Why is $f\left(x\right)=\sqrt{x}$ continuous?

vroos5p
2022-08-12
Answered

Why is $f\left(x\right)=\sqrt{x}$ continuous?

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asked 2022-08-22

What does continuous at a point mean?

asked 2022-07-13

How do you prove that the function $f\left(x\right)={(x+2{x}^{3})}^{4}$ is continuous at a =-1?

asked 2022-07-08

Definition: Here a neighborhood of the point $\overrightarrow{c}\in {\mathbb{R}}^{n}$ is the set ${N}_{r}(\overrightarrow{c})=\{\overrightarrow{x}\in {\mathbb{R}}^{n}:\Vert \overrightarrow{x}-\overrightarrow{c}{\Vert}_{2}<r\}$ for some $r>0$, and a subset $U\subset {\mathbb{R}}^{n}$ is said to be convex if $\overrightarrow{x},\overrightarrow{y}\in U,t\in [0,1]$ implies $t\overrightarrow{y}+(1-t)\overrightarrow{x}\in U$.

My Attempt: Suppose $\overrightarrow{x},\overrightarrow{y}\in {N}_{r}(\overrightarrow{c})$ and $t\in [0,1]$, we then have $\Vert \overrightarrow{x}-\overrightarrow{c}{\Vert}_{2}<r$ and $\Vert \overrightarrow{y}-\overrightarrow{c}{\Vert}_{2}<r$. Now,

$\Vert t\overrightarrow{y}+(1-t)\overrightarrow{x}-\overrightarrow{c}{\Vert}_{2}=\Vert t\overrightarrow{y}-t\overrightarrow{x}+\overrightarrow{x}-\overrightarrow{c}{\Vert}_{2}\le \Vert t\overrightarrow{y}-t\overrightarrow{x}{\Vert}_{2}+\Vert \overrightarrow{x}-\overrightarrow{c}{\Vert}_{2}<|t|\Vert \overrightarrow{y}-\overrightarrow{x}{\Vert}_{2}+r\le |t|(\Vert \overrightarrow{y}-\overrightarrow{c}{\Vert}_{2}+\Vert \overrightarrow{c}-\overrightarrow{x}{\Vert}_{2})+r<|t|(r+r)+r=(1+2|t|)r$

Screeching halt.

My Question: How should I make $\Vert t\overrightarrow{y}+(1-t)\overrightarrow{x}-\overrightarrow{c}{\Vert}_{2}$ less than $r$? My intuition is that I need the Cauchy-Schwarz inequality; but I am unsure where to apply it. Any hint would be greatly appreciated.

My Attempt: Suppose $\overrightarrow{x},\overrightarrow{y}\in {N}_{r}(\overrightarrow{c})$ and $t\in [0,1]$, we then have $\Vert \overrightarrow{x}-\overrightarrow{c}{\Vert}_{2}<r$ and $\Vert \overrightarrow{y}-\overrightarrow{c}{\Vert}_{2}<r$. Now,

$\Vert t\overrightarrow{y}+(1-t)\overrightarrow{x}-\overrightarrow{c}{\Vert}_{2}=\Vert t\overrightarrow{y}-t\overrightarrow{x}+\overrightarrow{x}-\overrightarrow{c}{\Vert}_{2}\le \Vert t\overrightarrow{y}-t\overrightarrow{x}{\Vert}_{2}+\Vert \overrightarrow{x}-\overrightarrow{c}{\Vert}_{2}<|t|\Vert \overrightarrow{y}-\overrightarrow{x}{\Vert}_{2}+r\le |t|(\Vert \overrightarrow{y}-\overrightarrow{c}{\Vert}_{2}+\Vert \overrightarrow{c}-\overrightarrow{x}{\Vert}_{2})+r<|t|(r+r)+r=(1+2|t|)r$

Screeching halt.

