## Proving the that \$ f: R rightarrow R: x rightarrow frac{1}{1+x^2} \$ is continuous

When I use the delta epsilon definition i get stuck with | (x-a)(x+a)|. I know that |x-a| is smaller than delta but i do not find how to make |x+a| small. Can anybody help me?

## algebraic topology – Induced group homormophism \$f_{*} : pi_1( S^1,p) rightarrow pi_1( S^1,f(p))\$ due to \$z mapsto z^2\$.

Let $$f: S^1 rightarrow S^1$$ s.t. $$z mapsto z^2$$. This map induces a group homormophism $$f_{*} : pi_1( S^1,p) rightarrow pi_1( S^1,f(p))$$ . How can we explicitly write this map? I know that the mapping is that a loop gets mapped to 2 loops. But I am not sure how to write it?

Morevover, is there any bijective map for which the iduced map between the fundamental group is not an isomorphism?

## real analysis – What is the inverse of \$G: (x,y) rightarrow (x, 1-ye^x+xe^y)\$

How can I calculate the inverse of
begin{align*} G: mathbb{R}^2 &rightarrow mathbb{R}^2\ (x,y) &mapsto (x, 1-ye^x+xe^y) end{align*}
I actually just need the inverse in a neighborhood of the point $$(0,1)$$, and I know that the inverse exists in that neighborhood because
$$(D_pG) = left(begin{matrix} 1 & 0\ -ye^x+e^y & xe^y-e^x end{matrix}right)_p implies (D_{(0,1)}G) = left(begin{matrix} 1 & 0\ e-1 & -1 end{matrix}right)$$
which is invertible, and by the inverse function theorem I know that there is $$G^{-1}:U subset mathbb{R}^2 rightarrow mathbb{R}^2$$ where $$G(0,1) in U$$ open set.

Here’s what I was able to achieve:

One thing to note is that when $$v rightarrow 0$$ we have that

$$G^{-1}(v) sim D_{(0,0)}G^{-1}(v) + G^{-1}(0,0)$$

But $$G^{-1}(0,0) = G^{-1}(G(0,1)) = (0,1)$$ and $$D_{(0,0)}G^{-1} = D_{G(0,1)}G^{-1} = (D_{(0,1)}G)^{-1}$$, therefore we can conclude that when $$v rightarrow 0$$ it follows that

$$G^{-1}((v_1,v_2)) sim (D_{(0,1)}G)^{-1}((v_1,v_2)) + (0,1) = (v_1,-v_1-v_2+v_1e+1)$$

Can someone please check my work and give me some directions?

Thanks!

## real analysis – \$epsilon,delta\$-proof for the limit of \$log sinh{(x^2)}-x^2\$ for \$x rightarrow infty\$

I want to use a $$epsilon,delta$$-proof for the existence and value for the limit of $$log sinh{(x^2)}-x^2$$
for $$x rightarrow infty$$.

Now, I know the definition for such proof to be $$forall epsilon > 0 exists c forall x > c: left | f(x)-L right |.

I am struggeling to approach this problem as I am not familiar with the method for such proofs. What if the limit as $$x rightarrow infty$$ is $$infty$$ then this definition would crumble, right?

## general topology – If f: \$mathbb{R}_{l} rightarrow S_{Omega}\$ is continuous, then f is not injective.

If f: $$mathbb{R}_{l} rightarrow S_{Omega}$$ is continuous, then f is not injective.

I’ve been trying to solve this problem for a few days, but I haven’t been able to see how can I do it. First, $$mathbb{R}_{l}$$ refers to $$mathbb{R}$$ with the lower limit topology (generated by intervals of the form $$(a,b)$$ ) and $$S_{Omega}$$ refers to the uncountable well-ordered set with the order topology (generated by (a,b) and rays). Until now, I’ve tried to use the fact that both of those topologic spaces are Hausdorff and suppose that f is injective to get a contradiction, but I don’t see where can I get it.

## Sufficient condition for uniform convergence of \$f_{epsilon}(x, cdot)\$ to \$f(x, cdot)\$ implies that \$f_{epsilon} rightarrow f\$ uniformly.

Consider functions $$f_{epsilon}: X times U rightarrow mathbb{R}$$ for all $$epsilon >0$$ and $$f: X times U rightarrow mathbb{R}$$. $$X subset mathbb{R}^n$$ and $$U subset mathbb{R}^m$$ are compact spaces.
Suppose that given $$x in X$$, $$f_{epsilon}(x, cdot)$$ and $$f(x, cdot)$$ are continuous (on $$U$$).

