Functional Limit

Let $A \subseteq R$ and $f : A \to R$ where $L \in R$ . We want $c$ to be a Limit Point of $A$ .

Limit of a function,

ε - δ

Version

$lim_{x \to c} f (x) = L$

means that $\forall ε > 0 \exists δ > 0 (0 < | x - c | < δ)$ and $x \in A$ implies that:

| f (x) - L | < ε

Example: Limit of a function

Consider $f (x) = 2 x^{2} + 5$ where $A \subseteq R$ . We want to show:

lim_{x \to 2} f (x) = 13

The scratch is:

\begin{aligned} | f (x) - 13 | & < ε \\ ⟺ | 2 x^{2} + 5 - 13 | & < ε \\ ⟺ | 2 x^{2} - 8 | & < ε \\ ⟺ 2 | x^{2} - 4 | & < ε \\ ⟺ 2 | x - 2 | | x + 2 | & < ε \end{aligned}

Now this will work if $0 < | x - 2 | < 1$ (we can make the left side as small as any number) so that implies $0 < | x + 2 | < 5$ so then the LHS will have a factor of $10$ , so that implies $δ = min {1, \frac{ε}{10}}$ . The proof highlights this.

Proof

Let $ε > 0$ .

Let $ε = min {1, \frac{ε}{10}}$ . Suppose $0 < | x - 2 | < δ$ . Note that since $| x - 2 | < δ \leq 1$ then $| x + 2 | < 5$ . Then:

\begin{aligned} | f (x) - 13 | & = | 2 x^{2} - 8 | \\ = 2 | x - 2 | | x + 2 | \\ < 2 \cdot δ \cdot 5 \\ \leq 2 \cdot \frac{ε}{10} \cdot 5 \\ = ε \end{aligned}

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