IVT, Continuity, Squeeze Theorem, and Sin x / x Limit Notes

IVT and Continuity: Core Ideas

Intermediate Value Theorem (IVT): If a function f is continuous on a closed interval [a,b] and y is any value between f(a) and f(b) (i.e., min{f(a),f(b)} ≤ y ≤ max{f(a),f(b)}), then there exists c ∈ [a,b] such that f(c) = y.
Special case for roots: If f(a) and f(b) have opposite signs (f(a)·f(b) ≤ 0) and f is continuous on [a,b], then there exists c ∈ [a,b] with f(c) = 0.
Practical use: To show there is a root in an interval, first verify continuity on the interval; then check a sign change (or that the target value lies between f(a) and f(b)).

Typical problem setup: Given a function f and an interval [a,b], use IVT to show existence of a root.
Key requirement: The function must be continuous on [a,b].
Issue in this example: The function f(x) = 2 tan(x) is not continuous on any interval that includes a vertical asymptote of tan(x) (where cos(x) = 0, i.e., x = π/2 + kπ).
Conclusion from the transcript: It is false to claim that IVT can guarantee at least one root in this case because continuity on the entire interval fails.
Takeaway: IVT cannot be applied to non-continuous functions on the interval; always check continuity first.

The idea discussed: To use IVT, you often look for a sign change or ensure a target value lies between f(a) and f(b).
If you try to assert a "gap" between f(a) and f(b) and pick a y-value that does not lie between them, IVT cannot guarantee a root.
Extension: The discussion also touches the Extreme Value Property: on a closed interval, a continuous function attains an absolute maximum and minimum (Extreme/Absolute Value Theorem).
Important contrast: If the function were not continuous, the function might fail to have a maximum or minimum on the interval.
Reminder from the transcript: In trig contexts, select radian measure; sin and cos are continuous everywhere, which supports straightforward limit computations via substitution.

Sine and cosine are continuous everywhere on the real line.
In trig limits, use radian measure for angles.
Direct substitution for continuous functions:
- ext{If } f ext{ is continuous at } c, ext{ then } \ \
  \,\lim_{x\to c} f(x) = f(c).
Example substitutions (from the transcript):
- \lim_{x\to 0} \sin x = \sin(0) = 0.
- \lim_{x\to 0} \cos x = \cos(0) = 1.
The transcript emphasizes that sin and cos are periodic; for sine and cosine, specific properties are recalled:
- Sine is odd: (\sin(-x) = -\sin(x)).
- Cosine is even: (\cos(-x) = \cos(x)).
- Periodicity: (\sin(x+2\pi) = \sin x), (\cos(x+2\pi) = \cos x).
A common mental model used in the lecture: when reasoning about limits near a point c, consider values of functions f, g, h that bracket g(x) and use a limit argument when the outer functions have the same limit (Squeeze Theorem).

Setup: Three functions f(x), g(x), h(x) with f(x) ≤ g(x) ≤ h(x) near c (for x near c, excluding c maybe).
If
\lim{x\to c} f(x) = \lim{x\to c} h(x) = L,
then
\lim_{x\to c} g(x) = L.
The transcript’s analogy uses three characters (Alex, Cecilia, Cameron) to illustrate the idea: if two outer limits converge to the same L, the middle one is squeezed to L as well.

Target: Evaluate \lim_{x\to 0} \frac{\sin x}{x}.
The straightforward substitution yields a 0/0 indeterminate form. The transcript notes to avoid just plugging in or using algebraic tricks like factoring or conjugates.
Standard approach (as per the transcript’s teaching): use the Squeeze Theorem with trigonometric inequalities rather than algebraic manipulations.
Key trig inequality (for x near 0, in radians):
- For x > 0:
  \cos x \le \frac{\sin x}{x} \le 1.
  This comes from the classic unit circle geometry:
- In the standard derivation one uses (\sin x \le x \le \tan x) for x in (0, π/2).
- Dividing by x sin x (positive in this range) yields:
  1 \le \frac{x}{\sin x} \le \frac{1}{\cos x} \Rightarrow \cos x \le \frac{\sin x}{x} \le 1.
- For x < 0: the same inequality holds by symmetry (cos is even, sin is odd).
Limits of the bounding functions as x → 0:
- (\lim_{x\to 0} \cos x = 1)
- The upper bound is constantly 1, so its limit is 1.
By the Squeeze Theorem, since
\cos x \le \frac{\sin x}{x} \le 1
and both bounds tend to 1 as x → 0, we obtain:
\lim_{x\to 0} \frac{\sin x}{x} = 1.
Note: The transcript mentions the preference for avoiding factoring or conjugate techniques in this particular discussion; the inequality/Squeeze approach is the highlighted method.
Result: \displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1.

IVT prerequisites recap:
- f must be continuous on the interval [a,b].
- The target value y must lie between f(a) and f(b) (or f(a) and f(b) must have opposite signs for a root).
Continuity and extreme values:
- On a closed interval [a,b], a continuous function attains a maximum and a minimum (Extreme/Maximum Value Theorem).
- If a function is not continuous on [a,b], it may fail to have a maximum or minimum.
Trig limits and radians:
- Trig limits rely on radian measure; all small-angle limit arguments are grounded in radians.
Key takeaways from the transcript's examples:
- Do not apply IVT if continuity is violated on the interval.
- Use the Squeeze Theorem to evaluate limits that yield indeterminate forms like 0/0, via bounding inequalities.
- For basic trig limits like sin x and cos x at 0, direct substitution works because of continuity: sin 0 = 0 and cos 0 = 1.
Quick reference formulas:
- IVT (general): If f is continuous on [a,b] and y is between f(a) and f(b), ∃ c ∈ [a,b] s.t. f(c) = y.
- For a root: if f(a)·f(b) ≤ 0, ∃ c ∈ [a,b] with f(c) = 0.
- Squeeze Theorem (informal): If f(x) ≤ g(x) ≤ h(x) and lim f = lim h = L, then lim g = L.
- Limit of sin x / x as x → 0: \lim_{x\to 0} \frac{\sin x}{x} = 1.