September 24th, 2026

Updated 1 Oct 2026Notes (PDF)

[0.0.11]Theorem(Monotone Convergence Theorem)#

Suppose $(x_n)$ is a monotonically increasing (resp. decreasing) sequence that is bounded above (resp. below). Then $x_n$ converges to $\sup\{x_n\}$ (resp. $\inf\{x_n\}$ ).

Proof.

Without loss of generality, suppose $x_n$ is monotonically increasing and bounded above. By completeness of $\bR$ , we are justified in letting $L = \sup\{x_n\}$ : we show $x_n \to L$ . Fix $\epsilon > 0$ . Since $L - \epsilon < L$ , there exists some $N \in \bZ_+$ such that $x_N > L - \epsilon$ , by definition of the supremum. Since $x_n$ is monotonically increasing, we have that

L - \epsilon < x_N \le x_n

for all $n \ge N$ . Since $L$ is an upper bound, note that $x_n < L + \epsilon$ for all $n \ge N$ as well. Thus $|x_n - L| < \epsilon$ for all $n \ge N$ , completing the proof.

[0.0.12]Theorem(Squeeze Theorem)#

Suppose $(a_n)$ and $(b_n)$ are sequences of reals that both converge to $L$ . Let $(x_n)$ be a sequence such that $a_n \le x_n \le b_n$ for every $n \in \bZ_+$ . Then $x_n \to L$ .

Proof.

Fix $\epsilon > 0$ . There is some $N \in \bZ_+$ such that $|a_n - L| < \epsilon/2$ and $|b_n - L| < \epsilon/2$ for every $n \ge N$ . Then note

L - \epsilon/2 < a_n \le x_n \le b_n < L + \epsilon/2,

and so $|x_n - L| \le \epsilon/2 < \epsilon$ for every $n \ge N$ , completing the proof.

All polynomials and rational functions are continuous on their domain. Elementary operations (addition, subtraction, multiplication, division, composition, etc.) of continuous functions preserve continuity. We omit the proof.

[0.0.13]Theorem(Intermediate Value Theorem)#

Suppose $f: [a, b] \to \bR$ is a continuous function. Suppose $y$ lies in between $f(a)$ and $f(b)$ : that is,

\min\{f(a), f(b)\} < y < \max\{f(a), f(b)\}.

Then there is some $c \in (a, b)$ such that $f(c) = y$ .

Proof.

Without loss of generality, suppose $f(a) < f(b)$ , and let $f(a) < y < f(b)$ . Define

S = \{x \in [a,b] : f(x) \le y\}.

Since $a \in S$ and $b$ is an upper bound for $S$ , we are justified in letting $c = \sup(S)$ . We show $f(c) = y$ .

Suppose $f(c) < y$ . By continuity of $f$ , there exists some $\delta > 0$ such that $|x-c| < \delta$ implies $f(x) < y$ . Since $c < b$ , we may choose some $x \in (c,b]$ with $|x-c| < \delta$ . Then $x \in S$ , contradicting that $c$ is an upper bound for $S$ .

Suppose instead $f(c) > y$ . By continuity of $f$ , there exists some $\delta > 0$ such that $|x-c| < \delta$ implies $f(x) > y$ . Since $c = \sup(S)$ , there exists some $x \in S$ such that $c-\delta < x \le c$ . But then $f(x) > y$ , contradicting $x \in S$ . Thus $f(c) = y$ .

If $f(a) > f(b)$ , we may symmetrically apply this proof to $-f$ . If $f(a) = f(b)$ , then the statement is vacuously true.

[0.0.14]Theorem(Stationary Point Theorem)#

Suppose $f: [a, b] \to \bR$ is a continuous function, and there is some $c \in (a, b)$ such that $f(c) > f(x)$ for every $x \in [a, b]$ . If $f$ is differentiable at $c$ , then $f'(c) = 0$ .

Proof.

Recall that

f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}.

Note that, for all $x \in [a, b]$ , we have $f(c) > f(x)$ , and so the numerator of the difference quotient is always negative. For every $x < c$ , note that $x - c \le 0$ , and so $f'(c) > 0$ . For every $x > c$ , note that $x - c > 0$ , and so $f'(c) \le 0$ . Thus $f'(c) = 0$ .