Master Continuity Equation with Plainmath

For a function $f:[-<!--\; -\; -->1,1]\to <!--\; \to \; -->R$, consider the following statements:
Statement 1: If
$‘<!--\; ‘\; -->\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}f\left(\frac{1}{n}\right)=f(0)=\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}f(-<!--\; -\; -->\frac{1}{n})‘<!--\; ‘\; -->,$
then $f$ is continuous at $x=0$
Statement 2: If $f$ is continuous at $x=0$, then
$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}f\left(\frac{1}{n}\right)=\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}f(-<!--\; -\; -->\frac{1}{n})=\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}f(e^{\frac{1}{n}}-<!--\; -\; -->1)=f(0)$
Then which of the above statements is/are true.

My Attempt:
I feel that $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}f\left(\frac{1}{n}\right)=f(0)$ is same as $\underset{x\to <!--\; \to \; -->0}{lim}f(x)=f(0)$, 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 $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}f\left(\frac{1}{n}\right)=f\left(\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{1}{n}\right)=f(0)$. By same logic we can prove $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}f(-<!--\; -\; -->\frac{1}{n})=\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}f(e^{\frac{1}{n}}-<!--\; -\; -->1)=f(0)$
Can there be counter-examples to what I am thinking

How do you prove that ${\displaystyle g\left(x\right)=\frac{1}{2}x}$ is continuous at ${\displaystyle x=\frac{1}{4}}$?

What does continuous at a point mean?

Find all continuous functions with the property that f(x)f(y)f(z) = 3, for every $x,y,z\in <!--\; \in \; -->R$, with x + y + z = 0.
I am quite new to new to functional equations and I have been stuck on this problem for quite a while.
I tried to somehow get and transform into a Cauchy equation but failed during the process.

How do you prove that the function ${\displaystyle f\left(x\right)=\frac{x-5}{|x-5|}}$ is continuous everywhere but x=5?

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

If $f(x)=\frac{1}{(x-<!--\; -\; -->2)(2x-<!--\; -\; -->5)}$ and $g(x)=\frac{1}{x^2}$ , we define f(g(x)) over the domain in which its defined , is it correct to say f(g(x)) is discontinuous at $+-<!--\; -\; -->\surd (2/5)$ ,0 and +-$1/\surd 2$ ? 0 because the input is going into g(x) hence as its not in domain of g(x) so it can be said its discontinuous at that point . and similarily other four points as its not in domain of f(g(x)) ? But if suppose we say h(x)=f(g(x)) will we say h(x) is discontinuous at those points (maybe different ones) where x is not defined after simplyfing the whole f(g(x)) to a rational function ?

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

Let X be a locally compact (not compact) Hausdorff space. Let $Y=X\cup <!--\; \cup \; -->\{0\}$ be the one point compactification of X. Now define a set $C_0(X)$ by:
$C_0(X):=\{f\in <!--\; \in \; -->C(X):f\text{ vanishes at infinity}\}.$
Now we say that $f\in <!--\; \in \; -->C(X)$ vanishes at infinity if for each $\mathrm{\epsilon <!--\; \epsilon \; -->}>0$ the set $\{\mathrm{\omega <!--\; \omega \; -->}\in <!--\; \in \; -->X:|f(\mathrm{\omega <!--\; \omega \; -->})|\ge <!--\; \ge \; -->\mathrm{\epsilon <!--\; \epsilon \; -->}\}$ is compact. Now consider $C(Y)$, the set of all continuous functions from Y to $\mathbb{C}$. Now there is a statement in the middle of a proof that, a function in $C(Y)$ belongs to $C_0(X)$ if and only if $f(0)=0.$ I am not able to prove this statement.

My approach: ′′${}^{″}{\Leftarrow <!--\; \Leftarrow \; -->}^{″}$′′ Let $f\in <!--\; \in \; -->C(Y)$ and $f(0)=0$. Nowf is continuous at 0. Then for each $\mathrm{\epsilon <!--\; \epsilon \; -->}>0$ there exists a neighborhood $U_0$ of 0 such that $|f(x)|<\mathrm{\epsilon <!--\; \epsilon \; -->}$ for all $x\in <!--\; \in \; -->U_0$. Now Y is compact and $U_0$ is open hence $Y\setminus <!--\; \setminus \; -->U_0$ is closed hence compact. Now I am not able to proceed from here and also the ′′${}^{″}{\Rightarrow <!--\; \Rightarrow \; -->}^{″}$′′ direction.

How do you prove that the function ${\displaystyle f\left(x\right)=\frac{x^2+x}{x}}$ is not continuous at a =0?

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

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

Write a conditional statement. Write the converse, inverse and contrapositve for your statement and determine the truth value of each. If a statement’s truth value is false, give a counterexample.

I am preparing for my exam and need help with the following task:
Let $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ be a continuous function with the estimation $x^2\le <!--\; \le \; -->f(x)$$\mathrm{\forall <!--\; \forall \; -->}x\in <!--\; \in \; -->\mathbb{R}$. Show that f takes on its absolute minima.
If a function is continuous, then $\underset{x\to <!--\; \to \; -->a}{lim}f(x)$=$f(a)$.
Our function has an absolute minima in $x_0$$\in <!--\; \in \; -->\mathbb{R}$, if $f(x)\ge <!--\; \ge \; -->f(x_0)$ for $x\in <!--\; \in \; -->\mathbb{R}$.
At first I thought the task is pretty easy. We learned how to prove that if $f:[a,b]\to <!--\; \to \; -->\mathbb{R}$ is continuous, then f has an absolute maxima and an absolute Minima in [a,b]. The Problem here is, that the domain of our function here is unbounded. Thats why I don't have any idea what I could and should use for the proof. We should probably use the estimation $x^2\le <!--\; \le \; -->f(x)$ $\mathrm{\forall <!--\; \forall \; -->}x\in <!--\; \in \; -->\mathbb{R}$. This gives us the information, that f is bounded from below with $f(x)\ge <!--\; \ge \; -->0$ $\mathrm{\forall <!--\; \forall \; -->}x\in <!--\; \in \; -->\mathbb{R}$. But how does this help me? And what else do we have?

