# Prove limit x tends to zero sin x/x=1 using sandwitch theorem?

2
by kkakhil32

2015-01-05T10:40:15+05:30

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

2015-01-05T21:54:21+05:30
Find the breadth of rectangular garden whose area is 400 Sq.m and length is 25m