If $$f_{epsilon}(x, cdot)$$ converges uniformly to $$f(x, cdot)$$, then does it imply that $$f_{epsilon} rightarrow f$$ uniformly?
I guess that from the uniform convergence of $$f_{epsilon}(x, cdot)$$ to $$f(x, cdot)$$,
we may choose $$overline{epsilon}(x) > 0$$ for any $$epsilon’ >0$$ such that $$|f_{epsilon(x)}(x, u) – f(x, u)| < epsilon’, forall epsilon(x) leq overline{epsilon}(x), forall (x, u) in X times U$$. Then, $$epsilon := min_{x in X}overline{epsilon}(x)$$ implies the desired result.

Actually, I’m not sure that such $$epsilon$$ exists. Is it correct? If not, are there any sufficient conditions to derive similar conclusions?

## complex numbers – Is \$T : mathbb{C} rightarrow mathbb{C}: a + bi mapsto overline{a + bi} = a – bi\$ a linear image?

So a question in my handbook for linear algebra is the following; Is the image T
$$T :mathbb{C} rightarrow mathbb{C}: a + bi mapsto overline{a + bi} = a – bi$$

an linear image if:

• $$mathbb{C}$$ is interpreted as a real vector space
• $$mathbb{C}$$ is interpreted as a complex vector space

I know how I would solve the question itself (I would check the axiom of linear image: if $$L: V rightarrow W$$ is a linear image then $$L(lambda_1v_1 + lambda_2v_2) = lambda_1L(v_1) + lambda_2L(v_2)$$) but I’m stuck at what they mean with the interpretation.

## real analysis – if \$lim _{xrightarrow infty}{f'(x)}=0\$ then does \$lim_{x rightarrow infty}{f(x)}\$ exist in the broad sense

Let $$f$$ be a differentiable a function in $$mathbb{R}$$, and let $$lim _{xrightarrow infty}{f'(x)}=0$$

Does $$lim_{x rightarrow infty}{f(x)}$$ exist in the broad sense?

I’m really lost here. This exercise is from a section on MVT, and intuitively it seems to be correct, but I can’t seem to find a lead. If someone could just give me a hint that would be great.

So far my best shot has been using Heine’s definition of the limit, but no dice.

## differential geometry – \$f: B^2 rightarrow B^2\$ be a continuous map such that \$f(x) = x\$ for every \$x in S^1\$, show \$f\$ is surjective

Let $$B^2 = {x in mathbb{R}^2; ||x|| leq 1}$$ be the unit disk inthe plane. Let $$f: B^2 rightarrow B^2$$ be a continuous map such that $$f(x) = x$$ for every $$x in S^1 = {x in mathbb{R}^2; ||x|| = 1}$$. Show that $$f$$ is surjective.

This my following attempt at a solution:

If $$f$$ isn’t surjective, then there exists a point $$y in B^2$$ such that $$f(x) – y$$ is never $$0$$ for all $$x in B^2$$. Thus $$frac{f(x) – y}{|f(x) – y|}$$ is well defined and exists on $$S^1$$.

however, I’m sure this is a dead end. Some hints would be greatly appreciated!

## \$lim_{n rightarrow infty} frac{(-1)^{n}sqrt{n}sin(n^{n})}{n+1}\$. Am I correct?

I have to find this limit:

begin{align} lim_{n rightarrow infty} frac{(-1)^{n}sqrt{n}sin(n^{n})}{n+1} end{align}

My attempt:

Since we know that $$-1leq sin (x) leq 1$$ for all $$x in mathbb{R}$$ we have:

begin{align} lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdot(-1)}{n+1}underbrace{leq}_{text{I have doubt if this is correct}}&lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(n^{n})}}{n+1} leq lim_{n rightarrow infty} frac{(1)^{n}cdotsqrt{n}cdot(1)}{n+1}\ \ lim_{n rightarrow infty} frac{(-1)^{n+1}cdotsqrt{n}}{n+1}leq&lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(n^{n})}}{n+1} leq lim_{n rightarrow infty} frac{sqrt{n}}{n+1} end{align}

My doubt is because of the $$(-1) terms$$. If everything is correct, then we have

begin{align} lim_{n rightarrow infty} frac{(-1)^{n+1}cdotsqrt{frac{1}{n}}}{1+frac{1}{n}}leq&lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(n^{n})}}{n+1} leq lim_{n rightarrow infty} frac{sqrt{frac{1}{n}}}{1+frac{1}{n}}\ \ Rightarrow 0 leq &lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(n^{n})}}{n+1} leq 0 \ \ therefore& lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(n^{n})}}{n+1}=0 end{align}

Am I correct? If I am not, how can I solve it? There are other ways to find this limit? I really appreciate your help!