What makes a function continuous at a point?

madeeha1d8
2022-09-23
Answered

What makes a function continuous at a point?

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embraci4i

Answered 2022-09-24
Author has **10** answers

Let $f\left(x\right)$ be a function defined in an interval $(a,b)$ and ${x}_{0}\in (a,b)$ a point of the interval.

Then the definition of continuity is that the limit of $f\left(x\right)$ as x approaches $x}_{0$ equals the value of $f\left(x\right)$ in $x}_{0$.

In symbols:

$\underset{x\to {x}_{0}}{lim}f\left(x\right)=f\left({x}_{0}\right)$

Based on the formal definition of limit, then, for every number $\epsilon >0$ we can find ${\delta}_{\epsilon}>0$ such that:

$\left|x-{x}_{0}\right|<{\delta}_{\epsilon}\Rightarrow \left|f\left(x\right)-f\left({x}_{0}\right)\right|<\epsilon$

This means that as x gets closer and closer to $x}_{0$ also $f\left(x\right)$ gets closer and closer to $f\left({x}_{0}\right)$ and thus the function is "smooth".

Then the definition of continuity is that the limit of $f\left(x\right)$ as x approaches $x}_{0$ equals the value of $f\left(x\right)$ in $x}_{0$.

In symbols:

$\underset{x\to {x}_{0}}{lim}f\left(x\right)=f\left({x}_{0}\right)$

Based on the formal definition of limit, then, for every number $\epsilon >0$ we can find ${\delta}_{\epsilon}>0$ such that:

$\left|x-{x}_{0}\right|<{\delta}_{\epsilon}\Rightarrow \left|f\left(x\right)-f\left({x}_{0}\right)\right|<\epsilon$

This means that as x gets closer and closer to $x}_{0$ also $f\left(x\right)$ gets closer and closer to $f\left({x}_{0}\right)$ and thus the function is "smooth".

asked 2022-05-03

I have the following question:

Given a polynom $p(x)$ I need to prove the the equation $\mathrm{tan}x=p(x)$ has at least one solution.

I defined the difference as a function $f(x)=\mathrm{tan}x-p(x)$ and I know that $f(x)$ is continues in any interval where $\mathrm{tan}x$ is defined, so I am trying to use the Intermediate value theorem, but I don't know how.

Given a polynom $p(x)$ I need to prove the the equation $\mathrm{tan}x=p(x)$ has at least one solution.

I defined the difference as a function $f(x)=\mathrm{tan}x-p(x)$ and I know that $f(x)$ is continues in any interval where $\mathrm{tan}x$ is defined, so I am trying to use the Intermediate value theorem, but I don't know how.

asked 2022-05-28

$f$ is continuous at ${x}_{0}$ and $f({x}_{0})<0$ then there exists a neighborhood of ${x}_{0}$ in which $f(x)<0$ .

Is it possible to prove this statement, and if so what is the best way to go about it? Contradiction, Induction, Deduction, etc... ?

Is it possible to prove this statement, and if so what is the best way to go about it? Contradiction, Induction, Deduction, etc... ?

asked 2022-07-08

Let $\text{}f\text{}$ be any function which is defined for all numbers. Show that $\text{}g(x)=f(x)+f(-x)\text{}$ is even.

$\begin{array}{}\text{(1)}& \mathrm{e}.\mathrm{g}.\text{}f(x)={x}^{2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}g(x)={x}^{2}+(-x{)}^{2}=2{x}^{2}\leftarrow \text{}\text{}\text{even}\end{array}$

$\begin{array}{}\text{(2)}& \mathrm{e}.\mathrm{g}.\text{}f(x)={x}^{3}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}g(x)={x}^{3}+(-x{)}^{3}=0\leftarrow \text{}\text{}\text{even}\end{array}$

$\begin{array}{}\text{(3)}& \mathrm{e}.\mathrm{g}.\text{}f(x)={x}^{3}+{x}^{2}\text{}\text{}\leftarrow \text{}\text{}\text{neither even nor odd}\end{array}$

$\begin{array}{}\text{(4)}& \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\text{}g(x)=({x}^{3}+{x}^{2})+(-{x}^{3}+{x}^{2})=2{x}^{2}\leftarrow \text{even}\end{array}$

But how it can be proven that the claim holds?

And needless to say, can I completely assume "for all numbers" in the problem statements belong to a set of complex numbers(handling imaginary numbers is also required in this problem)?

$\begin{array}{}\text{(1)}& \mathrm{e}.\mathrm{g}.\text{}f(x)={x}^{2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}g(x)={x}^{2}+(-x{)}^{2}=2{x}^{2}\leftarrow \text{}\text{}\text{even}\end{array}$

$\begin{array}{}\text{(2)}& \mathrm{e}.\mathrm{g}.\text{}f(x)={x}^{3}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}g(x)={x}^{3}+(-x{)}^{3}=0\leftarrow \text{}\text{}\text{even}\end{array}$

$\begin{array}{}\text{(3)}& \mathrm{e}.\mathrm{g}.\text{}f(x)={x}^{3}+{x}^{2}\text{}\text{}\leftarrow \text{}\text{}\text{neither even nor odd}\end{array}$

$\begin{array}{}\text{(4)}& \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\text{}g(x)=({x}^{3}+{x}^{2})+(-{x}^{3}+{x}^{2})=2{x}^{2}\leftarrow \text{even}\end{array}$

But how it can be proven that the claim holds?

And needless to say, can I completely assume "for all numbers" in the problem statements belong to a set of complex numbers(handling imaginary numbers is also required in this problem)?

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 2022-08-14

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 2022-10-27

Let us consider the identity function

$f:(\mathbb{R},d)\to (\mathbb{R},{d}_{usual})$

$f:x\to x$

Here we are considering $d(x,y)=|(x{)}^{3}-(y{)}^{3}|$

Is the function $f$ uniformly continuous on closed and bounded interval?

I am looking for an example of function $f$ which is not uniformly continuous on a closed and bounded interval but it is continuous.

$f:(\mathbb{R},d)\to (\mathbb{R},{d}_{usual})$

$f:x\to x$

Here we are considering $d(x,y)=|(x{)}^{3}-(y{)}^{3}|$

Is the function $f$ uniformly continuous on closed and bounded interval?

I am looking for an example of function $f$ which is not uniformly continuous on a closed and bounded interval but it is continuous.

asked 2022-06-01

Suppose $y(x)={x}^{2}$ is a solution of $y"+P(x){y}^{\prime}+Q(x)y=0$ on (0,1) where $P$ and $Q$ are continuous functions on (0,1). Can both $P$ and $Q$ be bounded functions? Justify your answer.

I have tried differentiating $y={x}^{2}$, ${y}^{\prime}=2x$ and ${y}^{\prime}/x=2$. Differentiating this once again yielded $y"-{y}^{\prime}/x=0$. So I don't think $P(x)$ can be bounded in (0,1). But Iam not so sure about whether this is correct or not. So it would be great if anyone has some ideas to share.

I have tried differentiating $y={x}^{2}$, ${y}^{\prime}=2x$ and ${y}^{\prime}/x=2$. Differentiating this once again yielded $y"-{y}^{\prime}/x=0$. So I don't think $P(x)$ can be bounded in (0,1). But Iam not so sure about whether this is correct or not. So it would be great if anyone has some ideas to share.