$$ | Combination of two sine limits | | $$\lim_{x \to 0} \frac{\tan ax}{\tan bx}$$ | $$\frac{a}{b}$$ | Combination of two tangent limits | Visible text: | Limit | Result | Notes | |-------|--------|-------| | | | Manipulation from basic theorem | | | | Because | | | | Combination of two sine limits | | | | Combination of two tangent limits | ### Trigonometric Combinations From the trigonometric ratio properties, we can derive several trigonometric combination properties: Component: MathContainer Children: ```math \lim_{x \to 0} \frac{\sin ax}{\sin bx} = \frac{a}{b} ``` ```math \lim_{x \to 0} \frac{\tan ax}{\sin bx} = \frac{a}{b} ``` ## Techniques for Solving Trigonometric Limits ### Substitution and Manipulation Techniques When facing complex trigonometric limits, we often need to manipulate expressions to use the fundamental theorem. **Calculate:** $$\lim_{x \to 0} \frac{\sin 2x}{3x}$$ Visible text: **Calculate:** We manipulate to obtain the standard form: Component: MathContainer Children: ```math \lim_{x \to 0} \frac{\sin 2x}{3x} = \lim_{x \to 0} \frac{2 \sin 2x}{2 \cdot 3x} = \lim_{x \to 0} \frac{2}{3} \cdot \frac{\sin 2x}{2x} ``` ```math = \frac{2}{3} \lim_{x \to 0} \frac{\sin 2x}{2x} = \frac{2}{3} \cdot 1 = \frac{2}{3} ``` ### Trigonometric Identity Techniques Often we need to use trigonometric identities to simplify expressions. Always ensure the function is defined at the point being approached. **Calculate:** $$\lim_{x \to \frac{\pi}{4}} \frac{\sin x + \cos x}{\sin x}$$ Visible text: **Calculate:** Since $$\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \neq 0$$, the denominator is not zero at $$x = \frac{\pi}{4}$$. Visible text: Since , the denominator is not zero at . Direct substitution: Component: MathContainer Children: ```math \lim_{x \to \frac{\pi}{4}} \frac{\sin x + \cos x}{\sin x} = \frac{\sin \frac{\pi}{4} + \cos \frac{\pi}{4}}{\sin \frac{\pi}{4}} ``` ```math = \frac{\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = \frac{2 \cdot \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = \frac{\sqrt{2}}{\frac{\sqrt{2}}{2}} = \sqrt{2} \cdot \frac{2}{\sqrt{2}} = 2 ``` ### Techniques for Special Forms For limits involving the form $$\frac{0}{0}$$, we need special techniques. Visible text: For limits involving the form , we need special techniques. **Calculate:** $$\lim_{x \to \frac{\pi}{2}} \frac{\sin^2(x - \frac{\pi}{2})}{x(x - \frac{\pi}{2})}$$ Visible text: **Calculate:** Let $$u = x - \frac{\pi}{2}$$, then when $$x \to \frac{\pi}{2}$$, we have $$u \to 0$$ and $$x = u + \frac{\pi}{2}$$. Visible text: Let , then when , we have and . Component: MathContainer Children: ```math \lim_{u \to 0} \frac{\sin^2 u}{(u + \frac{\pi}{2}) \cdot u} = \lim_{u \to 0} \frac{\sin^2 u}{u^2} \cdot \frac{u}{u + \frac{\pi}{2}} ``` ```math = \lim_{u \to 0} \left(\frac{\sin u}{u}\right)^2 \cdot \lim_{u \to 0} \frac{u}{u + \frac{\pi}{2}} ``` ```math = 1^2 \cdot \lim_{u \to 0} \frac{u}{\frac{\pi}{2} + u} = 1 \cdot \frac{0}{\frac{\pi}{2}} = 0 ``` ## Trigonometric Limits with Angle Identities ### Using Sum and Difference Formulas When dealing with trigonometric functions involving sum or difference of angles, we can use trigonometric identities. **Calculate:** $$\lim_{t \to 0} \frac{\cot 5t}{\cot 10t}$$ Visible text: **Calculate:** Using the definition of cotangent: Component: MathContainer Children: ```math \lim_{t \to 0} \frac{\cot 5t}{\cot 10t} = \lim_{t \to 0} \frac{\frac{\cos 5t}{\sin 5t}}{\frac{\cos 10t}{\sin 10t}} = \lim_{t \to 0} \frac{\cos 5t \sin 10t}{\sin 5t \cos 10t} ``` ```math = \lim_{t \to 0} \frac{\cos 5t}{\cos 10t} \cdot \frac{\sin 10t}{\sin 5t} = 1 \cdot \frac{10}{5} = 2 ``` ## Exercises 1. Calculate $$\lim_{x \to 0} \frac{\sin 3x}{4x}$$ 2. Calculate $$\lim_{x \to 0} \frac{\tan 2x}{\sin 5x}$$ 3. Calculate $$\lim_{x \to 0} \frac{1 - \cos 3x}{x^2}$$ 4. Calculate $$\lim_{x \to \frac{\pi}{6}} \frac{\sin x - \frac{1}{2}}{x - \frac{\pi}{6}}$$ 5. Calculate $$\lim_{x \to 0} \frac{\sin x \cos x}{x}$$ Visible text: 1. Calculate 2. Calculate 3. Calculate 4. Calculate 5. Calculate ### Answer Key 1. **Solution:** Use algebraic manipulation to obtain the standard form: ```math \lim_{x \to 0} \frac{\sin 3x}{4x} = \lim_{x \to 0} \frac{3 \sin 3x}{3 \cdot 4x} = \lim_{x \to 0} \frac{3}{4} \cdot \frac{\sin 3x}{3x} ``` ```math = \frac{3}{4} \lim_{x \to 0} \frac{\sin 3x}{3x} = \frac{3}{4} \cdot 1 = \frac{3}{4} ``` 2. **Solution:** Use trigonometric ratio properties: ```math \lim_{x \to 0} \frac{\tan 2x}{\sin 5x} = \lim_{x \to 0} \frac{\tan 2x}{2x} \cdot \frac{5x}{\sin 5x} \cdot \frac{2x}{5x} ``` ```math = \lim_{x \to 0} \frac{\tan 2x}{2x} \cdot \lim_{x \to 0} \frac{5x}{\sin 5x} \cdot \frac{2}{5} ``` ```math = 1 \cdot 1 \cdot \frac{2}{5} = \frac{2}{5} ``` 3. **Solution:** Use the identity $$1 - \cos ax = 2\sin^2\frac{ax}{2}$$: ```math \lim_{x \to 0} \frac{1 - \cos 3x}{x^2} = \lim_{x \to 0} \frac{2\sin^2\frac{3x}{2}}{x^2} ``` ```math = \lim_{x \to 0} 2 \cdot \left(\frac{\sin\frac{3x}{2}}{\frac{3x}{2}}\right)^2 \cdot \left(\frac{3}{2}\right)^2 ``` ```math = 2 \cdot 1^2 \cdot \frac{9}{4} = \frac{9}{2} ``` 4. **Solution:** Let $$h = x - \frac{\pi}{6}$$, then $$x = h + \frac{\pi}{6}$$ and when $$x \to \frac{\pi}{6}$$, we have $$h \to 0$$: ```math \lim_{h \to 0} \frac{\sin(h + \frac{\pi}{6}) - \frac{1}{2}}{h} ``` ```math = \lim_{h \to 0} \frac{\sin h \cos\frac{\pi}{6} + \cos h \sin\frac{\pi}{6} - \frac{1}{2}}{h} ``` ```math = \lim_{h \to 0} \frac{\sin h \cdot \frac{\sqrt{3}}{2} + \cos h \cdot \frac{1}{2} - \frac{1}{2}}{h} ``` ```math = \lim_{h \to 0} \frac{\frac{\sqrt{3}}{2}\sin h + \frac{1}{2}(\cos h - 1)}{h} ``` ```math = \frac{\sqrt{3}}{2} \lim_{h \to 0} \frac{\sin h}{h} + \frac{1}{2} \lim_{h \to 0} \frac{\cos h - 1}{h} ``` ```math = \frac{\sqrt{3}}{2} \cdot 1 + \frac{1}{2} \cdot 0 = \frac{\sqrt{3}}{2} ``` 5. **Solution:** Use the identity $$\sin x \cos x = \frac{1}{2}\sin 2x$$: ```math \lim_{x \to 0} \frac{\sin x \cos x}{x} = \lim_{x \to 0} \frac{\frac{1}{2}\sin 2x}{x} ``` ```math = \frac{1}{2} \lim_{x \to 0} \frac{\sin 2x}{x} = \frac{1}{2} \lim_{x \to 0} \frac{2 \sin 2x}{2x} ``` ```math = \frac{1}{2} \cdot 2 \lim_{x \to 0} \frac{\sin 2x}{2x} = 1 \cdot 1 = 1 ``` Visible text: 1. **Solution:** Use algebraic manipulation to obtain the standard form: 2. **Solution:** Use trigonometric ratio properties: 3. **Solution:** Use the identity : 4. **Solution:** Let , then and when , we have : 5. **Solution:** Use the identity :