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.

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: each term $\tfrac1n\ln\tfrac kn$ is at least $\int_{(k-1)/n}^{k/n}\ln x\,dx$ and, for $k < n$, at most $\int_{k/n}^{(k+1)/n}\ln x\,dx$ (the $k = n$ term is $0$). Summing, $-1 = \int_0^1\ln x\,dx \le \tfrac1n\sum_{k=1}^n\ln\tfrac kn \le \int_{1/n}^1\ln x\,dx \to -1$. 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 $\left|\cos x - \left(1 - \tfrac{x^2}{2}\right)\right| = \tfrac{|\cos\xi|}{24}x^4 \le \tfrac{x^4}{24}$.

Step 3 — the one-sided bound comes from stopping one term earlier: $\cos x = 1 - \tfrac{\cos\eta}{2}x^2$ for some $\eta$ between $0$ and $x$, and $\cos\eta \le 1$ gives $\cos x \ge 1 - \tfrac{x^2}{2}$ for EVERY real $x$.

Step 4 — so $1 - \tfrac{x^2}{2} \le \cos x \le 1 - \tfrac{x^2}{2} + \tfrac{x^4}{24}$ (the upper bound is the remainder estimate with its sign), and at $x = 1$ this pins $\cos 1 \approx 0.5403$ between $0.5$ and $0.5417$. The remainder term is a precise, citable error bar.

Graders look for three things: the $\delta$ is produced before the estimate, it depends only on $\varepsilon$ (and on the point, unless the claim is uniform), and the chain of inequalities actually reaches $\varepsilon$. Working backwards on scratch paper to FIND $\delta$ is fine; the written proof states it and verifies forwards. For $f(x) = x^2$ near $x_0$, the standard choice is $\delta = \min\left(1, \dfrac{\varepsilon}{2|x_0| + 1}\right)$, and the $\min$ with $1$ is what bounds the stray factor $|x + x_0|$.

The distinction decides what survives the limit. $f_n(x) = x^n$ on $[0,1]$ converges pointwise to a discontinuous limit, so the convergence cannot be uniform; uniform limits of continuous functions ARE continuous. Uniform convergence also licenses swapping limit with integral, and the Weierstrass $M$-test ($|f_n| \le M_n$ with $\sum M_n < \infty$) is the usual way to establish it for a series.

That is the reason so many Putnam statements say "continuous on $[0,1]$": the extreme value theorem hands you a maximiser to work with, and uniform continuity upgrades local estimates to global ones. On an open or unbounded domain both can fail — $f(x) = 1/x$ on $(0,1)$ is continuous, unbounded and not uniformly continuous — so the first question about any analysis problem is what the domain is.

Step 1 — pointwise limit: for each fixed $x > 0$, $f_n(x) = 0$ once $n > 1/x$, and $f_n(0) = 0$, so $f_n \to 0$ pointwise.

Step 2 — the integrals: $\int_0^1 f_n = n\cdot\tfrac1n = 1$ for every $n$.

Step 3 — so the limit of the integrals is $1$ while the integral of the limit is $0$. The mass escapes into a taller, narrower spike; uniform convergence (or domination) is exactly what forbids that, which is why the hypothesis must be stated, not assumed.