How do you use the Squeeze Theorem to find lim Sin(x)/x as x approaches zero?

1 Answer
Oct 19, 2015

For a non-rigorous proof, please see below.

Explanation:

For a positive central angle of x radians (0 < x < pi/2) (not degrees)
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Source:
commons.wikimedia.orgcommons.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.