Academics/CIS/Discrete Structures/Functions

$z=f(x,y)$
$f:X\to Y$
where X will be a cartesian product of sets

• like $R^n,n>1$

Domain is the set of the tuples inside $X$ for which $f:X\to Y$ is defined
Codomain will be the set of the values obtained when using the domain

$z=\sqrt{3x-y-2}$\

• Domain = $\{(x,y);y\le 3x-2\}$

Limits and Continuity

A $\underset{x\to a}{lim}f(x)=L$ can have its relation between $x$ and $f(x)$ measured by changes in distances $\delta$ and $ϵ$

• $|x-a|<\delta \to a-\delta <x<a+\delta$
• $|f(x)-L|<ϵ\to L-ϵ<f(x)<L+ϵ$

So $\underset{x\to a}{lim}f(x)=L$ means

For multivariable functions:

• distance from a point $(a,b)=||(x,y)-(a,b)||=||\tilde{x}-\tilde{a}||=\sqrt{(x-a{)}^2+(y-b{)}^2})$
• $0<||\tilde{x}-\tilde{a}||<\delta \to 0<(x-a{)}^2+(y-b{)}^2<{\delta }^2$
• $|f(x,y)-L|<ϵ$

Uniqueness of Limits

If $\underset{\tilde{x}\to \tilde{a}}{lim}f(\tilde{x})=L_1$ along one path and $\underset{\tilde{x}\to \tilde{a}}{lim}f(\tilde{x})=L_2$ along a different path then $\underset{\tilde{x}\to \mathrm{\infty }}{lim}\tilde{a}f(\tilde{x})$ does not exist

• paths can be any $y=mf(x)$ or $x=mf(y)$, since we don't know the relationship between the values of x and y

Only useful to prove that limit does not exist

Plug in Limits

A function $f:R^n\to R$ is continuous at $\tilde{a}\in R^n$ if

1. $f$ is defined at $\tilde{a}$
2. $\underset{\tilde{x}\to \tilde{a}}{lim}f(\tilde{x})$ exists
3. $\underset{\tilde{x}\to \tilde{a}}{lim}f(\tilde{x})=f(\tilde{a})$

So if $f$ is continuous, we can find $\underset{\tilde{x}\to \tilde{a}}{lim}f(\tilde{x})$ by just plugging in $\tilde{a}$

L'Hopital's Rule doesn't work in multivariables, but we can still:

• factor
• multiply by a conjugate
• common denominators

Squeeze Theorem

If $0\le |f(\tilde{x})-L|\le M(\tilde{x})\mathrm{\forall }\tilde{x}\ne \tilde{a}$ in some neighborhood of $\tilde{a}$ and $\underset{\tilde{x}\to \tilde{a}}{lim}M(\tilde{x})=0$ then $\underset{\tilde{x}\to \tilde{a}}{lim}f(\tilde{x})$ exists and equals $L$

• the mutivariable function $M(\tilde{x})$ whose limit is easy to calculate binds $f(\tilde{x})$ at 0

Can be used instead of the $\delta -ϵ$ proofs

Cannot prove that a limit does not exist