Master Differential Calculus Problems with Our Experts

Evaluate $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\sin }^2<!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}\sqrt{n^2+n}),n\in <!--\; \in \; -->\mathbb{N}$

Limit of $\underset{x\to <!--\; \to \; -->{\frac{5\pi }{2}}^{+}}{lim}\frac{5x-<!--\; -\; -->\tan <!--\; \; -->x}{\cos <!--\; \; -->x}$

I have to prove, without L'hopital rule, the following limit:
$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\sqrt{x}\sin <!--\; \; -->\frac{1}{x}=0$
I tried doing a variable change, setting $t=\frac{1}{x}$ and reaching the following:
$\underset{t\to <!--\; \to \; -->0}{lim}\sqrt{\frac{1}{t}}\sin <!--\; \; -->t$
But I can't prove neither.

Use continuity to evaluate the limit.
$\underset{x\to <!--\; \to \; -->3}{lim}\arctan <!--\; \; -->\left(\frac{x^2-<!--\; -\; -->9}{5x^2-<!--\; -\; -->15x}\right)$
Factored and got...
$\underset{x\to <!--\; \to \; -->3}{lim}\arctan <!--\; \; -->\left(\frac{x+3}{5x}\right)$

How to prove that ${\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}\cos <!--\; \; -->(\mathrm{\alpha <!--\; \alpha \; -->}t)e^{-<!--\; -\; -->\mathrm{\lambda <!--\; \lambda \; -->}t}\mathrm{d}t=\frac{\mathrm{\lambda <!--\; \lambda \; -->}}{{\mathrm{\alpha <!--\; \alpha \; -->}}^2+{\mathrm{\lambda <!--\; \lambda \; -->}}^2}$ using real methods?

What are the steps to solve this limit $\underset{x\to <!--\; \to \; -->0}{lim}\frac{\cos <!--\; \; -->(x+\mathrm{\pi <!--\; \pi \; -->}/2)}{x}$?

Why is $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}n\arctan <!--\; \; -->(\frac{x}{n})=x?$

Proving that $(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots <!--\; \cdots \; -->+\frac{1}{n!})$ has a limit

Consider $\underset{x\to <!--\; \to \; -->0}{lim}\frac{\sin <!--\; \; -->(\tan <!--\; \; -->x)}{\sin <!--\; \; -->x}$. The answer is 1. This is clear intuitively since $\tan <!--\; \; -->x\approx x$ for small x. How do you show this rigorously?

Why does $\underset{x\to <!--\; \to \; -->\mathrm{\pi <!--\; \pi \; -->}}{lim}\frac{\sin <!--\; \; -->(mx)}{\sin <!--\; \; -->(nx)}={(-<!--\; -\; -->1)}^{m-<!--\; -\; -->n}\frac{m}{n}$, for positive naturals m and n?

How do I find the $\underset{x\to <!--\; \to \; -->0}{lim}\frac{7x^3-<!--\; -\; -->4x^2}{\sin <!--\; \; -->(3x^2)}$ without using L’Hopital’s?

How to Find $\underset{x\to <!--\; \to \; -->0}{lim}{\left(\frac{\tan <!--\; \; -->x}{x}\right)}^{\frac{1}{x^2}}$

Solve limit $\underset{x\to <!--\; \to \; -->0}{lim}\frac{x-<!--\; -\; -->\sin <!--\; \; -->(x)}{(x\sin <!--\; \; -->(x){)}^{3/2}}$

How can we compute the following limit not using derivative?
$\underset{x\to <!--\; \to \; -->0}{lim}\frac{1-<!--\; -\; -->\cos <!--\; \; -->(3x)}{5x^2}$
I know $\underset{x\to <!--\; \to \; -->0}{lim}\frac{\cos <!--\; \; -->(x)-<!--\; -\; -->1}{x}=0$. But $\underset{x\to <!--\; \to \; -->0}{lim}\frac{1-<!--\; -\; -->\cos <!--\; \; -->(3x)}{5x^2}=\underset{x\to <!--\; \to \; -->0}{lim}\frac{\cos <!--\; \; -->(3x)-<!--\; -\; -->1}{3x}\frac{(-<!--\; -\; -->3)}{5x^2}=0.-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}$

For $n\ge <!--\; \ge \; -->1$, consider the sequence $u_n^3=-<!--\; -\; -->n(u_n+1)$. Evaluate
$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{u}_n$

Evaluate without L'Hopital: $\underset{x\to <!--\; \to \; -->3}{lim}(x-<!--\; -\; -->3)\csc <!--\; \; -->\mathrm{\pi <!--\; \pi \; -->}x$
I find the indeterminate form of 0 or $\frac{0}{0}$. The latter tells me that L'Hopital's is an option, but since we haven't seen derivatives yet I'm not allowed to used it.
Previously I already tried swapping the $\csc <!--\; \; -->\mathrm{\pi <!--\; \pi \; -->}x$ for $\frac{1}{\sin <!--\; \; -->\mathrm{\pi <!--\; \pi \; -->}x}$ but when doing this I can't seem to get rid of the sinus. I also believe that since the limit goes to 3 and not to 0, the $\underset{x\to <!--\; \to \; -->0}{lim}\frac{\sin <!--\; \; -->ax}{ax}=1$ rule is not an option either.
I tried making the $\sin <!--\; \; -->(\frac{\mathrm{\pi <!--\; \pi \; -->}}{2})$ so that =1, but without any success. Can anyone give me a hint?

