Some Requested Solutions and Help

1e) $\lim_{x\to0}e^{x\cos\left(e^{-1/x}\right)}$

It is enough to solve the following limit:

$\lim_{x\to0}x\cos\left(e^{-1/x}\right)$

Why is it enough to show this?

In any case, we observe that

$-|x|\leq x\cos\left(e^{-1/x}\right)\leq|x|$

Do you know why this is?

Continuing, we know that

$\lim_{x\to0}-|x|=0$ and $\lim_{x\to0}|x|=0$

Thus by the the squeeze theorem, we know that

$\lim_{x\to0}x\cos\left(e^{-1/x}\right)=0$

To complete the problem

$\lim_{x\to0}e^{x\cos\left(e^{-1/x}\right)}=e^{\lim_{x\to0}x\cos\left(e^{-1/x}\right)}=e^{0}=1$

6b) Find the absolute maximum value of $g(x)=\int_{0}^{x^{3}}\cos\left(t\right)dt$
on the interval $[0,\left(\frac{\pi}{2}\right)^{1/3}]$.

We would have already computed $g^{\prime}(x)$ from part a) but here
I'll treat the problem on it's own. We use the fundamental theorem
of calculus and the chain rule to obtain $g^{\prime}(x)=3x^{2}\cdot\cos\left(x^{3}\right)$.

In general: $F(x)=\int_{a}^{f(x)}g(t)dt$. Then $F^{\prime}(x)=g(f(x))\cdot f^{\prime}(x)$.

Or perhaps more generally: $F(x)=\int_{f(x)}^{g(x)}h(t)dt$. Then
to compute the derivate we should first rewrite $F(x)$.

$F(x)=\int_{f(x)}^{a}h(t)dt+\int_{a}^{g(x)}h(t)dt=-\int_{a}^{f(x)}h(t)dt+\int_{a}^{g(x)}h(t)dt$

$F^{\prime}(x)=-h(f(x))\cdot f^{\prime}(x)+h(g(x))\cdot g^{\prime}(x)$.

Why was finding the derivative so important in this problem?

The next step is to find the zeros of the derivative.

$g^{\prime}(x)=3x^{2}\cdot\cos\left(x^{3}\right)=0$ implies $x^{2}=0$
or $\cos\left(x^{3}\right)=0$.

Thus $x=0$ or $x^{3}=\frac{\pi}{2}+k\pi$ where $k\in\mathbb{Z}.$

Because $x\in[0,\left(\frac{\pi}{2}\right)^{1/3}]$ then we must have
$x=0$ or $x^{3}=\frac{\pi}{2}$.

Thus we have two critical points.

In general, we must remember to check the endpoints of the interval,
but in this problem, the critical points coincide with the endpoints.

Now we evaluate $g(x)$ at these critical points.

$g(0)=\int_{0}^{0}\cos tdt=0$

$g(\left(\frac{\pi}{2}\right)^{1/3})=\int_{0}^{\pi/2}\cos tdt=\sin t|_{0}^{\pi/2}=1$

We conclude the global maximum value is 1.

1d)

$\frac{\left(\sqrt{x^{2}+4}-x\right)\left(\sqrt{x^{2}+4}+x\right)}{\sqrt{x^{2}+4}+x}=\frac{\left(x^{2}+4\right)-x^{2}}{\sqrt{x^{2}+4}+x}=\frac{4}{\sqrt{x^{2}+4}+x}$

There seems to be an error with 2b on the second practice exam.

Arriving at $\lim_{x\to\infty}\frac{\ln(2x-1)-\ln(2x+2)}{1/(x+1)}=\lim_{x\to\infty}\frac{\frac{2}{2x-1}-\frac{2}{2x+2}}{\frac{-1}{(x+2)^{2}}}$

$=-2\cdot\lim_{x\to\infty}(x+2)^{2}\left(\frac{2x+2}{\left(2x-1\right)(2x+2)}-\frac{2x-1}{\left(2x-1\right)\left(2x+2\right)}\right)$

$=-2\cdot\lim_{x\to\infty}(x+2)^{2}\left(\frac{2-(-1)}{(2x-1)(2x+2)}\right)$

$=-2\cdot\lim_{x\to\infty}(x+2)^{2}\frac{3}{(2x-1)(2x+2)}=-2\cdot\frac{3}{4}=-\frac{3}{2}$