Lecture 24 - Intro to Functions

We want to find the limits of functions. What is:

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

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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Man this seems like a pain to do right? Do we have to do all of our Convergence of a Sequence definitions again but now for functions? Not quite. By the following theorem:

Sequential Criteria for Functional Limits

Let $A \subseteq R$ and $f : A \to R$ and $L \in R$ . Let $c$ be a Limit Point of $A$ .
The following are equivalent:

$lim_{x \to c} f (x) = L$
For all sequences $(x_{n}) \subseteq A - {c}$ with $x_{n} \to c$ , we have $f (x_{n}) \to L$ .

Proof

( $\Rightarrow$ ): Suppose $lim_{x \to c} f (x) = L$ . Then $\forall ε > 0 \exists δ > 0 (| x - c | < δ \to | f (x) - L | < ε)$ . Now let $(x_{n}) \subseteq A - {c}$ where $x_{n} \to c$ be arbitrary. We want to show $f (x_{n}) \to L$ . Let's use the definition of convergence of a sequence.

Let $ε > 0$ . By our given then $δ > 0$ such that $0 < | x - c | < δ \to | f (x) - L | < ε$ (where $x \in A$ ). Since $x_{n} \to c$ then $\exists N \in N$ where $\forall n \geq N (| x_{n} - c | < δ)$ . Now let's choose any $n \geq N$ , so then $0 < | x_{n} - c | < δ$ (the $x_{n} \neq c$ by construction). This gives us that $| f (x) - L | < ε$ thus completing the proof for convergence.

( $\Leftarrow$ ): We'll prove the contra-positive; namely that not (1) implies not (2). Now suppose that $\neg [lim_{x \to c} f (x) = L]$ . That implies that $\exists ε > 0 \forall δ > 0 \exists x \in A (0 < | x - c | < δ \land | f (x) - L | \geq ε)$ . We need to prove the opposite of (2); namely we need to construct this sequence such that $f (x_{n})$ doesn't converge to $L$ .

In particular, for $n \in N$ we can find some $x_{n} \in A - {c}$ such that $0 < | x_{n} - c | < \frac{1}{n} = δ$ such that then $| f (x_{n}) - L | \geq ε$ (see our given above). But then that implies that $f (x_{n}) ↛ L$ (it's the opposite of the convergence definition).

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This leads to some very useful corollaries:

Limit Laws (Algebraic Limit Theorem) for Functional Limits

Let $f, g : A \to R$ where $A \subseteq R$ and $c$ is a Limit Point of $A$ . Let $L, M \in R$ . If:
$lim_{x \to c} f (x) = L, lim_{x \to c} g (x) = M$
then:

$lim_{x \to c} [α f (x)] = α L$ for all $α \in R$ .
$lim_{x \to c} [f (x) + g (x)] = L + M$
$lim_{x \to c} [f (x) g (x)] = L M$
$lim_{x \to c} \frac{f (x)}{g (x)} = \frac{L}{M}$ provided $M \neq 0$ .

Proof

Use the Sequential Criterion For Functional Limits, but now using the Limit Laws (Algebraic Limit Theorem) for sequences for each.

☐

This is a corollary from the Sequential Criterion For Functional Limits:

Existence of Functional Limits

Let $f : A \to R$ where $A \subseteq R$ and $c$ is a Limit Point of $A$ .
If $\exists$ sequences $(x_{n}) \subseteq A - {c}$ and $(y_{n}) \subseteq A - {c}$ with $x_{n} \to c, y_{n} \to c$ but $(f (x_{n}))$ and $(f (y_{n}))$ do not converge to the same value. Then:
$lim_{x \to c} f (x)$
does not exist.

Namely this is the opposite of the Sequential Criterion For Functional Limits (2).

Example of Finding Functional Limits

Let $f (x) = \sin (\frac{1}{x})$ where $A = R - {0}$ .

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f(x) = sin(1/x)

A reasonable question to ask is if/what $lim_{x \to 0} f (x)$ is here (notice $0$ here is a limit point of the domain set). But notice that the sequence of points that intersect the graph at 0 on the right hand side make up a sequence where $f (x_{n}) \to 0$ . But we can do the same thing with the points that intersect the line $y = 1$ . Then $f (y_{n}) \to 1$ , so we have two sequences that converge to different limits, so then the limit of the original function must not exist.

More explicitly, let $n \in N$ . Then choose:
$x_{n} = \frac{1}{n π}, y_{n} = \frac{1}{\frac{π}{2} + 2 n π}$
Then notice that $x_{n}, y_{n} \to 0$ individually, but:
$f (x_{n}) = \sin (\frac{1}{\frac{1}{n π}}) = \sin (n π) = 0$
$f (y_{n}) = \sin (\frac{1}{\frac{1}{\frac{π}{2} + 2 n π}}) = \sin (\frac{π}{2} + 2 n π) = 1$
So since $f (x_{n}) \to 0$ and $f (y_{n}) \to 1$ then by our corrolary above then we get the limit does not exist.