This proof of this limit uses the Squeeze Theorem. However, getting things set up to use the
Squeeze Theorem can be a somewhat complex geometric argument that can be
difficult to follow so we’ll try to take it fairly slow.
Let’s start by assuming that . Since we are proving a limit that has it’s okay to assume that is not too large (i.e. ).
Also, by assuming that is positive we’re actually going to first
prove that the above limit is true if it is the righthand limit. As you’ll see if we can prove this then the
proof of the limit will be easy.
So, now that we’ve got our assumption on taken care of let’s start off with the unit
circle circumscribed by an octagon with a small slice marked out as shown
below.
Points A and C are the midpoints of their
respective sides on the octagon and are in fact tangent to the circle at that
point. We’ll call the point where
these two sides meet B.
From this figure we can see that the circumference of the
circle is less than the length of the octagon. This also means that if we look at the
slice of the figure marked out above then the length of the portion of the
circle included in the slice must be less than the length of the portion of
the octagon included in the slice.
Because we’re going to be doing most of our work on just
the slice of the figure let’s strip that out and look at just it. Here is a sketch of just the slice.
Now denote the portion of the circle by and the lengths of the two portion of the
octagon shown by and . Then by the observation about lengths we
made above we must have,
(1)
Next, extend the lines AB
and OC as shown below and call the
point that they meet D. The triangle now formed by AOD is a right triangle. All this is shown in the figure below.
The triangle BCD
is a right triangle with hypotenuse BD
and so we know . Also notice that . If we use these two facts in (1)
we get,
(2)
Next, as noted already the triangle AOD is a right triangle and so we can use a little right triangle
trigonometry to write . Also note that since it is nothing more than the radius of
the unit circle. Using this
information in (2)
gives,
(3)
The next thing that we need to recall is that the length
of a portion of a circle is given by the radius of the circle times the angle
that traces out the portion of the circle we’re trying to measure. For our portion this means that,
So, putting this into (3)
we see that,
or, if we do a little rearranging we get,
(4)
We’ll be coming back to (4)
in a bit. Let’s now add in a couple
more lines into our figure above.
Let’s connect A and C with a line and drop a line straight
down from C until it intersects AO at a right angle and let’s call the
intersection point E. This is all shown in the figure below.
Okay, the first thing to notice here is that,
(5)
Also note that triangle EOC is a right triangle with a hypotenuse of . Using some right triangle trig we can see
that,
Plugging this into (5)
and recalling that we get,
and with a little rewriting we get,
(6)
Okay, we’re almost done here. Putting (4)
and (6)
together we see that,
provided . Let’s also note that,
We are now set up to use the Squeeze Theorem. The only issue that we need to worry about
is that we are staying to the right of in our assumptions and so the best that the
Squeeze Theorem will tell us is,
So, we know that the limit is true if we are only working
with a righthand limit. However we
know that is an odd function and so,
In other words, if we approach zero from the left (i.e. negative ’s) then we’ll get the same values in
the function as if we’d approached zero from the right (i.e. positive ’s) and so,
We have now shown that the two onesided limits are the
same and so we must also have,
