## Answers

**First, we find the area of fig ist, 2nd and third**

**~Area of triangle in fig 1= 1/2*base*height = 1/2*OB*CD =1/2(1)Sinx (As OB=Radius=1unit and height=Sinx)****~ As sector of a circle has an area of the angle times the radius square divided by 2. So, area of fig 2nd= x(1)^2/2 (As here, angle is x and radius is 1)**

**~Area of triangle in fig third= 1/2*base*height=1/2*OB*AB=1/2(1)tanx [As Base OB=radius=1, and Height=tanx]**

**As it is clear from diagram that**

**Area fig 1 is ≤ Area of fig 2nd which is ≤ Area of fig third**

**implies Sinx/2 ≤ x/2 ≤ tanx/2**

**Multiplying by 2 we get**

**Sinx ≤ x ≤ tanx **

**Divide by Sinx,we have**

**1 ≤ x/sinx ≤ 1/Cosx**

**Taking reciprocal of the inequality, we have**

**1 ≥ Sinx/x ≥ Cosx**

**Taking Lim x approaches to 0 , we get**

**Limx-0 1 ≥ Limx-0 Sinx/x ≥ Limx-0 Cosx**

**1 ≥ 1 ≥ 1 {If 1 is ≥ a number which is ≥ to 1, then that number has to be 1 only, **

**Also acc to squeeze or sandwitch theorem which says if we have 3 functions, 1 function is always between the other 2 and those 2 functions squeeze into a certain point(here that point is 1 at limx-0) then the function in the middle has also must be squeezed to that point} **

**So when limx-0 1 and Limx-0 Cosx squeeze to 1 then acc to sandwitch theorem Limx-0 Sin/x will also be 1**