For a positive central angle of x radians (0 < x < pi/2) (not degrees)
Source:
commons.wikimedia.org
The geometric idea is that
"Area of "Delta KOA < "Area of " "Sector KOA" < "Area of "Delta LOA
"Area of "Delta KOA = 1/2(1)(sinx) \ \ \ (1/2"base"*"height")
"Area of " "Sector KOA" = 1/2 (1)^2 x \ \ \ (x is in radians)
"Area of "Delta LOA = 1/2tanx \ \ \ (AL = tanx)
So we have:
sinx/2 < x/2 < tanx/2
For small positive x, we have inx > 0 so we can multiply through by 2/sinx, to get
1 < x/sinx < 1/cosx
So
cosx < sinx/x < 1 for 0 < x < pi/2.
lim_(xrarr0^+) cosx = 1 and lim_(xrarr0^+) 1= 1
so lim_(xrarr0^+) sinx/x = 1
We also have, for these small x, sin(-x) = -sinx, so (-x)/sin(-x) = x/sinx and cos(-x) = cosx, so
cosx < sinx/x < 1 for -pi/2 < x < 0.
lim_(xrarr0^-) cosx = 1 and lim_(xrarr0^-) 1= 1
so lim_(xrarr0^-) sinx/x = 1
Since both one sided limits are 1, the limit is 1.
Note
This proof uses the fact that lim_(xrarr0)cosx = 1. That can also be stated "the cosine function is continuous at 0".
That fact can be proved from the fact that lim_(xrarr0) sinx = 0. (The sine function is continuous at 0.)
Which can be proved using the squeeze theorem in a argument rather like the one used above.
Furthermore: Using both of those facts we can show that the sine and cosine functions are continuous at every real number.