Master Calculus 1 Practice Problems with Expert Help

Proving sin(1/x) has an antiderivative
I would want to prove that the function defined as follows: $f(x)=\sin <!--\; \; -->(1/x)$ and $f(0)=0$ has an antiderivative on the entire $\mathbb{R}$ (well, I'm not sure if I haven't worded this awkwardly, but essentially I am trying to show that f is a derivative of some function that is differentiable in every $x\in <!--\; \in \; -->\mathbb{R}$).
First off, f is continuous on $\mathbb{R}\setminus <!--\; \setminus \; -->0$, hence it has an antiderivative for every $x\in <!--\; \in \; -->\mathbb{R}\setminus <!--\; \setminus \; -->0$. However, f is not continuous in $x=0$. $x=0$
Furthermore, I'm curious about the significance of $f(0)=0$. Suppose we redefine our function, and take $f(0)=c$, for some $c\in <!--\; \in \; -->\mathbb{R}$, $c\ne <!--\; \ne \; -->0$. Would then f have an antiderivative on $\mathbb{R}$?

So i have this limit:
$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}x(\sqrt{x^2+4}-<!--\; -\; -->\sqrt{x^2+2})$

Why do the antiderivatives of $y=1$ follow $e^x$?
If you integrate $y=1$, you get x. If you integrate that, you get $\frac{x^2}{2}$, following the power rule. If you continue this, integrating over and over, the antiderivatives are: 1, x, $\frac{x^2}{2}$, $\frac{x^3}{6}$, $\frac{x^4}{24}$, and so on. It follows the pattern of $\frac{x^n}{n!}$, which happens to be the infinite polynomial of $e^x$. Just by integrating $y=1$, you follow the $exp()$ function. Is there any intuitive way why this is?

How can I find a range of values of $x$ for which this approximation gives correct results rounded to $2$ decimal places?

Let
$s_n=\underset{j=1}{\overset{n}{\sum <!--\; \sum \; -->}}\underset{i=n+1}{\overset{+\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{\displaystyle \frac{1}{i^2+j^2}}$

What's wrong in my method to find values of a and b in the limit $\underset{x\to <!--\; \backslash rightarrow\; -->0}{lim}\frac{x(1+a\cos <!--\; \; -->x)-<!--\; -\; -->b\sin <!--\; \; -->x}{{\sin }^3<!--\; \; -->x}=1$

Antiderivative of $x^a{e}^{\left(bx\right)}$
I need to know the primitive function (Antiderivative) of this function:
$f(x)=x^a{e}^{\left(bx\right)}$ where a and b is a positive constants please how could I find the primitive of function ? is there any technique concerning this types?

How do you find the instantaneous rate of change of y=5x+9 at x=2

Function with compact support whose iterated antiderivatives also have compact support
Notation: If $f:<!--\; :\; -->\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ is continuous, let us denote $If:<!--\; :\; -->\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ its indefinite integral from 0, i.e., $(If)(x)={\int <!--\; \int \; -->}_0^{x}f(t)dt$, and iteratively $I^{k+1}f=I(I^{k}f)$.
Remark: If f is a continuous function with support contained in the open interval ]0,1[ then If has support contained in ]0,1[ iff $(If)(1)=0$.
Main question: Does there exist a $C^{\mathrm{\infty <!--\; \infty \; -->}}$ function f with support contained in the open interval ]0,1[ such that $I^{k}f$ has support contained in ]0,1[ for every $k\ge <!--\; \ge \; -->0$, or, equivalently, $(I^{k}f)(1)=0$ for all $k\ge <!--\; \ge \; -->0$?
Equivalent formulation: Does there exists a sequence $(f_k{)}_{k\in <!--\; \in \; -->\mathbb{Z}}$ of $C^{\mathrm{\infty <!--\; \infty \; -->}}$ functions each with support contained in the open interval ]0,1[, such that $f_{k-<!--\; -\; -->1}$ is the derivative of $f_k$?
Weaker question: Does there at least exist a continuous function f with the properties demanded in the main question?
Stronger question: Does there exist a $C^{\mathrm{\infty <!--\; \infty \; -->}}$ function f with compact support, whose Fourier transform vanishes identically on a nontrivial interval?
(A positive answer to the latter would imply a positive answer to the main question: rescale the function so its support is contained in ]0,1[, multiply it appropriately so its Fourier transform vanishes in a neighborhood of 0, and observe that the Fourier transform of $I^{k}f$ is, up to constants, ${\mathrm{\xi <!--\; \xi \; -->}}^k$ times that of f.)
Edit: Before someone points out that the identically zero function fits the bill, I should add that I want my functions to not vanish identically.

I need to find the following limit of a function of complex variable:
$\begin{array}{r}\underset{z\to <!--\; \backslash rightarrow\; -->i}{lim}\frac{\sqrt[3]{z+i}-<!--\; -\; -->\sqrt[3]{2i}}{\sqrt{z}-<!--\; -\; -->\sqrt{i}}\end{array}$

Say I have to calculate the limit
$\underset{x\to <!--\; \to \; -->0}{lim}{\displaystyle \frac{1-<!--\; -\; -->\cos <!--\; \; -->(x)}{3{\sin }^2<!--\; \; -->(x)}}$

To find the limit of $(1+\frac{1}{n^2})(1+\frac{2}{n^2})...(1+\frac{n}{n^2})$

Calculate $\underset{x\to <!--\; \to \; -->0^{+}}{lim}\frac{4^x-<!--\; -\; -->\sqrt{1-<!--\; -\; -->x}}{x^2\ln <!--\; \; -->(1+e^{1/x})}$ using L'Hospital's rule.

Integrate : $\int <!--\; \int \; -->\frac{1}{x^2\sqrt{2x-<!--\; -\; -->x^2}}dx$

Solve for $f(x)$
${\int <!--\; \int \; -->}_0^{x^3+1}\frac{f^{\prime }(t)}{1+f(t)}dt=\ln <!--\; \; -->(x)$

So far I got:
$\sqrt{2n+1}>\sqrt{2n-<!--\; -\; -->1}$, so $\sqrt{2n+1}-<!--\; -\; -->\sqrt{2n-<!--\; -\; -->1}>0$ for all $n>0$

I can't find a way to continue with this calculation of limit:
$\underset{x\to <!--\; \backslash rightarrow\; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{(\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}-<!--\; -\; -->\arctan <!--\; \; -->(x))}^{\frac{1}{\ln <!--\; \; -->(x)}}=\underset{x\to <!--\; \backslash rightarrow\; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{e}^{\frac{\ln <!--\; \; -->(\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}-<!--\; -\; -->\arctan <!--\; \; -->(x))}{\ln <!--\; \; -->(x)}}.$

How do you find the instantaneous rate of change of the function $f(x)=\frac{3}{x}$ when x=2?

Determine the value of the real number A so that a finite limit exists, and then compute the limit:
$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}(\sqrt{A{x}^2+2x}-5x)$