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The First Important Limit

The first important limit is one of the most fundamental and important limits in limit computation. It establishes the relationship between the sine function and a linear function.

Visual Illustration

The figure below shows a comparison of the graphs of $\sin x$ and $x$ , as well as their ratio $\frac{\sin x}{x}$ . From the figure, one can clearly see that as $x$ approaches 0, the graphs of $\sin x$ and $x$ nearly coincide, and their ratio tends to 1.

However, this is only an observation. Is it really true? We need to verify this limit through mathematical proof.

Geometric Figure

In the unit circle shown below, the area of the red triangle and the area of the blue sector become closer and closer as the angle decreases.

Angle28.6°

Now:

Let the angle be $x$
Let the radius of the circle be 1

Can you compute the area of the triangle and the area of the sector respectively?

Reference Answer

important limit

The area of the triangle is: $\displaystyle S_{\text{triangle}} = \frac{\sin x}{2}$

The area of the sector is: $\displaystyle S_{\text{sector}} = \frac{x}{2}$

From observation, when the angle approaches 0, the triangle and the sector almost coincide, and their ratio tends to 1. That is, the value of the limit of $\frac{\sin x}{x}$ as $x$ tends to 0 is 1.

However, this is only an observation. Is it really true? We need to verify this limit through mathematical proof.

If you forget the expression and value of this limit, you can recall it through this scenario.

The Limit Expression

The first important limit

$\lim_{x \to 0} \frac{\sin x}{x} = 1$

Geometric meaning: as the angle approaches 0, the ratio of the sine value to the angle approaches 1. In the unit circle, this means that when the angle is very small, the arc length is almost equal to the corresponding chord length.

Ideas for the Proof

There are many ways to prove this limit; below we introduce several common methods.

Proof by the Squeeze Theorem

One can prove that $\cos x < \frac{\sin x}{x} < 1$ (when $0 < x < \frac{\pi}{2}$ )
Since $\lim_{x \to 0} \cos x = 1$ and $\lim_{x \to 0} 1 = 1$
By the squeeze theorem, $\lim_{x \to 0} \frac{\sin x}{x} = 1$

Generalized Forms

Generalized forms of the first important limit

1. General Form

$\lim_{x \to 0} \frac{\sin kx}{x} = k$

Proof: $\lim_{x \to 0} \frac{\sin kx}{x} = \lim_{x \to 0} k \cdot \frac{\sin kx}{kx} = k \cdot 1 = k$

2. Composite Form

$\lim_{x \to 0} \frac{\sin f(x)}{f(x)} = 1$

where $\lim_{x \to 0} f(x) = 0$

3. Reciprocal Form

$\lim_{x \to 0} \frac{x}{\sin x} = 1$

Graph

Finally, let us look at the complete graph of $\frac{\sin x}{x}$ .

Beautiful and mysterious, as if it conceals the ultimate secret of the universe.

Geometric Proof

Through the geometric relations in the unit circle, one can intuitively understand the first important limit. As the angle $x$ tends to 0, the ratio of $\sin x$ to $x$ tends to 1.

Practice Problems

Exercise 1

Find the limit $\lim_{x \to 0} \frac{\sin 5x}{x}$ .

Reference Answer

important limit

Idea: Use the generalized form of the first important limit.

Detailed steps:

$\lim_{x \to 0} \frac{\sin 5x}{x} = \lim_{x \to 0} 5 \cdot \frac{\sin 5x}{5x}$
$= 5 \cdot \lim_{x \to 0} \frac{\sin 5x}{5x} = 5 \cdot 1 = 5$

Answer: The limit is 5.

Exercise 2

Find the limit $\lim_{x \to 0} \frac{\sin x^3}{x^3}$ .

Reference Answer

important limit

Idea: Use a change of variable and the first important limit.

Detailed steps:

Let $t = x^3$ ; then as $x \to 0$ , $t \to 0$
$\lim_{x \to 0} \frac{\sin x^3}{x^3} = \lim_{t \to 0} \frac{\sin t}{t} = 1$

Answer: The limit is 1.

Exercise 3

Find the limit $\lim_{x \to 0} \frac{\sin 4x}{\sin 2x}$ .

Reference Answer

important limit

Idea: Use the first important limit and equivalent infinitesimals.

Detailed steps:

As $x \to 0$ , $\sin 4x \sim 4x$ and $\sin 2x \sim 2x$
$\lim_{x \to 0} \frac{\sin 4x}{\sin 2x} = \lim_{x \to 0} \frac{4x}{2x} = 2$

Answer: The limit is 2.

Exercise 4

In the unit circle, when the angle $x$ is very small, what is the relationship between the arc length and the chord length? How is this related to the first important limit?

Reference Answer

important limit

Idea: Understand the relationship among arc length, chord length, and the first important limit from a geometric perspective.

