Limits, Continuity & Differentiability

Foundational Limit Definition

We write $\lim_{x \to a} f(x) = L$ if for every $\epsilon > 0$, there exists $\delta > 0$ such that $0 < |x - a| < \delta \implies |f(x) - L| < \epsilon$. The limit exists if and only if the Left Hand Limit (LHL) and Right Hand Limit (RHL) are finite and equal: $$\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L$$

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1. Standard Fundamental Limits

$$\begin{array}{rl}\underset{x\to 0}{lim}\frac{\sin x}{x}& =1,\underset{x\to 0}{lim}\frac{\tan x}{x}=1,\underset{x\to 0}{lim}\frac{1-\cos x}{x^2}=\frac{1}{2}\\ \underset{x\to 0}{lim}\frac{e^x-1}{x}& =1,\underset{x\to 0}{lim}\frac{\ln (1+x)}{x}=1,\underset{x\to 0}{lim}\frac{(1+x{)}^n-1}{x}=n\end{array}$$

Indeterminate Form $1^{\mathrm{\infty }}$

If $\underset{x\to a}{lim}f(x)=1$ and $\underset{x\to a}{lim}g(x)=\mathrm{\infty }$, then:

$$\underset{x\to a}{lim}[f(x){]}^{g(x)}=e^{\underset{x\to a}{lim}g(x)[f(x)-1]}$$

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2. Taylor & Maclaurin Series Expansions (Speed Secret)

For resolving challenging $0/0$ limits where L'Hôpital requires 4+ tedious derivative iterations:

$$\begin{array}{rl}\sin x& =x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots \\ \cos x& =1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots \\ \tan x& =x+\frac{x^3}{3}+\frac{2x^5}{15}+\dots \\ e^x& =1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots \\ \ln (1+x)& =x-\frac{x^2}{2}+\frac{x^3}{3}-\dots (-1<x\le 1)\end{array}$$

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3. Continuity & Intermediate Value Theorem

A function $f(x)$ is continuous at $x=a$ iff:

$$\underset{x\to a}{lim}f(x)=f(a)⟺\underset{h\to 0}{lim}f(a-h)=\underset{h\to 0}{lim}f(a+h)=f(a)$$

Intermediate Value Theorem (IVT)

If $f: [a, b] \to \mathbb{R}$ is continuous and $f(a) \neq f(b)$, then for every value $u$ between $f(a)$ and $f(b)$, there exists at least one $c \in (a, b)$ such that $f(c) = u$.
Bolzano's Corollary: If $f(a) \cdot f(b) < 0$, then $f(x) = 0$ has at least one real root in $(a, b)$.

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4. Differentiability from First Principles

$f(x)$ is differentiable at $x=a$ iff the derivative limit exists:

$$f^{\prime }(a)=\underset{h\to 0}{lim}\frac{f(a+h)-f(a)}{h}$$

• Left Derivative (LHD): $f^{\prime }(a^{-})=\underset{h\to 0^{+}}{lim}\frac{f(a-h)-f(a)}{-h}$
• Right Derivative (RHD): $f^{\prime }(a^{+})=\underset{h\to 0^{+}}{lim}\frac{f(a+h)-f(a)}{h}$

$$\text{Differentiability at }a⟺\text{LHD}=\text{RHD}\in \mathbb{R}$$

Crucial Hierarchy

$$\text{Differentiability} \implies \text{Continuity} \implies \text{Limit Existence}$$ The converses are FALSE. $f(x) = |x|$ is continuous at $x = 0$ with $\text{LHD} = -1 \neq \text{RHD} = +1$, so it is non-differentiable at $0$.