What are the three conditions for continuity at a point?

timberwuf8r
2022-10-07
Answered

What are the three conditions for continuity at a point?

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Dakota Duarte

Answered 2022-10-08
Author has **7** answers

A function f(x) is continuous at a point (a,b) if and only if:

1. f(a) is defined;

2. $\underset{x\to a}{lim}f\left(x\right)$ is defined; and

3. $\underset{x\to a}{lim}f\left(x\right)=b$

1. f(a) is defined;

2. $\underset{x\to a}{lim}f\left(x\right)$ is defined; and

3. $\underset{x\to a}{lim}f\left(x\right)=b$

asked 2022-08-14

How do you prove by definition that the function $f\left(x\right)={x}^{2}\mathrm{sin}\left(\frac{1}{x}\right)$ is continuous at x=0?

asked 2022-05-10

As part of a larger question regarding proving a subset is not connected, so I have $S=\{(x,y)\in {\mathbb{R}}^{2}:xy=1\}$ and I have said this is not connected as the function below:

$f(x)=\{\begin{array}{ll}1& x>0\\ 0& x<0\end{array}$

is a surjective, continuous function that maps $f:S\to \{0,1\}$

So I now want to show $f(x)$ is continuous, I have attempted to use $li{m}_{x\to a}f(x)=f(a)$ but I can't work out how to get this written out and what values of $x$ I need to take? I'm also doubting whether this is continuous but for all values of $x$ would not include 0 as it cannot take this value?

$f(x)=\{\begin{array}{ll}1& x>0\\ 0& x<0\end{array}$

is a surjective, continuous function that maps $f:S\to \{0,1\}$

So I now want to show $f(x)$ is continuous, I have attempted to use $li{m}_{x\to a}f(x)=f(a)$ but I can't work out how to get this written out and what values of $x$ I need to take? I'm also doubting whether this is continuous but for all values of $x$ would not include 0 as it cannot take this value?

asked 2022-08-29

For a function $f:[-1,1]\to R$, consider the following statements:

Statement 1: If

$\u2018\underset{n\to \mathrm{\infty}}{lim}f\left(\frac{1}{n}\right)=f(0)=\underset{n\to \mathrm{\infty}}{lim}f(-\frac{1}{n})\u2018,$

then $f$ is continuous at $x=0$

Statement 2: If $f$ is continuous at $x=0$, then

$\underset{n\to \mathrm{\infty}}{lim}f\left(\frac{1}{n}\right)=\underset{n\to \mathrm{\infty}}{lim}f(-\frac{1}{n})=\underset{n\to \mathrm{\infty}}{lim}f({e}^{\frac{1}{n}}-1)=f(0)$

Then which of the above statements is/are true.

My Attempt:

I feel that $\underset{n\to \mathrm{\infty}}{lim}f\left(\frac{1}{n}\right)=f(0)$ is same as $\underset{x\to 0}{lim}f(x)=f(0)$, so $f$ should be continuous at $x=0$. So statement 1 must be true.

In statement 2, since $f$ is continuous at $x=0$ we have $\underset{n\to \mathrm{\infty}}{lim}f\left(\frac{1}{n}\right)=f\left(\underset{n\to \mathrm{\infty}}{lim}\frac{1}{n}\right)=f(0)$. By same logic we can prove $\underset{n\to \mathrm{\infty}}{lim}f(-\frac{1}{n})=\underset{n\to \mathrm{\infty}}{lim}f({e}^{\frac{1}{n}}-1)=f(0)$

Can there be counter-examples to what I am thinking

Statement 1: If

$\u2018\underset{n\to \mathrm{\infty}}{lim}f\left(\frac{1}{n}\right)=f(0)=\underset{n\to \mathrm{\infty}}{lim}f(-\frac{1}{n})\u2018,$

then $f$ is continuous at $x=0$

Statement 2: If $f$ is continuous at $x=0$, then

$\underset{n\to \mathrm{\infty}}{lim}f\left(\frac{1}{n}\right)=\underset{n\to \mathrm{\infty}}{lim}f(-\frac{1}{n})=\underset{n\to \mathrm{\infty}}{lim}f({e}^{\frac{1}{n}}-1)=f(0)$

Then which of the above statements is/are true.

