Notes - Analysis II HT23, Function limits
Flashcards
Can you write the definition of $f : E \to \mathbb{R}$ tending towards $L$ at some point $p$?
$\lim _ {x \to p} f(x)=L$ if and only if
\[\forall\varepsilon>0 \text{ } \exists \delta > 0 \text{ s.t. } \forall x \in E \text{ } (0 < |x-p| < \delta \implies |f(x)-L| < \varepsilon)\]What useful result links function limits and the convergence of sequences?
If $f : E \to \mathbb{R}$, $E \subseteq \mathbb{R}$ and $p$ is a limit point of $E$, then
$f(x) \to L$ as $x \to p$
if and only if
For all sequences $(p _ n)$, $p _ n \in E$, $p _ n \ne p$, $p _ n \to p$, $f(p _ n) \to L$ as $n \to \infty$.
Becausse if $f : E \to \mathbb{R}$, $E \subseteq \mathbb{R}$ and $p$ is a limit point of $E$, then
$f(x) \to L$ as $x \to p$
if and only if
For all sequences $(p _ n)$, $p _ n \in E$, $p _ n \ne p$, $p _ n \to p$, $f(p _ n) \to L$ as $n \to \infty$.
what is a useful tool for working out if $f(x)$ tends towards a limit?
$f(x) \to L$ as $x \to p$
For all sequences $(p _ n)$, $p _ n \in E$, $p _ n \ne p$, $p _ n \to p$, $f(p _ n) \to L$ as $n \to \infty$.
Showing two different sequences $f(p _ n)$ and $f(q _ n)$ tend towards different limits.
Can you state what it means for a function $f : E \to \mathbb{R}$ to tend to $\infty$ at $x = p$ (read carefully)?
What restriction of $f : E \to \mathbb{R}$ are you checking when calculating $\lim _ {x \to p^+} f(x)$?
Can you simplify the condition
\[x > p \text{ and } 0 <|x-p|<\delta\]
?
Can you state what it means for $f(x) \to L$ as $x \to \infty$ (read carefully)?
Can you state what it means for $f(x) \to L$ as $x \to -\infty$ (read carefully)?
What condition do you need for $f : E \to \mathbb{R}$ so the definition of $f(x) \to L$ as $x \to \infty$ or $x\to -\infty$ to make sense?
It can’t be bounded above/below, or vacously true.
What’s an easy way to prove the AOL holds for function limits as well?
Convert the function limits into the limits of sequences.
Which of the theorems in the AOL cannot be proved directly by comparison to the AOL for sequences?
Why can’t you prove that if $f(x) \to A$ and $g(x) \to B$ as $x \to p$ then
\[\frac{f(x)}{g(x)} \to \frac{A}{B}\]
by direct conversion to sequences?
Because the domain of $f/g$ might be different from $f$ and $g$.
Can you state the result about the function limit of $g \circ f$ as $x \to p$?
Suppose $f : E \to \mathbb{R}$ and $g : E’ \to \mathbb{R}$ with $f(E) \subseteq E’$. Assume
- $p$ is a limit point of $E$
- $\lim _ {x\to p} f(x) = q \in \mathbb{R}$
- $f(x) \ne q$ for all $x \in E \backslash \{p\}$
Then $q$ is a limit point of $E’$. If in addition
- $\lim _ {y \to q} g(y) = L \in \mathbb{R} \cup \{\pm \infty \}$
then $\lim _ {x\to p} g(f(x)) = L$.
When proving that if $f : E \to \mathbb{R}$, $E \subseteq \mathbb{R}$ and $p$ is a limit point of $E$, then
$f(x) \to L$ as $x \to p$
if and only if
For all sequences $(p _ n)$, $p _ n \in E$, $p _ n \ne p$, $p _ n \to p$, $f(p _ n) \to L$ as $n \to \infty$.
how do you tackle the ‘$\impliedby$’ direction?
$f(x) \to L$ as $x \to p$
For all sequences $(p _ n)$, $p _ n \in E$, $p _ n \ne p$, $p _ n \to p$, $f(p _ n) \to L$ as $n \to \infty$.
Take the contrapositive, $f(x) \not\to L \implies \exists (p _ n) \ldots$
When proving that if $f : E \to \mathbb{R}$, $E \subseteq \mathbb{R}$ and $p$ is a limit point of $E$, then
$f(x) \to L$ as $x \to p$
if and only if
For all sequences $(p _ n)$, $p _ n \in E$, $p _ n \ne p$, $p _ n \to p$, $f(p _ n) \to L$ as $n \to \infty$.
you handle the second condition by taking the contrapositive, and therefore know that $f(x) \not\to L$ as $x \to p$, i.e.
\[\exists \varepsilon > 0:\forall \delta > 0 : \exists x\in E \text{ s.t. } (0 < |x - p| < \delta \text{ and } |f(x) - L| \ge \varepsilon)\]
how do you construct the $n$-th term $p _ n$ of a sequence?
$f(x) \to L$ as $x \to p$
For all sequences $(p _ n)$, $p _ n \in E$, $p _ n \ne p$, $p _ n \to p$, $f(p _ n) \to L$ as $n \to \infty$.
