thanks.

2. Prove that between two zeros of $e^x\, \sin(x)=1$ there is a zeros of $e^x\,\cos(x)=-1$

Let: $f(x) = e^x\, \sin(x) - 1$

First observe that:

$e^x\,\cos(x)+1=f'(x)-f(x)$,

Now if $x_1$ and $x_2$ are consecutive zeros of $f(x)$ then the signs of $f'(x_1)$ and $f'(x_2)$ would be different (the other possibility that one or both of these is zero does not occur as that would imply that for one ,or both, of the zeros that $\cos(x)=\sin(x) \Rightarrow x=\pi/4+n\pi$, but such a point cannot be a zero of $f(x)$).

But the sign of $f'(x_1)$ and $f'(x_2)$ being different, implies that $e^x\,\cos(x)+1$ is positive at one root and negative at the other, so as $e^x\,\cos(x)+1$ is continuous, by the intermediate value theorem must be zero at some point between $x_1$ and $x_2$. which proves the result.

RonL

3. Find where the function:

$f(x)=\left( x+\frac{1}{x} \right) ^x$

is an increasing and where it is a decreasing function.

First we restrict attention to $x>0$, as $f(x)$ is complex or undefined otherwise.

Rewrite $f(x)$ as: $f(x)=e^{x\,\ln(x+1/x)}$

Now $f(x)$ is differentiable on $(0,\, \infty)$and its derivative is:

$\frac{df}{dx}=\left[\frac{x^2}{x^2+1}+\ln(x+1/x) \right] e^{x\,\ln(x+1/x)}$.

But the term in the square brackets is always positive, as is the term with the exponential. Therefore $f(x)$ is increasing on $(0, \infty)$.

RonL

4. First you probably mean,
$\lim_{x\to 0^+}x^{x^x}$
You can write,
$x^{\exp(x\ln x)}$
Thus,
$\exp(\ln x\exp(x\ln x))$
Let us look at the term inside,
$\ln x\exp(x\ln x)$
Now,
$\lim_{x\to 0^+}\ln x=-\infty$
And,
$\lim_{x\to 0^+}\exp (x\ln x)=1$
By using the L'Hopital rule inside and then applying the limit composite rule.
Thus,
$\lim_{x\to 0^+}\ln x\exp (x\ln x)=-\infty$ by the extended reals.

Thus,*)
$\lim_{x\to -\infty}\exp x=0$
If we subtract 1 from this limit we have -1 as the answer to the limit question.

*)I am note sure about this. Because I employed the limit composite rule for infinite limits (limits that do not exists). Thus, it might not work.