You want to find $a$ so that $$\lim_{x\to 0}\frac{ax^2}{\sin^2x}=3\;;\tag{1}$$ recall that $\lim_{x\to 0}\frac{\sin x}x=1$, and write
$$3=\lim_{x\to 0}\frac{ax^2}{\sin^2x}=a\lim_{x\to 0}\frac{x^2}{\sin^2x}=a\left(\lim_{x\to 0}\frac{x}{\sin x}\right)^2=a(1)^2=a\;.$$
You asked whether this was okay:
$\frac{ax^2}{sin^2(x)}=3 \to ax^2=3sin^2(x) \to a=3\frac{sin^2(x)}{x^2}$, take the limit $\lim_{x\to0}3\frac{sin^2(x)}{x^2} = 3$, so $a=3$
It simply isn’t true that $\frac{ax^2}{\sin^2 x}=3$: the limit on the lefthand side is essential. This means that you can’t perform the algebraic manipulations that you used. You could argue that if $(1)$ holds, then clearly $a\ne 0$, so dividing by $3a$ yields $$\frac1a=\lim_{x\to 0}\frac{x^2}{3\sin^2 x}=\frac13\left(\lim_{x\to 0}\frac{x}{\sin x}\right)^2=\frac13\;,$$ because $a$ and $3$ are constants and can be moved in and out of the limit.
What’s really going on here is a little more complicated, though: you don’t actually know at this point that you want to solve $(1)$ for $a$.
You know that $$ax^2-5x^4\le f(x)\le ax^2\;,$$ and for $x$ close to but not equal to $0$ you know that $\sin^2x>0$. Those are the only $x$ of interest when we compute a limit as $x\to 0$, so we may divide through by $\sin^2x$ to get
$$\frac{ax^2-5x^4}{\sin^2x}\le\frac{f(x)}{\sin^2x}\le\frac{ax^2}{\sin^2x}\;.$$
In the limit as $x\to 0$ we want the middle term to be $3$, so we must have
$$\lim_{x\to 0}\frac{ax^2-5x^4}{\sin^2x}\le3\le\lim_{x\to 0}\frac{ax^2}{\sin^2x}\;.\tag{2}$$
As we’ve seen, $$\lim_{x\to 0}\frac{ax^2}{\sin^2x}=a\lim_{x\to 0}\left(\frac{x}{\sin x}\right)^2=a(1)^2=a\;,$$ so $(2)$ requires that $a\ge 3$. On the other hand,
$$\lim_{x\to 0}\frac{ax^2-5x^4}{\sin^2x}=\lim_{x\to 0}\frac{ax^2}{\sin^2x}-5\lim_{x\to 0}x^2\left(\frac{x}{\sin x}\right)^2=a-5(0)(1)^2=a\;,$$
so $(2)$ also requires that $a\le 3$, and it follows that $a=3$.
For any positive, real $k$ there is a $c$ such that $kx + c \gt \ln(x)$ (a proof of this is below using derivatives). If you divide each side by $x$, you can conclude that your function is less than $k + c/x$.
So we have
$$
k + \frac{c}{x} \gt\frac{ \ln(x)}{x} \gt 0
$$
Remember that this is really a load of squeezing inequalities, one for each choice of $k$, with a fitting $c$ to go with it.
In the limit as $x \to \infty$, this becomes
$$
k \geq \lim_{x \to \infty}\frac{\ln(x)}{x} \geq 0
$$
Since $k$ could be any positive, real number, we have the result we want.
Proof of the existence of $c$:
Let's set $x_0 = \frac{1}{k}$. I claim that any $c$ greater than
$$
c' = \ln(x_0) - kx_0
$$
works. We see that $kx + c'$ is tangent to $\ln(x)$ at $x = x_0$, since they have the same functional value and the same derivative. They also do not intersect at any other point since at any point before $x_0$, $\ln(x)$ has greater derivative, and at any point after $kx + c'$ has the greater derivative.
Therefore, any $c\gt c'$ will result in a line $kx + c$ which is strictly greater than $\ln(x)$ for all positive $x$.
Best Answer
Hint: Try showing that $$ 0 < \frac{n!}{n^n} = \boldsymbol{\frac{1}{n}} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdots \frac{n-1}{n} \cdot \frac{n}{n} < \frac{1}{n} $$ for $n \ge 3$.