Calculus 1: limits using squeeze or sandwich theorem, verification. I have the next problem:
Suppose that a function $g$ satisfies that $-1\leq g(x) \leq 1$ for all $x\geq 0$.
Calculate $\lim_{x\to\infty}\dfrac{x+g(2x)}{g(3x)+4x}$  
So: $$\lim_{x\to\infty}\dfrac{x+g(2x)}{g(3x)+4x}=\lim_{x\to\infty}\dfrac{\frac{x+g(2x)}{x}}{\frac{g(3x)+4x}{x}}=\lim_{x\to\infty}\dfrac{1+\frac{g(2x)}{x}}{\frac{g(3x)}{x}+4}$$
And because $-1\leq g(x) \leq 1$ then $\frac{g(2x)}{x}\to 0$ and $\frac{g(3x)}{x}\to 0$ as $x\to \infty$, therefore:
$$\lim_{x\to\infty}\dfrac{1+\frac{g(2x)}{x}}{\frac{g(3x)}{x}+4}=\frac{1}{4}$$
But I don't know in which part I am using the squeeze/sandwich theorem, this is the second bullet of an exercise and the first bullet is just to enunciate the theorem.
 A: It is straightforward that
$$-1\le g(2x) \le 1 \implies -\frac{1}{x}\le \frac{g(2x)}{x} \le \frac{1}{x}\quad\text{ for }x>0.$$
Then, since $$\lim_{x\to \infty}{-\frac{1}{x}}=0=\lim_{x\to\infty}\frac{1}{x},$$
the squeeze theorem implies that $\displaystyle{\lim_{x\to \infty}\frac{g(2x)}{x}}$ exists and also it is equal to $0$.
Similarly, as a consequence of the squeeze theorem we get
$$\lim_{x\to \infty}\frac{g(3x)}{x}=0.$$
A: Get rid of $g(x)$ entirely.  $-1\le g(x) \le 1$.  So $x - 1 \le x + g(2x) \le x + 1$.
Likewise. $4x - 1 \le g(3x) + 4x \le 4x + 1$.
So $\dfrac{x - 1}{4x + 1} \le \dfrac{x + g(2x)}{g(3x) + 4x} \le \dfrac{x + 1}{4x - 1}$
So $1/4 = \lim\dfrac{x-1}{4x+1} \le \lim \text{wholeness} \le \lim \dfrac{x + 1}{4x - 1} = \dfrac14$.
Your method worked too.  You used $\frac{-1}x \le \frac{g(2x)}x < \frac1x$ and $\frac{-1}x \le \frac{g(3x)}x < \frac1x$ for the squeeze theorem twice.  It also works.
A: You can get the limit with a single application of the squeeze theorem.
For the numerator at $x>1$, 
$$0<1-\frac{1}{x}\le 1+\frac{g(2x)}{x} \le 1+\frac{1}{x}$$
and for the denominator at $x>1$,
$$0<4-\frac{1}{x} \le \frac{g(3x)}{x}+4 \le 4+\frac{1}{x}$$
So for all $x>1$
$$\frac{1-\frac{1}{x}}{4+\frac{1}{x}} \le \frac{1+\frac{g(2x)}{x}}{\frac{g(3x)}{x}+4} \le \frac{1+\frac{1}{x}}{4-\frac{1}{x}}$$
and $\lim_{x\to\infty}{\frac{1-\frac{1}{x}}{4+\frac{1}{x}}}=\frac{1}{4}$ while $\lim_{x\to\infty}{\frac{1+\frac{1}{x}}{4-\frac{1}{x}}}=\frac{1}{4}$, so by the squeeze theorem
$$\lim_{x\to\infty}{\frac{1+\frac{g(2x)}{x}}{\frac{g(3x)}{x}+4}}=\frac{1}{4}$$ 
