# $\displaystyle \lim_{x \,\to\, 0} \dfrac{\sin{x}}{x}$ Rule

## Formula

$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{x}}{x}}$ $\,=\,$ $1$

### Introduction

The limit of the ratio of sine of an angle to the same angle is equal to one as the angle of a right triangle approaches zero. It is a most useful math property while finding the limit of any function in which the trigonometric function sine is involved. So, let us understand this limit rule in mathematical form.

1. Denote an angle of a right triangle by a variable $x$.
2. The ratio of the length of opposite side to the length of hypotenuse is represented by the sine of angle $x$. The sine of angle $x$ is written as $\sin{x}$ in trigonometry.

The ratio of the sine function $\sin{x}$ to the angle of a right-angled triangle $x$ is written as follows in mathematics.

$\dfrac{\sin{x}}{x}$

The limit of this rational function as the angle $x$ is closer to zero, is mathematically written as follows in calculus.

$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{x}}{x}}$

According to the calculus, the limit of the quotient sine of angle $x$ divided by the angle $x$ is one as the angle of a right triangle $x$ tends to zero.

$\therefore\,\,\,\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{x}}{x}}$ $\,=\,$ $1$

#### Other form

The trigonometric limit rule in sine function is also popularly written in the following form in mathematics.

$\displaystyle \large \lim_{\theta \,\to\, 0}{\normalsize \dfrac{\sin{\theta}}{\theta}}$ $\,=\,$ $1$

#### Proofs

There are two ways to prove this limit of trigonometric function property in mathematics.

##### Relation between Sine and Angle

It is derived on the basis of close relation between $\sin{x}$ function and angle $x$ as the angle $x$ closer to zero.

##### Taylor (or) Maclaurin Series Method

It can also be derived by the expansion of $\sin{x}$ function as per Taylor (or) Maclaurin series.

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