How do you prove that the function $g\left(x\right)=\frac{{x}^{3}}{x}$ is continuous everywhere but x=0?

abrigairaic
2022-08-08
Answered

How do you prove that the function $g\left(x\right)=\frac{{x}^{3}}{x}$ is continuous everywhere but x=0?

You can still ask an expert for help

Royce Golden

Answered 2022-08-09
Author has **12** answers

There are several ways to prove continuity, they differ in key ideas and mathematical formality.

The function f is continuous at some point c of its domain if the limit of f(x) as x approaches c through the domain of f exists and is equal to f(c). In mathematical notation, this is written as

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

In detail this means three conditions: 1) first, f has to be defined at c. 2) Second, the limit on the left hand side of that equation has to exist. 3) Third, the value of this limit must equal f(c).

For our function: $f\left(x\right)=\frac{{x}^{3}}{x}$, any value we replace we have a value, but zero, which will give $\frac{0}{0}$, it is undefined. Therefore, it is not continuous.

Nonetheless, as you can see from the graph below, this function is continuous on 0 , since $f\left(x\right)={x}^{2}$. It is just the parabola masked-out.

But the ideas herein can be used to prove that a true not continuous function is not continuous in a point, e.g. $\frac{1}{x}$, apply the ideas herein for 0, it goes to infinity.

graph{x^3/x [-2.5, 2.5, -1.25, 1.25]}

The function f is continuous at some point c of its domain if the limit of f(x) as x approaches c through the domain of f exists and is equal to f(c). In mathematical notation, this is written as

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

In detail this means three conditions: 1) first, f has to be defined at c. 2) Second, the limit on the left hand side of that equation has to exist. 3) Third, the value of this limit must equal f(c).

For our function: $f\left(x\right)=\frac{{x}^{3}}{x}$, any value we replace we have a value, but zero, which will give $\frac{0}{0}$, it is undefined. Therefore, it is not continuous.

Nonetheless, as you can see from the graph below, this function is continuous on 0 , since $f\left(x\right)={x}^{2}$. It is just the parabola masked-out.

But the ideas herein can be used to prove that a true not continuous function is not continuous in a point, e.g. $\frac{1}{x}$, apply the ideas herein for 0, it goes to infinity.

graph{x^3/x [-2.5, 2.5, -1.25, 1.25]}

asked 2022-06-15

Let $X$ be a non-empty set and let ${x}_{0}\in X$. The topology $T$ is defined by the collection of subsets $U\subset X$ such that either $U=\mathrm{\varnothing}$ or $U\ni {x}_{0}$.

Is it true that if $f:X\to X$ is continuous, does it follow that $f({x}_{0})={x}_{0}$ with respect to the topology $T$?

I under stand that the converse is true, that is for any function $f:X\to X$ with $f({x}_{0})={x}_{0}$ is continuous with respect to the topology T but i don't know how to prove if this case is true.

Is it true that if $f:X\to X$ is continuous, does it follow that $f({x}_{0})={x}_{0}$ with respect to the topology $T$?

I under stand that the converse is true, that is for any function $f:X\to X$ with $f({x}_{0})={x}_{0}$ is continuous with respect to the topology T but i don't know how to prove if this case is true.

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 2022-09-18

Let

$f(x)=\{\begin{array}{lcc}\frac{{z}^{3}-1}{{z}^{2}+z+1}& if& |z|\ne 1\\ \\ \frac{-1+i\sqrt{3}}{2}& if& |z|=1\end{array}$

is $f$ continous in ${z}_{0}=\frac{1+\sqrt{3}i}{2}$

I think to $f$ is not continous at ${z}_{0}$, i try using the sequence criterion for continuity searching a sequence $\{{z}_{n}\}$ such that $\underset{n\to \mathrm{\infty}}{lim}{z}_{n}={z}_{0}$ but $\underset{n\to \mathrm{\infty}}{lim}f({z}_{n})\ne f({z}_{0})$. but i cant find that sequence, ill be very grateful for any hint or help to solve my problem.

$f(x)=\{\begin{array}{lcc}\frac{{z}^{3}-1}{{z}^{2}+z+1}& if& |z|\ne 1\\ \\ \frac{-1+i\sqrt{3}}{2}& if& |z|=1\end{array}$

is $f$ continous in ${z}_{0}=\frac{1+\sqrt{3}i}{2}$

I think to $f$ is not continous at ${z}_{0}$, i try using the sequence criterion for continuity searching a sequence $\{{z}_{n}\}$ such that $\underset{n\to \mathrm{\infty}}{lim}{z}_{n}={z}_{0}$ but $\underset{n\to \mathrm{\infty}}{lim}f({z}_{n})\ne f({z}_{0})$. but i cant find that sequence, ill be very grateful for any hint or help to solve my problem.

asked 2022-07-02

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 2022-06-29

Let $\mathrm{\Omega}=\mathbb{C}\mathrm{\setminus}[-1,1]$, i.e. deleting ``the line'' only, is there a function $f:\mathrm{\Omega}\to \mathbb{C}$ such that $f$ satisfies $f(z{)}^{2}=1-{z}^{2}$ and is continuous on this region?

My guess is that such a function would exist, but requires a piecewise definition. A candidate solution I have been working on is $f(z)={e}^{\frac{1}{2}\mathrm{log}(1-{z}^{2})}$. The problem with this solution is having the domain, as I realize it is possible to have $1-{z}^{2}>1$, where my solution is well-defined. Is there a way to work around this or should I try something else?

My guess is that such a function would exist, but requires a piecewise definition. A candidate solution I have been working on is $f(z)={e}^{\frac{1}{2}\mathrm{log}(1-{z}^{2})}$. The problem with this solution is having the domain, as I realize it is possible to have $1-{z}^{2}>1$, where my solution is well-defined. Is there a way to work around this or should I try something else?

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 2022-07-12

Let be $X$ a topological space and we suppose that $E$ and $F$ are homeomorphic though a map $f$ from $F$ to $E$. So I ask to me if $E$ is open/closed when $F$ is open/closed but unfortunatley I was not able to prove or to disprove this so that I thought to put a specific question where I ask some clarification: in particular if the result is generally false I'd like to know if it can be true we additional hypotesis, e.g. Hausdorff separability, First Countability, Metric Topology, etc....