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.

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