Chapter 4

An easy recap.

Limit definitions

Definition · Limit of a function#

Let

$$\begin{array}{ll} \bullet\ E \subseteq \mathbb{R}^m & \bullet\ \mathbf{x} \in \mathbb{R}^m \\[0.5em] \bullet\ f : E \setminus \left\{ \mathbf{x} \right\} \to \mathbb{R}^p & \bullet\ \mathbf{a} \in \mathbb{R}^p \end{array}$$

Then $f(\mathbf{y}) \to \mathbf{a}$ as $\mathbf{y} \to \mathbf{x}$ iff for any $\epsilon > 0$, there exists $\delta > 0$ such that

$$\mathbf{y} \in E : \| \mathbf{y} - \mathbf{x} \| < \delta \implies \| f(\mathbf{y}) - \mathbf{a} \| < \epsilon.$$

Proposition from Exercise

Let $E,\mathbf{x}, f, \mathbf{a}$ be as in the above definition.

Define $\tilde{f} : E \to \mathbb{R}^p$ by

$$\tilde{f}(\mathbf{y}) = \begin{cases} f(y) & \text{ if } \mathbf{y \neq x} \\ \mathbf{a} & \text{ if } \mathbf{y=x} \end{cases}$$

Then, $\tilde{f}(\mathbf{y}) \to \mathbf{a}$ as $\mathbf{y \to x}$ iff $\tilde{f}$ is continuous at $\mathbf{x}$.

Bookwork: Composition of Limits

Lemma · Composition of limits#

If,

$$k(h) \to k_L \text{ as } h \to h_L \\ q(k) \to q_L \text{ as } k \to k_L$$

Then,

$$q(k(h)) \to q_L \text{ as } h \to h_L$$

Proof of Composition of limits#

This could be proven via the above Exercise proposition and using results about continuity of compositions.

We can just prove it directly. The second assumption gives us,

$$\forall \epsilon > 0 \exists \delta > 0 : \lVert k-k_L \rVert < \delta \implies \lVert q-q_L \rVert < \epsilon$$

The first assumption gives us

$$\exists \nu > 0 : \lVert h-h_L \rVert < \nu \implies \lVert k - k_L \rVert < \delta$$

Hence

$$\exists \nu > 0 : \lVert h - h_L \rVert < \nu \implies \lVert q(k(h)) - q_L \rVert < \epsilon$$