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 $1$?
 A: 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.
A: It would be incorrect if $x$ were negative (since then you'd have $\sqrt{4x^2} = 2|x| \ne 2x$) but limits as $x\to+\infty$ depend only on the values of the function when $x$ is positive, so this is correct.
\begin{align}
\lim_{x\to\infty} \frac {2x}{\sqrt{4x^2+1}} & = \lim_{x\to\infty} \sqrt{\frac{4x^2}{4x^2+1}} \\[12pt]
& = \sqrt{ \lim_{x\to\infty} \frac{4x^2}{4x^2+1} } & & \text{because the square root function is continuous} \\[10pt]
& = \text{etc.}
\end{align}
In some contexts the justification of the second step by citing continuity of the square root function would be obligatory.
A: It's fine. If you want slightly more rigor,
$$ \lim_{x \to \infty} \frac{2x}{\sqrt{4x^2 - 1}} = \sqrt{ \lim_{x \to \infty} \frac{4x^2}{4x^2 - 1}} = \sqrt{ \lim_{x \to \infty} \left(1 +  \frac{1}{4x^2 - 1}\right)} = 1. $$
A: For fun:
Let $x >0.$
$\dfrac{2x}{\sqrt{4x^2 +8x +1}} \lt \dfrac{2x}{\sqrt{4x^2+1} }\lt $
$\dfrac{2x}{\sqrt{4x^2}}$.
$\dfrac{2x}{\sqrt{(2x+1)^2}}\lt \dfrac{2x}{\sqrt{4x^2+1}} \lt $
$\dfrac{2x}{\sqrt{4x^2}}.$
Note : $\sqrt{x^2} = |x| =x,$ since $x>0.$
$\dfrac{2x}{\sqrt{(2x+1)^2}} = \dfrac{2x}{2x+1}= 
\dfrac{2}{2+1/x} $.
We have:
$\dfrac {2}{2+1/x} \lt \dfrac {2x}{\sqrt{4x^2 +1}} \lt \dfrac{2x}{2x}=1$.
The limit is?
A: Yes it is completely correct in this case,since for $x>0$ $\sqrt{4x^2}=2x$ , and equivalent to get out $2x$ from the denominator.
