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Theorem of Limit

Untitled

Theorem:
To prove:
  lim     sin θ = 1        where θ is measured in radian                                                                                       θ→0      θ

Let  O be the centre of the circle and r be the radius. AP is an arc that subtends an angle θ at centre O. Let PQ be the tangent of the circle at P. BA is produced and joined at Q.

Join PA and draw PR perpendicular to BA. Now,

Area of △AOP ≤ Area of sector AOP ≤ Area of △POQ
Now,
Area of △AOP = ½ OA. PR       [ OA = r ]
                           = ½ r. PR         
                           = ½ r2sinθ        [ PR= rsinθ ]

Area of sector AOP = ½ r2θ

Area of △POQ = ½ OP. PQ       [ OP = r ]
                          = ½ r. PQ
                          = ½ r2tanθ         [ PQ = r tanθ ]


∴   Area of △AOP  ≤  Area of sector AOP  ≤  Area of △POQ
           ½ r2sinθ  ≤  ½ r2 θ  ≤   ½ r2 tanθ 
     Dividing by sinθ on all sides,
                  1      ≤       θ     ≤        1                                                                                                                                                             sinθ           cosθ
                  1       ≥        sinθ        ≥        cosθ                                                                                                                                                θ
Applying     lim   ,                                                                                                                                                                       θ→0

 lim   1   ≥   lim    sinθ      ≥       lim    cosθ                                                                                                   θ→0          θ→0    θ                 θ→0

  1         ≥                  lim    sinθ         ≥    1
                                θ→0     θ

∴  lim     sinθ         =   1
   θ→0      θ