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

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Jan 2, 2018

Answer:

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:
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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.

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