How does this limit with a square root behave as it approaches infinity? $$
\lim_{x\to \infty}\frac{2x}{\sqrt{4x^2+1}}
$$

Can I substitute $\sqrt{4x^2}$ for $2x$ in the denominator to show that this limit approaches $1$?
Yes, dividing both the numerator and the denominator by $2x$ is fine, and converting $2x$ into $\sqrt{4x^2}$ is fine. The reason that this is fine is that as $x\to \infty$, we know that (eventually) $x$ and therefore $2x$ is strictly positive, and you can always do those things with positive numbers.

So it is not the case that $\frac{2x}{\sqrt{4x^2 + 1}} = \frac{1}{\sqrt{1+\frac{1}{4x^2}}}$. It is, however, the case that they are equal once $x$ is sufficiently close to $\infty$, and for our limit that's all we care about.
Yes it is completely correct in this case,since for $x&gt;0$ $\sqrt{4x^2}=2x$ , and equivalent to get out $2x$ from the denominator.