Limits, Continuity & the Big Theorems

The first reflex for any strange limit: rewrite it as a difference quotient, a Riemann sum, or $\exp(\text{log})$. The second: Taylor expand — $\tfrac{\sin x - x}{x^3} \to -\tfrac16$ because $\sin x = x - \tfrac{x^3}{6} + \cdots$, and $\lim(\cos x)^{1/x^2} = e^{-1/2}$ because $\ln\cos x \approx -\tfrac{x^2}{2}$. L'Hôpital is the tool of last resort: it is slow and it hides the structure.

Existence of roots is IVT; uniqueness is monotonicity (a derivative that keeps one sign) — $x = \cos x$ has exactly one solution, $e^x = 3x$ exactly two. MVT converts a statement about VALUES into one about a DERIVATIVE at some unknown point; Rolle is the case $f(a) = f(b)$, which proves that between two roots of $f$ lies a root of $f'$ (so a polynomial with $n$ real roots has $n - 1$ real critical points). Taylor with the Lagrange remainder gives INEQUALITIES: $\cos x \ge 1 - \tfrac{x^2}{2}$, $e^x \ge 1 + x$, $\ln(1+x) \le x$ — each a one-line remainder-sign argument.

$f' = 3x^2 - 3$ vanishes at $\pm1$; $f(-1) = 3 > 0$ and $f(1) = -1 < 0$ with $f \to \mp\infty$ at $\mp\infty$: three sign changes, hence three real roots (and a cubic has at most three). They are $2\cos\tfrac{2\pi k}{9}$-type numbers — not needed for the count.

Rolle: $f$ attains a max and min on $[a,b]$ (extreme value theorem). If both occur at endpoints then $f$ is constant and any $c$ works; otherwise an interior extremum has $f'(c) = 0$ (Fermat: the difference quotients from the two sides have opposite signs). MVT: apply Rolle to $g(x) = f(x) - \tfrac{f(b) - f(a)}{b - a}(x - a)$, which has $g(a) = g(b) = f(a)$; $g'(c) = 0$ gives $f'(c) = \tfrac{f(b)-f(a)}{b-a}$. $\blacksquare$ Significance: the MVT is how every inequality between a function and its derivative is proved, and how Taylor's remainder is derived (apply it to the right auxiliary function).

$\ln\tfrac{(n!)^{1/n}}{n} = \tfrac1n\sum_{k=1}^n\ln k - \ln n = \tfrac1n\sum_{k=1}^n\ln\tfrac kn$, a Riemann sum for $\int_0^1\ln x\,dx$. The integrand is unbounded at $0$, so justify: the sum is a right-endpoint sum of a monotone function, hence squeezed between $\int_0^1\ln x\,dx = -1$ and $\int_{1/n}^1\ln x\,dx + \tfrac1n\ln\tfrac1n \to -1$. Exponentiate: the limit is $e^{-1}$. $\blacksquare$ Significance: "log, Riemann sum, exponentiate" handles every limit of a geometric-mean shape — and handling the endpoint singularity honestly is what earns the $10$.

Step 1 — MVT on $\ln$ over $[1, 1 + x]$: $\ln(1 + x) - \ln 1 = \tfrac1c\cdot x$ for some $c \in (1, 1 + x)$, and $\tfrac1c > \tfrac{1}{1+x}$, giving $\ln(1+x) > \tfrac{x}{1+x}$.

Step 2 — the same MVT also gives $\ln(1+x) < x$ (since $\tfrac1c < 1$).

Step 3 — squeeze with $x = \tfrac1n$: $\tfrac{1/n}{1 + 1/n} < \ln(1 + \tfrac1n) < \tfrac1n$, multiply by $n$: $\tfrac{n}{n+1} < n\ln(1 + \tfrac1n) < 1$, so the limit is $1$.

Step 4 — exponentiate to recover $(1 + \tfrac1n)^n \to e$, now PROVED rather than assumed.

Step 1 — Taylor at $0$ to order $3$ with Lagrange remainder: $\cos x = 1 - \tfrac{x^2}{2} + 0\cdot x^3 + \tfrac{\cos^{(4)}(\xi)}{4!}x^4$ for some $\xi$ between $0$ and $x$.

Step 2 — $\cos^{(4)} = \cos$, and $|\cos\xi| \le 1$, so the remainder is at most $\tfrac{x^4}{24}$ in absolute value.

Step 3 — in fact $\cos\xi \ge 0$ is not guaranteed, but the BOUND is; and since the next omitted term is $+\tfrac{x^4}{24}\cos\xi$, we also get $\cos x \ge 1 - \tfrac{x^2}{2}$ whenever $\cos\xi \ge 0$ — e.g. for $|x| \le \tfrac\pi2$. The remainder term is a precise, citable error bar.