$x$ is a literal but represents angle of a right angled triangle and the sine function is represented by $\sin{x}$.

The ratio of $\sin{x}$ to $x$ is expressed as $\dfrac{\sin{x}}{x}$. The value of ratio of $\sin{x}$ to $x$ as $x$ approaches $0$ is written in the following mathematical form.

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

As the angle tends to zero, a special close relation between angle and sine of angle is revealed. In this special case, the value of sine function is approximately equal to the angle. Observe the following table to understand it.

Angle $(x)$ in Radian | $\sin{x}$ |
---|---|

$0$ | $0$ |

$0.1$ | $0.099833417$ $\approx 0.1$ |

$0.02$ | $0.019998667$ $\approx 0.02$ |

$0.003$ | $0.002999996$ $\approx 0.03$ |

$0.0004$ | $0.0004$ |

$0.00005$ | $0.00005$ |

$0.000123$ | $0.000123$ |

$0.0006583$ | $0.0006583$ |

$0.00027595$ | $0.00027595$ |

$0.000039178$ | $0.000039178$ |

Therefore, $\sin{x} \approx x$ as $x$ approaches zero.

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

$\require{cancel} \implies \displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin{x}}{x} = \large \lim_{x \,\to\, 0} \normalsize \dfrac{\cancel{x}}{\cancel{x}}$

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

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

Therefore, it is proved that the value of ratio of sin function to angle is equal to one as the angle approaches zero.

List of most recently solved mathematics problems.

Jul 04, 2018

Limit (Calculus)

Evaluate $\displaystyle \large \lim_{x \,\to\, \tan^{-1}{3}} \normalsize {\dfrac{\tan^2{x}-2\tan{x}-3}{\tan^2{x}-4\tan{x}+3}}$

Jun 23, 2018

Limit (Calculus)

Evaluate $\displaystyle \large \lim_{x \to 0} \normalsize \dfrac{e^{x^2}-\cos{x}}{x^2}$

Jun 22, 2018

Integral Calculus

Evaluate $\displaystyle \int \dfrac{1+\cos{4x}}{\cot{x}-\tan{x}} dx$

Jun 21, 2018

Limit

Evaluate $\displaystyle \large \lim_{x \to \infty} \normalsize {\sqrt{x^2+x+1}-\sqrt{x^2+1}}$

Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers.
Know more

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.