Geometric Proof of a Limit

Date: 12/06/2003 at 19:15:19
From: Jackos
Subject: Limits

Can you prove that lim[x->0](sinx)/x = 1 without using L'Hopital's 
rule?

Date: 12/06/2003 at 21:32:49
From: Doctor Rob
Subject: Re: Limits

Thanks for writing to Ask Dr. Math, Jackos!

Here's a nice geometric proof.

Draw a unit circle with center at the origin O.  From the origin, draw 
a radius which makes an acute angle of t radians with the positive 
x-axis.  Then 0 < t < Pi/2.  It intersects the circle at the point 
P(cos[t],sin[t]).  Draw the tangent line x = 1.  Extend the radius to 
meet that tangent at the point R(1,tan[t]).  Connect P to Q(1,0).

Consider the areas of the triangle OPQ, the sector OPQ of the circle, 
and the triangle OQR.  The area of triangle OPQ is (1/2)*1*1*sin[t]
(half the product of two sides times the sine of the included angle).  
The area of sector OPQ is (1/2)*1^2*t (half the square of the radius 
times the central angle measure).  The area of the triangle OQR is 
(1/2)*1*tan[t] (half the base times the altitude).

Obviously by containment, those three areas satisfy

   A(triangle OPQ) < A(sector OPQ) < A(triangle OQR)

      (1/2)*sin[t] < (1/2)*t       < (1/2)*tan[t]

            sin[t] < t             < sin[t]/cos[t]

Rearranging these inequalities, we have the general relationship

   cos[t] < sin[t]/t < 1,  0 < t < Pi/2.

Also, if -Pi/2 < t < 0, then 0 < -t < Pi/2, so

   cos[-t] < sin[-t]/(-t) < 1,  -Pi/2 < t < 0

    cos[t] < sin[t]/t     < 1,  -Pi/2 < t < 0.

Thus the inequality holds for all nonzero values of t between -Pi/2 
and Pi/2.  Now take the limit as t -> 0.

   1 = lim cos[t] <= lim sin[t]/t <= lim 1 = 1,
       t->0          t->0            t->0

so

   lim sin[t]/t = 1.
   t->0

There's another proof which uses the infinite Maclaurin series

   sin(t) = t - t^3/3! + t^5/5! - t^7/7! + ...

Feel free to write again if I can help further.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/