Is $3^x \lt 1 + 2^x + 3^x \lt 3 \cdot 3^x$ right? Is $3^x \lt 1 + 2^x + 3^x \lt 3 \cdot 3^x$ right?
This is from my lecture notes which is used to solve: 


But when $x = 0$, $(1 + 2^x + 3^x = 3) \gt (3^0 = 1)$? The thing is how do I choose which what expression should go on the left & right side? 
 A: When $x=0$, the left side $3^0=1$, the center is $3$ as you say, and the right side is $3\cdot 3^0=3 \cdot 1=3$ so the center and right sides are equal.  But you want this for large $x$, so could restrict the range to $x \gt 1$, say.
A: The inequality is, for $x>0$
$$3^x<1+2^x+3^x<3\cdot 3^x$$
When $x=0$, you obtain $1<1+1+1\le 3\cdot 1$ (I think you have a type in the last part of the post.
But, this is of no concern.  You are trying to find $\lim_{x\rightarrow\infty} (1+2^x+3^x)^{4/x}$. Since the limit is being taken at infinity, you are concerned only with values of $x$ that are large. Large in particular means eventually $x$ is positive, and you can use the inequality above (or, rather, the inequality following it in your post).
The Theorem being used is the following:
Suppose the inequality
$$\tag{1}
f(x)\le g(x)\le h(x)
$$
holds for all $x\ge a$, for some real number $a$.
If $\lim\limits_{x\rightarrow\infty} f(x)$ and  $\lim\limits_{x\rightarrow\infty} h(x)$ exist and are equal with common value $L$, then  $\lim\limits_{x\rightarrow\infty} g(x)$ 
exists and is equal to $L$.
So the Squeeze Theorem is valid whenever the required inequality holds for $x$ sufficiently large in the case where the limit  is taken at infinity.
(If the limit is taken at $b$, then the inequality need onlly holds for $x$ near $b$.)
A: Firstly, your lecture notes says "For all $x>0$", so you can't choose $0$. In fact:
Let $I$ be an interval having the point $a$ as a limit point. Let $f$, $g$, and $h$ be functions defined on $I$, except possibly at a itself. Suppose that for every $x$ in $I$ not equal to a, we have:
$$g(x)\leq f(x)\leq h(x)$$
also suppose that
$$\lim_{x\to a}g(x)=\lim_{x\to a}h(x)=L$$
Then we have
$$\lim_{x\to a}f(x)=L$$
So it doesn't necessary have to be "For all $x>0$". If you know Your limit points is $+\infty$ so all you need is to be sure that some relation
$$g(x)\leq f(x)\leq h(x)$$
holds starting with some $x$.
