L’Hôpital’s Rule (0/0 form, one-sided): Let a<ba<ba<b, and let f,g:[a,b)→Rf,g:[a,b)\to\mathbb{R}f,g:[a,b)→R be continuous on [a,b)[a,b)[a,b) and differentiable on (a,b)(a,b)(a,b). Assume: f(a)=g(a)=0f(a)=g(a)=0f(a)=g(a)=0, g′(x)≠0g'(x)\neq 0g′(x)=0 for all x∈(a,b)x\in(a,b)x∈(a,b), the limit L=lim⁡x→a+f′(x)g′(x)L=\lim_{x\to a^+}\frac{f'(x)}{g'(x)}L=limx→a+​g′(x)f′(x)​ exists in R∪{±∞}\mathbb{R}\cup\{\pm\infty\}R∪{±∞}. Then the limit lim⁡x→a+f(x)g(x)\lim_{x\to a^+}\frac{f(x)}{g(x)}limx→a+​g(x)f(x)​ exists and equals LLL: lim⁡x→a+f(x)g(x)=lim⁡x→a+f′(x)g′(x)=L. \lim_{x\to a^+}\frac{f(x)}{g(x)}=\lim_{x\to a^+}\frac{f'(x)}{g'(x)}=L. limx→a+​g(x)f(x)​=limx→a+​g′(x)f′(x)​=L. ...

Let f:U⊆Rn→Rf:U\subseteq \mathbb{R}^n\to\mathbb{R}f:U⊆Rn→R and g:U→Rmg:U\to\mathbb{R}^mg:U→Rm be differentiable, and consider maximizing/minimizing f(x)f(x)f(x) subject to the constraint g(x)=0g(x)=0g(x)=0. A point x∗∈Ux^\ast\in Ux∗∈U satisfies the Lagrange multiplier condition if g(x∗)=0g(x^\ast)=0g(x∗)=0, and there exists λ∈Rm\lambda\in\mathbb{R}^mλ∈Rm such that ∇f(x∗)=Dg(x∗)Tλ.\nabla f(x^\ast)=Dg(x^\ast)^{\mathsf T}\lambda.∇f(x∗)=Dg(x∗)Tλ. Geometrically, this says the gradient of fff at an extremum is orthogonal to the tangent space of the constraint set (when that tangent space is well-defined). Under regularity hypotheses (e.g., Dg(x∗)Dg(x^\ast)Dg(x∗) has rank mmm), this condition is necessary for constrained extrema. ...

Lagrange multipliers theorem: Let U⊆RnU\subseteq\mathbb{R}^nU⊆Rn be open, let f:U→Rf:U\to\mathbb{R}f:U→R and g:U→Rmg:U\to\mathbb{R}^mg:U→Rm be C1C^1C1, and define the constraint set C={x∈U:g(x)=0}. C=\{x\in U: g(x)=0\}. C={x∈U:g(x)=0}. Suppose x∗∈Cx^\ast\in Cx∗∈C is a local maximizer or minimizer of fff restricted to CCC, and assume the constraint is regular at x∗x^\astx∗: rank⁡Dg(x∗)=m. \operatorname{rank} Dg(x^\ast)=m. rankDg(x∗)=m. Then there exists λ∈Rm\lambda\in\mathbb{R}^mλ∈Rm such that ∇f(x∗)=Dg(x∗)Tλ. \nabla f(x^\ast)=Dg(x^\ast)^{\mathsf T}\lambda. ∇f(x∗)=Dg(x∗)Tλ. ...

Least Upper Bound Theorem: If E⊆RE\subseteq \mathbb{R}E⊆R is nonempty and bounded above , then sup⁡E\sup EsupE exists in R\mathbb{R}R. This theorem is the working form of completeness : it guarantees the existence of optimal bounds and is used to prove convergence of monotone sequences , the existence of limits, and many properties of continuous functions. Proof sketch (optional): This is typically taken as an axiom (completeness) or proved from an equivalent completeness formulation (e.g., Cauchy completeness or nested intervals). The main idea is that the “gap-free” nature of R\mathbb{R}R forces a least upper bound to exist. ...