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 $\stackrel{¯<!--\; ¯\; -->}{D}(0,r)$ 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 $\underset{z\to <!--\; \to \; -->0}{lim}\frac{f(z)}{z}=f^{\prime }(0)$ 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!

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

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....

$f,g:<!--\; :\; -->\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ continuous functions such that $f(g(x))=x$. Prove that $g(f(x))=x$.
Since $f$ is surjective, we basically have to prove that $f$ is injective. For $g$ is clearly, but how can I prove that $f$ is injective ? Please, help me!

Definition: Here a neighborhood of the point $\stackrel{\to <!--\; \to \; -->}{c}\in <!--\; \in \; -->{\mathbb{R}}^n$ is the set $N_r(\stackrel{\to <!--\; \to \; -->}{c})=\{\stackrel{\to <!--\; \to \; -->}{x}\in <!--\; \in \; -->{\mathbb{R}}^n:\Vert <!--\; \Vert \; -->\stackrel{\to <!--\; \to \; -->}{x}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{c}{\Vert <!--\; \Vert \; -->}_2<r\}$ for some $r>0$, and a subset $U\subset <!--\; \subset \; -->{\mathbb{R}}^n$ is said to be convex if $\stackrel{\to <!--\; \to \; -->}{x},\stackrel{\to <!--\; \to \; -->}{y}\in <!--\; \in \; -->U,t\in <!--\; \in \; -->[0,1]$ implies $t\stackrel{\to <!--\; \to \; -->}{y}+(1-<!--\; -\; -->t)\stackrel{\to <!--\; \to \; -->}{x}\in <!--\; \in \; -->U$.

My Attempt: Suppose $\stackrel{\to <!--\; \to \; -->}{x},\stackrel{\to <!--\; \to \; -->}{y}\in <!--\; \in \; -->N_r(\stackrel{\to <!--\; \to \; -->}{c})$ and $t\in <!--\; \in \; -->[0,1]$, we then have $\Vert <!--\; \Vert \; -->\stackrel{\to <!--\; \to \; -->}{x}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{c}{\Vert <!--\; \Vert \; -->}_2<r$ and $\Vert <!--\; \Vert \; -->\stackrel{\to <!--\; \to \; -->}{y}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{c}{\Vert <!--\; \Vert \; -->}_2<r$. Now,
$\Vert <!--\; \Vert \; -->t\stackrel{\to <!--\; \to \; -->}{y}+(1-<!--\; -\; -->t)\stackrel{\to <!--\; \to \; -->}{x}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{c}{\Vert <!--\; \Vert \; -->}_2=\Vert <!--\; \Vert \; -->t\stackrel{\to <!--\; \to \; -->}{y}-<!--\; -\; -->t\stackrel{\to <!--\; \to \; -->}{x}+\stackrel{\to <!--\; \to \; -->}{x}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{c}{\Vert <!--\; \Vert \; -->}_2\le <!--\; \le \; -->\Vert <!--\; \Vert \; -->t\stackrel{\to <!--\; \to \; -->}{y}-<!--\; -\; -->t\stackrel{\to <!--\; \to \; -->}{x}{\Vert <!--\; \Vert \; -->}_2+\Vert <!--\; \Vert \; -->\stackrel{\to <!--\; \to \; -->}{x}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{c}{\Vert <!--\; \Vert \; -->}_2<|t|\Vert <!--\; \Vert \; -->\stackrel{\to <!--\; \to \; -->}{y}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{x}{\Vert <!--\; \Vert \; -->}_2+r\le <!--\; \le \; -->|t|(\Vert <!--\; \Vert \; -->\stackrel{\to <!--\; \to \; -->}{y}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{c}{\Vert <!--\; \Vert \; -->}_2+\Vert <!--\; \Vert \; -->\stackrel{\to <!--\; \to \; -->}{c}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{x}{\Vert <!--\; \Vert \; -->}_2)+r<|t|(r+r)+r=(1+2|t|)r$
Screeching halt.

My Question: How should I make $\Vert <!--\; \Vert \; -->t\stackrel{\to <!--\; \to \; -->}{y}+(1-<!--\; -\; -->t)\stackrel{\to <!--\; \to \; -->}{x}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{c}{\Vert <!--\; \Vert \; -->}_2$ 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.

Let $f$ be any function which is defined for all numbers. Show that $g(x)=f(x)+f(-<!--\; -\; -->x)$ is even.

$\begin{array}{}\text{(1)}& \mathrm{e}.\mathrm{g}.<!--\; \; -->f(x)=x^2⟹<!--\; ⟹\; -->g(x)=x^2+(-<!--\; -\; -->x{)}^2=2x^2\leftarrow <!--\; \leftarrow \; -->\text{even}\end{array}$

$\begin{array}{}\text{(2)}& \mathrm{e}.\mathrm{g}.<!--\; \; -->f(x)=x^3⟹<!--\; ⟹\; -->g(x)=x^3+(-<!--\; -\; -->x{)}^3=0\leftarrow <!--\; \leftarrow \; -->\text{even}\end{array}$

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

$\begin{array}{}\text{(4)}& ⟹<!--\; ⟹\; -->g(x)=(x^3+x^2)+(-<!--\; -\; -->x^3+x^2)=2x^2\leftarrow <!--\; \leftarrow \; -->\text{even}\end{array}$

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)?