The limit as t --> 0, of, sin(t)/t

[danbaron]

What is the limit as t --> 0, of, sin(t)/t?

("t --> 0", means, "t approaches 0")

We can't just plug in 0 for t in sin(t)/t, because, we get, 0/0.

But, we can do the calculation.

t is the angle shown in the picture (see the attached file, "test.jpg"). By definition, cos(t) equals length 0A divided by length 0P. By definition, sin(t) equals length AP divided by length 0P. In this case, length 0P equals 1, so, cos(t) equals length 0A, and sin(t) equals length AP.

We need to be able to determine the area of a sector (with center at 0) of a circle.

We know that the area of a circle is pi*r^2. And, the area of a sector is proportional to the angle, t, so,

area of sector / (pi*r^2) = t/(2*pi) (so, when t = 2*pi, the ratio is 1).

Therefore,

area of sector = (t/2*pi) * pi*r^2, = t*r^2/2.

{
How do we know that the area, A, of a circle is pi*r^2? We can do it like this. Divide the circle into n isosceles triangles. For instance, if n equals 4, then we will have circumscribed the circle with a square. The length of the square's side, l, will equal the base, b, of each of the 4 triangles. The circle's radius will equal the height, h, of each of the 4 triangles. So for n = 4, our approximation of A will be 4 times the area of one of the triangles, equaling, 4*b*h/2, or, 4*l*r/2, where the perimeter of the square equals 4*l. If, instead, n equals 5, we will have circumscribed the circle with a regular (with equal sides) pentagon. As n increases, we circumscribe the circle with regular polygons having more and more sides. We let n become very big. Then, it is reasonable to assume that the perimeter of the polygon, n*l, approaches the circle's circumference, 2*pi*r, and we get that A = pi*r^2.

You could ask, how do we know the circumference of a circle equals 2*pi*r? If you think about it, the circumference of any circle can be written as r times "something" (no matter what the size of the circle, the proportion between r and "something" doesn't change), or, as the diameter, d, times "something" divided by 2. It turns out that, the "something", is 2*pi. There are lots of ways to determine pi's value.
}

The area of sector 0AC is,

t*cos(t)^2/2.

The area of triangle 0AP is,

cos(t)*sin(t).

The area of sector 0BP is,

t/2.

If you look at the picture, you see that,

area of sector 0AC < area of triangle 0AP < area of sector 0BP, so,

t*cos(t)^2/2 < cos(t)*sin(t) < t/2.

We want to turn the center term in the inequality into sin(t)/t, so we multiply the inequality by, 2/(t*cos(t)), and we get,

cos(t) < sin(t)/t < 1/cos(t).

From the picture, the limit as t --> 0, of cos(t), is 1.

Therefore, the limit as t --> 0, of 1/cos(t), is 1.

Since, sin(t)/t is squeezed between cos(t) and 1/cos(t), the limit as t --> 0, of sin(t)/t, must also be 1.

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Squeeze Principle

If g(x) <= f(x) <= h(x) and

limit as x --> a of g(x) = L = limit as x --> a of h(x), then

limit as x --> a of f(x) = L.

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Bonus calculation.

What is the limit as t --> 0, of, (1-cos(t))/t?

Again, we can't just plug in 0 for t in (1-cos(t))/t, because, we again get 0/0.

(1-cos(t))/t = ((1-cos(t))/t)*((1+cos(t))/(1+cos(t))),

= (1-cos(t)^2)/(t*(1+cos(t)) = sin(t)^2/(t*(1+cos(t)),

= (sin(t)/t)*(sin(t)/(1+cos(t)).

Therefore,

the limit as t --> 0, of, (1-cos(t))/t = the limit as t --> 0, of, (sin(t)/t)*(sin(t)/(1+cos(t)).

From above,

the limit as t --> 0, of, (sin(t)/t) = 1.

Now, we get,

the limit as t --> 0, of, (1-cos(t))/t = the limit as t --> 0, of, 1*(sin(t)/(1+cos(t)),

= 1*(0/(1+1) = 0.

It means that as t --> 0, 1-cos(t) becomes much smaller than t.