Lebesgue criterion for Riemann integrability: Let f:[a,b]→Rf:[a,b]\to\mathbb{R}f:[a,b]→R be bounded , and let D⊆[a,b]D\subseteq[a,b]D⊆[a,b] be the set of points where fff is discontinuous. Then fff is Riemann integrable on [a,b][a,b][a,b] if and only if DDD has measure zero ; i.e., for every ε>0\varepsilon>0ε>0 there exists a countable collection of open intervals {Ij}\{I_j\}{Ij​} such that D⊆⋃j=1∞Ijand∑j=1∞∣Ij∣<ε. D\subseteq \bigcup_{j=1}^\infty I_j \quad\text{and}\quad \sum_{j=1}^\infty |I_j|<\varepsilon. D⊆⋃j=1∞​Ij​and∑j=1∞​∣Ij​∣<ε. ...

Lebesgue Number Lemma: Let (X,d)(X,d)(X,d) be a compact metric space and let U\mathcal{U}U be an open cover of XXX. Then there exists δ>0\delta>0δ>0 (a Lebesgue number for U\mathcal{U}U) such that for every x∈Xx\in Xx∈X, B(x,δ)⊆Ufor some U∈U.B(x,\delta)\subseteq U \quad \text{for some } U\in\mathcal{U}.B(x,δ)⊆Ufor some U∈U. This lemma is used to pass from pointwise local control to uniform control on compact sets (e.g., in proofs of uniform continuity and partitions of unity in more advanced settings). ...

Refinement lemma (used for Lebesgue numbers): Let (X,d)(X,d)(X,d) be a metric space , let K⊆XK\subseteq XK⊆X be compact , and let U\mathcal{U}U be an open cover of KKK. For each x∈Kx\in Kx∈K, choose Ux∈UU_x\in\mathcal{U}Ux​∈U with x∈Uxx\in U_xx∈Ux​. Since UxU_xUx​ is open, there exists rx>0r_x>0rx​>0 such that B(x,rx)⊆Ux. B(x,r_x)\subseteq U_x. B(x,rx​)⊆Ux​. Then there exist points x1,…,xN∈Kx_1,\dots,x_N\in Kx1​,…,xN​∈K such that K⊆⋃i=1NB ⁣(xi,rxi2). K\subseteq \bigcup_{i=1}^N B\!\left(x_i,\frac{r_{x_i}}{2}\right). K⊆⋃i=1N​B(xi​,2rxi​​​). In particular, if δ=min⁡1≤i≤Nrxi2\delta=\min_{1\le i\le N} \frac{r_{x_i}}{2}δ=min1≤i≤N​2rxi​​​, then δ>0\delta>0δ>0 and for every x∈Kx\in Kx∈K there exists U∈UU\in\mathcal{U}U∈U with B(x,δ)⊆UB(x,\delta)\subseteq UB(x,δ)⊆U. ...

Limit Comparison Test: Let an>0a_n>0an​>0 and bn>0b_n>0bn​>0. If lim⁡n→∞anbn=L\lim_{n\to\infty}\frac{a_n}{b_n}=Llimn→∞​bn​an​​=L with 0<L<∞0<L<\infty0<L<∞, then ∑an\sum a_n∑an​ converges if and only if ∑bn\sum b_n∑bn​ converges. ...

Let (an)(a_n)(an​) be a sequence in the extended real line [−∞,∞][-\infty,\infty][−∞,∞]. Define the tail infima in:=inf⁡{ak:k≥n}∈[−∞,∞].i_n := \inf\{a_k : k\ge n\}\in[-\infty,\infty].in​:=inf{ak​:k≥n}∈[−∞,∞]. Then the limit inferior of (an)(a_n)(an​) is lim inf⁡n→∞an:=lim⁡n→∞in,\liminf_{n\to\infty} a_n := \lim_{n\to\infty} i_n,n→∞liminf​an​:=n→∞lim​in​, where the limit exists in [−∞,∞][-\infty,\infty][−∞,∞] because (in)(i_n)(in​) is nondecreasing. ...

Let (X,dX)(X,d_X)(X,dX​) and (Y,dY)(Y,d_Y)(Y,dY​) be metric spaces , let E⊆XE\subseteq XE⊆X, let x0∈Xx_0\in Xx0​∈X be a limit point of EEE (meaning every neighborhood of x0x_0x0​ meets E∖{x0}E\setminus\{x_0\}E∖{x0​}), and let f:E→Yf:E\to Yf:E→Y. We say that f(x)f(x)f(x) tends to L∈YL\in YL∈Y as x→x0x\to x_0x→x0​, written ...