# limit of $\frac{sinx}{x}$ as $x$ approaches 0

###### Theorem 1.

$$\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{x}=1$$ |

$$ for $$.

###### Proof.

First, let $$. Then $$. Note also that

$$ | (1) |

Multiplying both of this inequality^{} by $\mathrm{cos}x$ yields

$$ | (2) |

By this theorem (http://planetmath.org/ComparisonOfSinThetaAndThetaNearTheta0),

$$ | (3) |

$$ | (4) |

Dividing by $x$ yields

$$ | (5) |

Now let $$. Then $$. Plugging $-x$ into inequality (5) gives

$$ | (6) |

Since $\mathrm{cos}$ is an even function^{} and $\mathrm{sin}$ is an odd function, we have

$$ | (7) |

Therefore, inequality (5) holds for all real $x$ with $$.

Since $\mathrm{cos}$ is continuous^{}, $\underset{x\to 0}{lim}\mathrm{cos}x=\mathrm{cos}0=1$. Thus,

$$1=\underset{x\to 0}{lim}\mathrm{cos}x\le \underset{x\to 0}{lim}\frac{\mathrm{sin}x}{x}\le \underset{x\to 0}{lim}1=1.$$ | (8) |

By the squeeze theorem, it follows that $\underset{x\to 0}{lim}{\displaystyle \frac{\mathrm{sin}x}{x}}=1$. ∎

Note that the above limit is also valid if $x$ is considered as a complex variable.

Title | limit of $\frac{sinx}{x}$ as $x$ approaches 0 |
---|---|

Canonical name | LimitOfdisplaystylefracsinXxAsXApproaches0 |

Date of creation | 2013-03-22 16:58:45 |

Last modified on | 2013-03-22 16:58:45 |

Owner | Wkbj79 (1863) |

Last modified by | Wkbj79 (1863) |

Numerical id | 10 |

Author | Wkbj79 (1863) |

Entry type | Theorem |

Classification | msc 26A06 |

Classification | msc 26A03 |

Related topic | ComparisonOfSinThetaAndThetaNearTheta0 |

Related topic | SincFunction |

Related topic | DerivativesOfSinXAndCosX |

Related topic | DerivativesOfSineAndCosine |