Master Differential Calculus Problems with Our Experts

Using the Squeeze Theorem to evaluate $\underset{x\to <!--\; \to \; -->0}{lim}\sqrt[3]{x}\cos <!--\; \; -->(\ln <!--\; \; -->(x^4))$
Would this just be a simple application of the Squeeze Theorem? (I can't use L'hopital or Taylor polynomials.)
$-<!--\; -\; -->\sqrt[3]{x}\le <!--\; \le \; -->\sqrt[3]{x}\cos <!--\; \; -->(\ln <!--\; \; -->(x^4))\le <!--\; \le \; -->\sqrt[3]{x}$
The limit from both sides is 0, so the middle limit must also be 0. Is that right?

Given :
$S_n=\underset{k=1}{\overset{n}{\sum <!--\; \sum \; -->}}\frac{1}{k}\text{ , }{T}_n={\int <!--\; \int \; -->}_1^{n+1}\frac{1}{x}dx$
Find a value of n so that $S_n\ge <!--\; \ge \; -->10$

A function is said to be continuous on an interval when the function is defined at every point on that interval and undergoes no interruptions, jumps, or breaks. If some function f(x) satisfies these criteria from x=a to x=b, for example, we say that f(x) is continuous on the interval [a, b]. The brackets mean that the interval is closed -- that it includes the endpoints a and b. In other words, that the interval is defined as a ≤ x ≤ b. An open interval (a, b), on the other hand, would not include endpoints a and b, and would be defined as a < x < b.

So, if a function is continuous on an open interval $(a,b)$ or even semi-open such as $[a,b]$, $(a,b]$ is it bounded?

limit of $x/\sin <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}x)$ as x approaches zero?

Finding the limit of $\frac{\sin <!--\; \; -->(2x)}{8x}$

What is $\underset{x\to <!--\; \to \; -->0}{lim}\frac{x\cos <!--\; \; -->x-<!--\; -\; -->\sin <!--\; \; -->x}{x^2\sin <!--\; \; -->x}$?

Evaluate $\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}(\cos <!--\; \; -->\sqrt{x}-<!--\; -\; -->\cos <!--\; \; -->\sqrt{x-<!--\; -\; -->1})$

I am trying to prove that
$\underset{N\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{1}{2N}\underset{k=1}{\overset{N}{\sum <!--\; \sum \; -->}}[\cos <!--\; \; -->((x-<!--\; -\; -->kL)\cdot <!--\; \cdot \; -->q)+\cos <!--\; \; -->((x+kL)\cdot <!--\; \cdot \; -->q)]=0$
for every $x,L\in <!--\; \in \; -->{\mathbb{R}}^{+}$ with $x\le <!--\; \le \; -->L$ and for every $q\ne <!--\; \ne \; -->\frac{2n\mathrm{\pi <!--\; \pi \; -->}}{L}$, $n\in <!--\; \in \; -->\mathbb{Z}$.

Geometric (Trigonometric) inequality $\frac{(a+b+c{)}^3}{3abc}\le <!--\; \le \; -->1+\frac{4R}{r}$

Compute the inverse function of the following polynomial on $[0,1]$?
$f(x)=\mathrm{\alpha <!--\; \alpha \; -->}{x}^3-<!--\; -\; -->2\mathrm{\alpha <!--\; \alpha \; -->}{x}^2+(\mathrm{\alpha <!--\; \alpha \; -->}+1)x$
(where $\mathrm{\alpha <!--\; \alpha \; -->}$ is within $]0,3[$)

Let $H_1$ and $H_2$ be Hilbert spaces, let $T\in <!--\; \in \; -->B(H_1,H_2)$. Suppose that $\mathrm{Ker}<!--\; \; -->T$ is finite-dimensional and that $\mathrm{Im}<!--\; \; -->T$ is closed in $H_2$. Prove that $\mathrm{Ker}<!--\; \; -->(T+K)$ is finite-dimensional for each $K\in <!--\; \in \; -->K(H_1,H_2)$.

1. Define Hilbert space as direct sum of two complemented subspaces
2. For compact operator use Hilbert Schmidt decomposition

Main idea is to prove that intersection of Spectrum of such $T and compact K is finite

How to find the limit of $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}(n\sin <!--\; \; -->\frac{n\mathrm{\pi <!--\; \pi \; -->}}{2}\cos <!--\; \; -->\frac{1}{n})$

Limit: $\underset{x\to <!--\; \to \; -->0}{lim}\frac{\tan <!--\; \; -->3x}{\sin <!--\; \; -->2x}$

I have found out that
$\underset{x\to <!--\; \to \; -->0}{lim}\frac{\sin <!--\; \; -->x}{x}=1$
I want to know the behaviour of the graph as x approaches $\pm <!--\; \pm \; -->\mathrm{\infty <!--\; \infty \; -->}$.

$\frac{1}{x}$ and its inverse breaking a rule of inverse functions?

How to prove this?
$\underset{h\to <!--\; \to \; -->0}{lim}\frac{\sin <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}+h)-<!--\; -\; -->\sin <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})}{\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}+h)-<!--\; -\; -->\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})}=-<!--\; -\; -->\frac{1}{\tan <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})}$

how to find $\underset{x\to <!--\; \to \; -->0}{lim}{\sin }^2<!--\; \; -->(\frac{1}{x}){\sin }^2<!--\; \; -->x$

Use the limit rule to find $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{n}^2(1-<!--\; -\; -->\cos <!--\; \; -->(\frac{1}{n}))$