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
- 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`
- 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?
- 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`.
- 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`).