Evaluate limit $\underset{x\to <!--\; \to \; -->0}{lim}\frac{\ln <!--\; \; -->(\tan <!--\; \; -->2x)}{\ln <!--\; \; -->(\tan <!--\; \; -->3x)}$

I want to calculate the limit of
$\underset{x\to <!--\; \to \; -->0}{lim}\frac{1-<!--\; -\; -->\cosh <!--\; \; -->(x^5)}{x^6{\sin }^2<!--\; \; -->(x^2)}$

While proving that the quotient space ${\stackrel{¯<!--\; ¯\; -->}{B}}^n/S^{n-<!--\; -\; -->1}$ is homeomorphic to $S^n$, I needed to construct a continuous function $p:{\stackrel{¯<!--\; ¯\; -->}{B}}^n\to <!--\; \to \; -->S^n$. I figured that by fixing two points $R=(0,0,\dots <!--\; \dots \; -->,r)$ and $L=(0,0,\dots <!--\; \dots \; -->,-<!--\; -\; -->r)$ in the closed $n$-ball ${\stackrel{¯<!--\; ¯\; -->}{B}}^n$ of radius $r$, I could then use the function
$p(x)=\{\begin{array}{l}(x_1,\dots <!--\; \dots \; -->,x_n,\sqrt{r^2-<!--\; -\; -->||x|{|}^2}):(x_n-<!--\; -\; -->r{)}^2\le <!--\; \le \; -->(x_n+r{)}^2\\ (x_1,\dots <!--\; \dots \; -->,x_n,-<!--\; -\; -->\sqrt{r^2-<!--\; -\; -->||x|{|}^2}):(x_n+r{)}^2<(x_n-<!--\; -\; -->r{)}^2\end{array}$
(where $||x||$ is the standard norm in ${\mathbb{R}}^n$) to determine the sign of the $(n+1)$th component of the image of $x\in <!--\; \in \; -->{\stackrel{¯<!--\; ¯\; -->}{B}}^n$. However, knowing that the closed unit interval can be mapped to the circle by the parametrization $x\mapsto <!--\; \mapsto \; -->(\cos <!--\; \; -->(2\mathrm{\pi <!--\; \pi \; -->}x),\sin <!--\; \; -->(2\mathrm{\pi <!--\; \pi \; -->}x))$, I can't help but to wonder whether some similar construction could be made from ${\stackrel{¯<!--\; ¯\; -->}{B}}^n$ to $S^n$?

That is, do you happen to know some nice/comfortable continuous mappings from a general closed $n$-ball to $n$-sphere, similar to the parametrization of a unit circle by a unit interval?

Theorem: Let $(X,d_X)$ and $Y(d_Y)$ be metric spaces. A function $f:X\to <!--\; \to \; -->Y$ is continuous if and only if for every open $U\subset <!--\; \subset \; -->Y$, then
$f^{-<!--\; -\; -->1}(U)=\{x\in <!--\; \in \; -->X:f(x)\in <!--\; \in \; -->U\}$
is open in $X$.


If I want to show that the function $f:[0,1]\cup <!--\; \cup \; -->[2,4]\subset <!--\; \subset \; -->\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ defined by
$f(x)=\{\begin{array}{ll}1,& x\in <!--\; \in \; -->[0,1];\\ 2,& x\in <!--\; \in \; -->[2,4].\end{array}$
is continuous, then I need to consider any open set $U\subset <!--\; \subset \; -->\mathbb{R}$ and show that
$f^{-<!--\; -\; -->1}(U)=\{x\in <!--\; \in \; -->[0,1]\cup <!--\; \cup \; -->[2,4]:f(x)\in <!--\; \in \; -->U\}$
is open.

I understand that if $1\notin <!--\; \notin \; -->U$ and $2\notin <!--\; \notin \; -->U$, then $f^{-<!--\; -\; -->1}(U)=\mathrm{\varnothing <!--\; \varnothing \; -->}$. But I don't understand what happens when only $1\in <!--\; \in \; -->U$, or only $2\in <!--\; \in \; -->U$, or $1,2\in <!--\; \in \; -->U$.

If only $1\in <!--\; \in \; -->U$, then is it correct to say that $f^{-<!--\; -\; -->1}(U)$=[0,1]? If only $2\in <!--\; \in \; -->U$, then is it correct to say that $f^{-<!--\; -\; -->1}(U)$=[2,4]? But I am really confused since these are closed intervals. I am lost.

Can someone help me to find $f^{-<!--\; -\; -->1}(U)$ for those cases? Any clues or hints will be appreciated.