My Question: How should I make $\Vert t\overrightarrow{y}+(1-t)\overrightarrow{x}-\overrightarrow{c}{\Vert}_{2}$ less than $r$? My intuition is that I need the Cauchy-Schwarz inequality; but I am unsure where to apply it. Any hint would be greatly appreciated.

asked 2022-06-13

Let $h(t):[0,1]\to \mathbb{C}$ be a continuous function. Prove that $f:\mathbb{C}\setminus [0,1]\to \mathbb{C}$ defined by $f(z)={\int}_{0}^{1}\frac{h(t)}{z-t}$ is holomorphic.

My attempt:

$\underset{z\to 0}{lim}\frac{f(z+{z}_{0})-f({z}_{0})}{z}=\underset{z\to 0}{lim}\frac{{\int}_{0}^{1}\frac{h(t)}{z+{z}_{0}-t}-{\int}_{0}^{1}\frac{h(t)}{{z}_{0}-t}}{}=\phantom{\rule{0ex}{0ex}}\underset{z\to 0}{lim}\frac{{\int}_{0}^{1}(\frac{({z}_{0}-t)h(t)}{(z+{z}_{0}-t)({z}_{0}-t)}-\frac{(z+{z}_{0}-t)h(t)}{({z}_{0}-t)(z+{z}_{0}-t)})dt}{z}=\underset{z\to 0}{lim}\frac{{\int}_{0}^{1}\frac{-zh(t)}{(z+{z}_{0}-t)({z}_{0}-t)}dt}{z}=\phantom{\rule{0ex}{0ex}}-\underset{z\to 0}{lim}{\int}_{0}^{1}\frac{h(t)}{(z+{z}_{0}-t)({z}_{0}-t)}dt=-{\int}_{0}^{1}\frac{h(t)}{({z}_{0}-t{)}^{2}}dt$

Where I'm not sure about the last equality.

Is my solution correct? Any help would be appreciated.

My attempt:

$\underset{z\to 0}{lim}\frac{f(z+{z}_{0})-f({z}_{0})}{z}=\underset{z\to 0}{lim}\frac{{\int}_{0}^{1}\frac{h(t)}{z+{z}_{0}-t}-{\int}_{0}^{1}\frac{h(t)}{{z}_{0}-t}}{}=\phantom{\rule{0ex}{0ex}}\underset{z\to 0}{lim}\frac{{\int}_{0}^{1}(\frac{({z}_{0}-t)h(t)}{(z+{z}_{0}-t)({z}_{0}-t)}-\frac{(z+{z}_{0}-t)h(t)}{({z}_{0}-t)(z+{z}_{0}-t)})dt}{z}=\underset{z\to 0}{lim}\frac{{\int}_{0}^{1}\frac{-zh(t)}{(z+{z}_{0}-t)({z}_{0}-t)}dt}{z}=\phantom{\rule{0ex}{0ex}}-\underset{z\to 0}{lim}{\int}_{0}^{1}\frac{h(t)}{(z+{z}_{0}-t)({z}_{0}-t)}dt=-{\int}_{0}^{1}\frac{h(t)}{({z}_{0}-t{)}^{2}}dt$

Where I'm not sure about the last equality.

Is my solution correct? Any help would be appreciated.

asked 2022-05-28

I think it's obvious but I cannot show it explicitly.

If a function $f$ is continuous and positive definite, that is, $f(0)=0$ and $f(x)>0$ for all nonzero $x$.

Then, how to show that $\underset{t\to \mathrm{\infty}}{lim}f({x}_{t})=0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\underset{t\to \mathrm{\infty}}{lim}{x}_{t}=0$?

If a function $f$ is continuous and positive definite, that is, $f(0)=0$ and $f(x)>0$ for all nonzero $x$.

Then, how to show that $\underset{t\to \mathrm{\infty}}{lim}f({x}_{t})=0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\underset{t\to \mathrm{\infty}}{lim}{x}_{t}=0$?

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 2022-08-09

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