Question

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

Answer 1

For a non-rigorous proof, please see below.

Explanation:

For a positive central angle of `x` radians (`0 < x < pi/2`) (not degrees)

Source:

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.