Fix $\varepsilon > 0$ and take the $x$ in the case where $\delta = \frac{1}{n}$.
When proving that if $g(x) \to B$ as $x \to p$ where $B \ne 0$, what value of $\varepsilon$ do you take in the definition of a function limit in order to show $g(x)$ is non-zero in an interval containing $p$?
\[\lim_{x \to \infty} \frac{2x^2 + 1}{3x^2 + 7} = \lim_{x\to\infty} \frac{2 + \frac{1}{x^2}\space}{3+\frac{7}{x^2}\space}\]
How can you use the rule about the limits of compositions of functions to turn this into a limit towards $0$?
Let $y = \frac{1}{x}$, $y \to 0$ as $x \to \infty$. Then
\[\lim_{y\to0} \frac{2+y^2}{3+7y^2} = \frac{2}{3}\]Quickly prove the equivalence of the two following definitions:
\[\forall \varepsilon > 0 \text { } \exists\delta > 0 \text { s.t. } \forall x \in E \text{ } 0 < |x - p| < \delta \implies |f(x) - L| < \varepsilon\]
and
\[\forall(p_n), p_n \to p, p_n \ne p, p_n \in E \implies f(p_n) \to L\]
- $\implies$: Suppose $(p _ n)$ is a sequence as above. We aim to show $\forall \varepsilon > 0 \exists N \in \mathbb N \text{ s.t. } \vert f(p _ n) - f(p) \vert < \varepsilon$. This is immediate from fixing $\varepsilon > 0$ and taking the $\varepsilon$ as $\delta$ in the definition of $p _ n \to p$, then we have $\forall n \ge N, 0 < \vert x - p \vert < \delta \implies \vert f(p _ n) - f(p) \vert < \varepsilon$.
- $\impliedby$: Take contrapositive and use it to construct a sequence where everything is true, but $f(p _ n) \not\to L$.
What sequence $q _ n$ do you construct in the proof for the following to show that $q$ is a limit point of $E’$??
Suppose $f : E \to \mathbb{R}$ and $g : E’ \to \mathbb{R}$ with $f(E) \subseteq E’$. Assume
- $p$ is a limit point of $E$
- $\lim _ {x\to p} f(x) = q \in \mathbb{R}$
- $f(x) \ne q$ for all $x \in E \backslash \{p\}$
Then $q$ is a limit point of $E’$. If in addition
- $\lim _ {y \to q} g(y) = L \in \mathbb{R} \cup \{\pm \infty \}$
then $\lim _ {x\to p} g(f(x)) = L$.
Let $q _ n = f(p _ n)$, which satisfies all requirements.
When proving the following, what common proof technique involving compositions do you use?
Suppose $f : E \to \mathbb{R}$ and $g : E’ \to \mathbb{R}$ with $f(E) \subseteq E’$. Assume
- $p$ is a limit point of $E$
- $\lim _ {x\to p} f(x) = q \in \mathbb{R}$
- $f(x) \ne q$ for all $x \in E \backslash \{p\}$
Then $q$ is a limit point of $E’$. If in addition
- $\lim _ {y \to q} g(y) = L \in \mathbb{R} \cup \{\pm \infty \}$
then $\lim _ {x\to p} g(f(x)) = L$.
Taking the $\delta$ you get out of one definition of convergence as the $\varepsilon$ in another.
Proofs
Prove function limits are unique?
Todo?
Prove that if $f : E \to \mathbb{R}$, $E \subseteq \mathbb{R}$ and $p$ is a limit point of $E$, then
$f(x) \to L$ as $x \to p$
if and only if
For all sequences $(p _ n)$, $p _ n \in E$, $p _ n \ne p$, $p _ n \to p$, $f(p _ n) \to L$ as $n \to \infty$.
$f(x) \to L$ as $x \to p$
For all sequences $(p _ n)$, $p _ n \in E$, $p _ n \ne p$, $p _ n \to p$, $f(p _ n) \to L$ as $n \to \infty$.
Todo?
Prove, by converting to sequences, the AOL for function limits.
Todo.
Prove that if $f(x) \to A$ and $g(x) \to B$ as $x \to p$ where $f : E \to \mathbb{R}$ and $p$ is a limit point, then
\[\frac{f(x)}{g(x)} \to \frac{A}{B}, \text{ } B \ne 0\]
Todo?
Prove that if $g(x) \to B$ as $x \to p$ and $B \ne 0$, then there exists some interval containing $p$ that is non-zero.
Todo?
Prove the following. Assume $f : E \to \mathbb{R}$ and $g : E’ \to \mathbb{R}$ with $f(E) \subseteq E’$. Further assume
- $p$ is a limit point of $E$
- $\lim _ {x\to p} f(x) = q \in \mathbb{R}$
- $f(x) \ne q$ for all $x \in E \backslash \{p\}$
Then $q$ is a limit point of $E’$. If in addition
- $\lim _ {y \to q} g(y) = L \in \mathbb{R} \cup \{\pm \infty \}$
then $\lim _ {x\to p} g(f(x)) = L$.
Todo?
- Proof $\lim _ {x\to p} f(x) = L \iff \lim _ {x \to p^+}f(x) = L \text{ and } \lim _ {x\to p^-} f(x) = L$.