My Attempt:

I feel that $\underset{n\to \mathrm{\infty}}{lim}f\left(\frac{1}{n}\right)=f(0)$ is same as $\underset{x\to 0}{lim}f(x)=f(0)$, so $f$ should be continuous at $x=0$. So statement 1 must be true.

In statement 2, since $f$ is continuous at $x=0$ we have $\underset{n\to \mathrm{\infty}}{lim}f\left(\frac{1}{n}\right)=f\left(\underset{n\to \mathrm{\infty}}{lim}\frac{1}{n}\right)=f(0)$. By same logic we can prove $\underset{n\to \mathrm{\infty}}{lim}f(-\frac{1}{n})=\underset{n\to \mathrm{\infty}}{lim}f({e}^{\frac{1}{n}}-1)=f(0)$

Can there be counter-examples to what I am thinking

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 2022-05-14

Theorem: Let $(X,{d}_{X})$ and $Y({d}_{Y})$ be metric spaces. A function $f:X\to Y$ is continuous if and only if for every open $U\subset Y$, then

${f}^{-1}(U)=\{x\in X:f(x)\in U\}$

is open in $X$.

If I want to show that the function $f:[0,1]\cup [2,4]\subset \mathbb{R}\to \mathbb{R}$ defined by

$f(x)=\{\begin{array}{ll}1,& x\in [0,1];\\ 2,& x\in [2,4].\end{array}$

is continuous, then I need to consider any open set $U\subset \mathbb{R}$ and show that

${f}^{-1}(U)=\{x\in [0,1]\cup [2,4]:f(x)\in U\}$

is open.

I understand that if $1\notin U$ and $2\notin U$, then ${f}^{-1}(U)=\mathrm{\varnothing}$. But I don't understand what happens when only $1\in U$, or only $2\in U$, or $1,2\in U$.

If only $1\in U$, then is it correct to say that ${f}^{-1}(U)$=[0,1]? If only $2\in U$, then is it correct to say that ${f}^{-1}(U)$=[2,4]? But I am really confused since these are closed intervals. I am lost.

Can someone help me to find ${f}^{-1}(U)$ for those cases? Any clues or hints will be appreciated.

${f}^{-1}(U)=\{x\in X:f(x)\in U\}$

is open in $X$.

If I want to show that the function $f:[0,1]\cup [2,4]\subset \mathbb{R}\to \mathbb{R}$ defined by

$f(x)=\{\begin{array}{ll}1,& x\in [0,1];\\ 2,& x\in [2,4].\end{array}$

is continuous, then I need to consider any open set $U\subset \mathbb{R}$ and show that

${f}^{-1}(U)=\{x\in [0,1]\cup [2,4]:f(x)\in U\}$

is open.

I understand that if $1\notin U$ and $2\notin U$, then ${f}^{-1}(U)=\mathrm{\varnothing}$. But I don't understand what happens when only $1\in U$, or only $2\in U$, or $1,2\in U$.

If only $1\in U$, then is it correct to say that ${f}^{-1}(U)$=[0,1]? If only $2\in U$, then is it correct to say that ${f}^{-1}(U)$=[2,4]? But I am really confused since these are closed intervals. I am lost.

Can someone help me to find ${f}^{-1}(U)$ for those cases? Any clues or hints will be appreciated.

asked 2022-09-26

For a continuous function (let's say f(x)) at a point x=c, is f(c) the limit of the function as x tends to c?

asked 2022-08-09

Is $f\left(x\right)=\frac{{x}^{2}-9}{x-3}$ continuous at x=3?