Wednesday, 18 June 2014

calculus - Why does limxrightarrow0fracsin(x)x=1?




I am learning about the derivative function of ddx[sin(x)]=cos(x).



The proof stated: From limx0sin(x)x=1...



I realized I don't know why, so I wanted to learn why part is true first before moving on. But unfortunately I don't have the complete note for this proof.





  1. It started with a unit circle, and then drew a triangle at (1,tan(θ))

  2. It show the area of the big triangle is tanθ2

  3. It show the area is greater than the sector, which is θ2
    Here is my question, how does this "section" of the circle equal to θ2? (It looks like a pizza slice).

  4. From there, it stated the area of the smaller triangle is sin(θ)2. I understand this part. Since the area of the triangle is 12(base×height).


  5. Then they multiply each expression by 2sin(θ) to get
    1cos(θ)θsin(θ)1




And the incomplete notes ended here, I am not sure how the teacher go the conclusion limx0sin(x)x=1. I thought it might be something to do with reversing the inequality... Is the answer obvious from this point? And how does step #3 calculation works?



Answer



Draw the circle of radius 1 centered at (0,0) in the Cartesian plane.



Let θ be the length of the arc from (1,0) to a point on the circle. The radian measure of the corresponding angle is θ and the height of the endpoint of the arc above the coordinate axis is sinθ.



Now look at what happens when θ is infinitesimally small. The length of the arc is θ and the height is also θ, since that infinitely small part of the circle looks like a vertical line (you're looking at the neighborhood of (1,0) under a microscope).



Since θ and sinθ are the same when θ is infinitesimally small, it follows that sinθθ=1 when θ is infinitesimally small.



That is how Leonhard Euler viewed the matter in the 18th century.




Why does the sector of the circle have area θ/2?



The whole circle has area πr2=π12=π. The fraction of the circle in the sector is
arccircumference=θ2π.
So the area is
θ2ππ=θ2.


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