## The limit of `(sin(theta))/theta` as `theta --> 0`

HELP

One of the most important trigonometry limits is the fact that `lim_(theta-->0) (sin(theta))/theta = 1`. The graph of `(sin(theta))/theta` is shown above, and as expected, the graph is not defined at `theta = 0`.

#### Explore

1. Confirm by looking at the graph above, and zooming in if necessary (shift + scroll wheel), that indeed, it appears that `lim_(theta-->0) (sin(theta))/theta = 1`
2. Now check the box next to "Show squeezing functions." Again, confirm by examining the graph above that it appears that [`cos^2 (theta) < (sin(theta))/theta < 1`] for `-pi/2 < theta < pi/2` (and `theta != 0`). If `theta = 0` does the inequality still hold?
3. Prove the next two statements, carefully citing any rules of limits and/or continuity that are needed.
• `lim_(theta-->0) 1 = 1`.
• `lim_(theta-->0) cos^2 (theta) = 1`.
4. Informally justify the following statement: If we can show that `cos^2 (theta) < (sin(theta))/theta < 1` for `-pi/2 < theta < pi/2` (and `theta != 0`), then `lim_(theta-->0) (sin(theta))/theta = 1`.

#### Proof

We now prove that `cos^2 (theta) < (sin(theta))/theta < 1` for `-pi/2 < theta < pi/2` (and `theta != 0`).

1. Consider the graph above. You can move the blue point on the unit circle to change the value of `theta`. Oberve that the `x`-value of the blue point is `cos(theta)` and the `y`-value of the blue point is `sin(theta)`.
2. Click the top checkbox to show the "small sector". Do you see how the area is computed? Recall that the area of a sector is `1/2 theta r^2` where `r` is the radius of the circle.
3. Click the other checkboxes and confirm that the area formulas are correct.
4. By comparing areas, it's clear that for `0 < theta < pi/2`, [`1/2 theta cos^2 (theta) < 1/2 sin(theta) < 1/2 theta`]
5. Multiply through by 2, [`theta cos^2 (theta) < sin(theta) < theta`]
6. Divide through by (positive) `theta`, [`cos^2 (theta) < (sin(theta))/theta < 1`]
7. Finally, we consider the case where, `-pi/2 < theta < 0`. We see
• `cos(-theta) = cos(theta)`, so `cos^2 (-theta) = cos^2 (theta)`, and
• `(sin(-theta))/-theta = (-sin(theta))/-theta = (sin(theta))/theta`,
and the previous string of inequalities still holds.
8. Thus, for `-pi/2 < theta < pi/2` (and `theta != 0`), [`cos^2 (theta) < (sin(theta))/theta < 1`] as needed.