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Homelimit of $\displaystyle \frac{\sin x}{x}$ as $x$ approaches 0

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

###### Theorem 1.

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

Note that this entry uses the result $x<\tan x$ for $\displaystyle 0<x<\frac{\pi}{2}$. I have received word that a proof of this fact which uses no calculus will be supplied on PlanetMath. Please let me know when this is completed so that I can edit this entry accordingly.

###### Proof.

First, let $\displaystyle 0<x<\frac{\pi}{2}$. Then $0<\cos x<1$. Note also that

$x<\tan x.$ | (1) |

Multiplying both sides of this inequality by $\cos x$ yields

$x\cos x<\sin x.$ | (2) |

By this theorem,

$\sin x<x.$ | (3) |

Combining inequalities (2) and (3) gives

$x\cos x<\sin x<x.$ | (4) |

Dividing by $x$ yields

$\cos x<\frac{\sin x}{x}<1.$ | (5) |

Now let $\displaystyle\frac{-\pi}{2}<x<0$. Then $\displaystyle 0<-x<\frac{\pi}{2}$. Plugging $-x$ into inequality (5) gives

$\cos(-x)<\frac{\sin(-x)}{-x}<1.$ | (6) |

Since $\cos$ is an even function and $\sin$ is an odd function, we have

$\cos x<\frac{-\sin x}{-x}<1.$ | (7) |

Since $\cos$ is continuous, $\displaystyle\lim_{{x\to 0}}\cos x=\cos 0=1$. Thus,

$1=\lim_{{x\to 0}}\cos x\leq\lim_{{x\to 0}}\frac{\sin x}{x}\leq\lim_{{x\to 0}}1% =1.$ | (8) |

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

## Mathematics Subject Classification

26A06*no label found*26A03

*no label found*

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## Recent Activity

new correction: Error in proof of Proposition 2 by alex2907

Jun 24

new question: A good question by Ron Castillo

Jun 23

new question: A trascendental number. by Ron Castillo

## Comments

## Simpler proof

Since sin(x)/x is an even function, the proof for the theorem could be made more elegant just considering the case x>0. In other words, it would suffice to prove that the limit is 0 for x->0+.