Detailed steps:

In the unit circle, the arc length corresponding to the angle $x$ is $x$ (in radians)
The corresponding chord length is $2\sin(\frac{x}{2})$
When $x$ is very small, $\sin(\frac{x}{2}) \approx \frac{x}{2}$ , so the chord length $\approx 2 \cdot \frac{x}{2} = x$
This shows that for very small angles, the arc length is approximately equal to the chord length
This is exactly the geometric basis of the first important limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$

Answer: For very small angles, the arc length is approximately equal to the chord length, which provides a geometric explanation of the first important limit.

Exercise 5

Find the limit $\lim_{x \to 0} \frac{\sin 3x}{x}$ .

Reference Answer

important limit

Idea: Use the generalized form of the first important limit.

Detailed steps:

$\lim_{x \to 0} \frac{\sin 3x}{x} = \lim_{x \to 0} 3 \cdot \frac{\sin 3x}{3x}$
$= 3 \cdot \lim_{x \to 0} \frac{\sin 3x}{3x} = 3 \cdot 1 = 3$

Answer: The limit is 3.

Exercise 6

Find the limit $\lim_{x \to 0} \frac{\sin x^2}{x^2}$ .

Reference Answer

important limit

Idea: Use a change of variable and the first important limit.

Detailed steps:

Let $t = x^2$ ; then as $x \to 0$ , $t \to 0$
$\lim_{x \to 0} \frac{\sin x^2}{x^2} = \lim_{t \to 0} \frac{\sin t}{t} = 1$

Answer: The limit is 1.

Exercise 7

Find the limit $\lim_{x \to 0} \frac{\sin 2x}{\sin 3x}$ .

Reference Answer

important limit

Idea: Use the first important limit and equivalent infinitesimals.

Detailed steps:

As $x \to 0$ , $\sin 2x \sim 2x$ and $\sin 3x \sim 3x$
$\lim_{x \to 0} \frac{\sin 2x}{\sin 3x} = \lim_{x \to 0} \frac{2x \cdot \frac{\sin 2x}{2x}}{3x \cdot \frac{\sin 3x}{3x}} = \frac{2}{3}$

Answer: The limit is $\frac{2}{3}$ .

Exercise 8

Use the method of comparing areas to prove the first important limit. Explain concretely how, in the unit circle, when the angle $x$ is very small, the relationship between the sector area and the triangle area derives $\lim_{x \to 0} \frac{\sin x}{x} = 1$ .

Reference Answer

important limit

Idea: Prove the first important limit by comparing the sector area and the triangle area in the unit circle.

Detailed steps:

Geometric construction:
- In the unit circle, take a sector with angle $x$
- The sector is enclosed by two radii and an arc
- Inside the sector there is a triangle, enclosed by two radii and a chord
Area computation:
- Sector area: $S_{\text{sector}} = \frac{1}{2} \cdot 1^2 \cdot x = \frac{x}{2}$
- Triangle area: $S_{\text{triangle}} = \frac{1}{2} \cdot 1 \cdot 1 \cdot \sin x = \frac{\sin x}{2}$
Area relationship:
- When $x$ is very small, the sector area and the triangle area are very close
- That is: $\frac{x}{2} \approx \frac{\sin x}{2}$
- Therefore: $x \approx \sin x$
Limit derivation:
- As $x \to 0$ , $\frac{\sin x}{x} \approx 1$
- Hence: $\lim_{x \to 0} \frac{\sin x}{x} = 1$

Answer: By comparing the sector area and the triangle area in the unit circle, the two areas become equal as the angle $x$ tends to 0, thereby proving $\lim_{x \to 0} \frac{\sin x}{x} = 1$ .

Summary

Symbols Used in This Article

符号说明

符号	类型	读音/说明	在本文中的含义
$\sin$	数学符号	sine function	A trigonometric function
$\lim$	数学符号	limit	The limit of a function or sequence
$\to$	数学符号	tends to	A variable tending to some value
$\pi$	希腊字母	Pi	The ratio of a circle’s circumference to its diameter

中英对照

中文术语	英文术语	音标	说明
第一个重要极限	first important limit	/fɜːst ɪmˈpɔːtənt ˈlɪmɪt/	One of the most fundamental limits in limit computation
正弦函数	sine function	/saɪn ˈfʌŋkʃən/	A trigonometric function, denoted $\sin x$
单位圆	unit circle	/ˈjuːnɪt ˈsɜːkəl/	A circle of radius 1
扇形	sector	/ˈsektə/	The region in a circle enclosed by two radii and an arc
三角形	triangle	/ˈtraɪæŋɡəl/	A figure enclosed by three sides
夹逼准则	squeeze theorem	/skwiːz ˈθɪərəm/	A method of evaluating limits by constructing inequalities

Important Limits

The Second Important Limit

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The First Important Limit

Visual Illustration

Geometric Figure

The Limit Expression

Ideas for the Proof

Proof by the Squeeze Theorem

Generalized Forms

1. General Form

2. Composite Form

3. Reciprocal Form

Graph

Geometric Proof

Practice Problems

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Exercise 6

Exercise 7

Exercise 8

Summary

Symbols Used in This Article

中英对照

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