What is the continuity of $f\left(t\right)=3-\sqrt{9-{t}^{2}}$

Payton George
2022-10-12
Answered

What is the continuity of $f\left(t\right)=3-\sqrt{9-{t}^{2}}$

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Warkallent8

Answered 2022-10-13
Author has **16** answers

$f\left(t\right)=3-\sqrt{9-{t}^{2}}$ has domain $[-3,3]$

For a in $(-3,3)$, $\underset{t\to a}{lim}f\left(t\right)=f\left(a\right)$ because

$\underset{t\to a}{lim}(3-\sqrt{9-{t}^{2}})=3-\underset{t\to a}{lim}\sqrt{9-{t}^{2}}$

$=3-\sqrt{\underset{t\to a}{lim}(9-{t}^{2})}=3-\sqrt{9-\underset{t\to a}{lim}{t}^{2}})$

$=3-\sqrt{9-{a}^{2}}=f\left(a\right)$

So f is continuous on (−3,3).

Similar reasoning will show that

$\underset{t\to -{3}^{+}}{lim}f\left(t\right)=f(-3)$ and

$\underset{t\to {3}^{-}}{lim}f\left(t\right)=f\left(3\right)$

So f is continuous on [−3,3].

For a in $(-3,3)$, $\underset{t\to a}{lim}f\left(t\right)=f\left(a\right)$ because

$\underset{t\to a}{lim}(3-\sqrt{9-{t}^{2}})=3-\underset{t\to a}{lim}\sqrt{9-{t}^{2}}$

$=3-\sqrt{\underset{t\to a}{lim}(9-{t}^{2})}=3-\sqrt{9-\underset{t\to a}{lim}{t}^{2}})$

$=3-\sqrt{9-{a}^{2}}=f\left(a\right)$

So f is continuous on (−3,3).

Similar reasoning will show that

$\underset{t\to -{3}^{+}}{lim}f\left(t\right)=f(-3)$ and

$\underset{t\to {3}^{-}}{lim}f\left(t\right)=f\left(3\right)$

So f is continuous on [−3,3].

asked 2022-06-22

Let ${H}_{1}$ and ${H}_{2}$ be Hilbert spaces, let $T\in B({H}_{1},{H}_{2})$. Suppose that $\mathrm{Ker}T$ is finite-dimensional and that $\mathrm{Im}T$ is closed in ${H}_{2}$. Prove that $\mathrm{Ker}(T+K)$ is finite-dimensional for each $K\in K({H}_{1},{H}_{2})$.

1. Define Hilbert space as direct sum of two complemented subspaces

2. For compact operator use Hilbert Schmidt decomposition

Main idea is to prove that intersection of Spectrum of such $T and compact K is finite

1. Define Hilbert space as direct sum of two complemented subspaces

2. For compact operator use Hilbert Schmidt decomposition

Main idea is to prove that intersection of Spectrum of such $T and compact K is finite

asked 2022-05-31

I have seen it stated that for an open subset $Y\subseteq X$ such that $X$ is a compact Hausdorff space we get an identification of the ${C}^{\ast}$-algebras : $C(X\setminus Y)\cong C(X)/{C}_{0}(Y)$.

I suppose that this relies on the Tietze extension theorem but I fail to connect the dots.

How do I realize this?

I suppose that this relies on the Tietze extension theorem but I fail to connect the dots.

How do I realize this?

asked 2022-07-17

Define $f:D(0,1)->C$ to be holomorphic such that $f(0)=0$. I want to extend $\frac{f(z)}{z}$ to be continuous on $\overline{D}(0,r)$ for arbitrary $0<r<1$.

My initial guess was to define:

$g(z)=\frac{f(z)}{z}$ for non-zero $z$ but extend $g(z)$ to be 0 for $z=0$.

But if I take $\underset{z\to 0}{lim}\frac{f(z)}{z}={f}^{\prime}(0)$ using L'Hopital's rule. So I think this should be the expression for $g(0)$? I wonder if:

i) L'hopital's rule is applicable to complex functions, and,

ii)Am I taking the limit incorrectly? I also do not see a use for $f(0)=0$ in my deduction so I'm rather perplexed.

Any help would be appreciated!

My initial guess was to define:

$g(z)=\frac{f(z)}{z}$ for non-zero $z$ but extend $g(z)$ to be 0 for $z=0$.

But if I take $\underset{z\to 0}{lim}\frac{f(z)}{z}={f}^{\prime}(0)$ using L'Hopital's rule. So I think this should be the expression for $g(0)$? I wonder if:

i) L'hopital's rule is applicable to complex functions, and,

ii)Am I taking the limit incorrectly? I also do not see a use for $f(0)=0$ in my deduction so I'm rather perplexed.

Any help would be appreciated!

asked 2022-07-01

In real functions, do we have a notion of one-sided measure theoretic limits? I want to define them with the following:

$\underset{x\to {c}^{+}}{lim}f(x)=L$

iff

$\mathrm{\forall}\u03f5,\mathrm{\exists}\delta >0,J:=f((c,c+\delta )),\mu (J\cap {B}_{\u03f5}(L))=\mu (J)$

$\underset{x\to {c}^{+}}{lim}f(x)=L$

iff

$\mathrm{\forall}\u03f5,\mathrm{\exists}\delta >0,J:=f((c,c+\delta )),\mu (J\cap {B}_{\u03f5}(L))=\mu (J)$

asked 2022-05-01

I understand that I have to use the discontinuity criterion:

Theorem: Let $S\subset \mathbb{R}$, let $f:S\to \mathbb{R}$ and let $c\in S$.

Then f is discontinuous at c if and only if there exists a sequence $\{{x}_{n}\}$ in S such that $lim({x}_{n})=c$, but the sequence $\{f({x}_{n})\}$ does not converge to $f(c)$.

I tried splitting up the function as $f(x)={x}^{2}$ when $x=\frac{p}{q}$, for $p,q\in \mathbb{Z}$ but I do not know how to define for irrational numbers

I also don't know how to think of a sequence show the limit exists but it does not converge...

Theorem: Let $S\subset \mathbb{R}$, let $f:S\to \mathbb{R}$ and let $c\in S$.

Then f is discontinuous at c if and only if there exists a sequence $\{{x}_{n}\}$ in S such that $lim({x}_{n})=c$, but the sequence $\{f({x}_{n})\}$ does not converge to $f(c)$.

I tried splitting up the function as $f(x)={x}^{2}$ when $x=\frac{p}{q}$, for $p,q\in \mathbb{Z}$ but I do not know how to define for irrational numbers

I also don't know how to think of a sequence show the limit exists but it does not converge...

asked 2022-11-17

What does continuity mean?

asked 2022-10-30

So I was studying topology and I came across the next theorem:

A function $f:X\to Y$ is continous iff for every $\u03f5>0$ there is $\delta >0$ such that if ${d}_{x}(x,y)<\delta $ then ${d}_{y}(f(x),f(y))<\u03f5$.

Since every distance in the uniform topology is at most 1, the theorem will always be true for any function from ${R}^{\omega}$ to ${R}^{\omega}$ with the uniform topology, am I wrong?

A function $f:X\to Y$ is continous iff for every $\u03f5>0$ there is $\delta >0$ such that if ${d}_{x}(x,y)<\delta $ then ${d}_{y}(f(x),f(y))<\u03f5$.

Since every distance in the uniform topology is at most 1, the theorem will always be true for any function from ${R}^{\omega}$ to ${R}^{\omega}$ with the uniform topology, am I wrong?